SCHAUM'S OUTLINE SERIES THEORV AND PROBLEMS OF
SCHAVM'S OUTLINE OF
THEORY AXD PROBLEMS OF -v
LINEAR
ALGEBRA
BY
SEYMOUR LIPSCHUTZ,
Ph.D.
Associate Professor of Mathematics
Temple University
SCHAIJM'S OIJTUl^E SERIES McGRAW-HILL BOOK COMPANY New
York, St. Louis, San Francisco, Toronto, Sydney
•ed
Copyright © 1968 by McGraw-Hill, Inc. All Rights Reserved. Printed in the United States of America. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.
37989
8910 8HSH 754321
liX)mOM
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Preface Linear algebra has in recent years become an essential part of the mathematical background required of mathematicians, engineers, physicists and other scientists. This requirement reflects the importance and wide applications of the subject matter. This book is designed for use as a textbook for a formal course in linear algebra or as a supplement to all current standard texts. It aims to present an introduction to linear algebra which will be found helpful to all readers regardless of their fields of specialization. More material has been included than can be covered in most first courses. This has been done to make the book more flexible, to provide a useful book of reference, and to stimulate further interest in the subject.
Each chapter begins with clear statements of pertinent definitions, principles and theorems together with illustrative and other descriptive material. This is followed by graded sets of solved and supplementary problems. The solved problems serve to illustrate and amplify the theory, bring into sharp focus those fine points without which the student continually feels himself on unsafe ground, and provide the repetition of basic principles so vital to effective learning. Numerous proofs of theorems are included among the solved problems. The supplementary problems serve as a complete review of the material of each chapter.
The
three chapters treat of vectors in Euclidean space, linear equations and These provide the motivation and basic computational tools for the abstract treatment of vector spaces and linear mappings which follow. A chapter on eigenvalues and eigenvectors, preceded by determinants, gives conditions for representing a linear operator by a diagonal matrix. This naturally leads to the study of various canonical forms, specifically the triangular, Jordan and rational canonical forms. In the last chapter, on inner product spaces, the spectral theorem for symmetric operators is obtained and is applied to the diagonalization of real quadratic forms. For completeness, the appendices include sections on sets and relations, algebraic structures and polynomials over a field. first
matrices.
wish to thank many friends and colleagues, especially Dr. Martin Silverstein and Tsang, for invaluable suggestions and critical review of the manuscript. also want to express my gratitude to Daniel Schaum and Nicola Monti for their very I
Dr. I
Hwa
helpful cooperation.
Seymour Lipschutz Temple University January, 1968
CONTENTS Page Chapter
1
VECTORS IN Introduction.
product.
Chapter
2
R"
AND
C"
1
Vectors in R«.
Norm
Vector addition and scalar multiplication. and distance in R". Complex numbers. Vectors in C«.
Dot
LINEAR EQUATIONS Introduction.
18
Linear equation.
System of linear equations. Solution of a system of linear equations. Solution of a homogeneous system of linear equations.
Chapter
3
MATRICES
35
Introduction.
Matrices. Matrix addition and scalar multiplication. Matrix multiplication. Transpose. Matrices and systems of linear equations. Echelon matrices. Row equivalence and elementary row operations. Square matrices. Algebra of square matrices. Invertible matrices. Block matrices.
Chapter
Chapter
4
5
VECTOR SPACES AND SUBSPACES Introduction.
Examples of vector
linear spans.
Row
BASIS
63
Subspaces.
Linear combinations, space of a matrix. Sums and direct sums. spaces.
AND DIMENSION
86
Introduction. Linear dependence. Basis and dimension. Dimension and subspaces. Rank of a matrix. Applications to linear equations. Coordinates.
Chapter
B
LINEAR MAPPINGS
121
Mappings. Linear mappings. Kernel and image of a linear mapping. Singular and nonsingular mappings. Linear mappings and systems of linear equations. Operations with linear mappings. Algebra of linear operators. Invertible operators.
Chapter
7
MATRICES AND LINEAR OPERATORS Matrix representation of a linear operator. Matrices and linear mappings.
Introduction. Similarity.
Chapter
8
150
Change of
basis.
DETERMINANTS Introduction.
171
Permutations.
Determinant. Properties of determinants. Minors and cofactors. Classical adjoint. Applications to linear equations. Determinant of a linear operator. Multilinearity and determinants.
Chapter
9
EIGENVALUES AND EIGENVECTORS Introduction.
Polynomials of matrices and linear operators. Eigenvalues and eigenvectors. Diagonalization and eigenvectors. Characteristio polynomial, Cayley-Hamilton theorem. Minimum polynomial. Characteristic and minimum polynomials of linear operators.
197
CONTENTS Page Chapter
10
CANONICAL FORMS
222
Invariance. Invariant direct-sum decomPrimary decomposition. Nilpotent operators, Jordan canonical positions. form. Cyclic subspaces. Rational canonical form. Quotient spaces. Introduction.
Chapter
11
Triangular form.
LINEAR FUNCTION ALS AND THE DUAL SPACE
249
Linear functionals and the dual space. Dual basis. Second dual Annihilators. Transpose of a linear mapping.
Introduction. space.
Chapter
12
BILINEAR, QUADRATIC
AND HERMITIAN FORMS
261
Bilinear forms. Bilinear forms and matrices. Alternating bilinear forms. Symmetric bilinear forms, quadratic forms. Real symmetric bilinear forms.
Law
Chapter
IB
Hermitian forms.
of inertia.
INNER PRODUCT SPACES
279
Cauchy-Schwarz inequality. OrthogoGram-Schmidt orthogonalization process. Linear nality. Orthonormal sets. functionals and adjoint operators. Analogy between A(V) and C, special operators. Orthogonal and unitary operators. Orthogonal and unitary matrices. Change of orthonormal basis. Positive operators. Diagonalization and canonical forms in Euclidean spaces. Diagonalization and canonical forms in Introduction.
Inner product spaces.
Spectral theorem.
unitary spaces.
Appendix A
SETS AND RELATIONS Sets,
elements.
Set
operations.
315 Product
sets.
Relations.
Equivalence
relations.
Appendix B
ALGEBRAIC STRUCTURES Introduction.
AppendixC
320
Rings, integral domains and
fields.
Modules.
POLYNOMIALS OVER A FIELD Introduction.
INDEX
Groups.
Ring of polynomials.
Notation.
327 Divisibility.
Factorization.
331
chapter
Vectors
in R^
and
1
C
INTRODUCTION In various physical applications there appear certain quantities, such as temperature and speed, which possess only "magnitude". These can be represented by real numbers and are called scalars. On the other hand, there are also quantities, such as force and velocity, which possess both "magnitude" and "direction". These quantities can be represented by arrows (having appropriate lengths and directions and emanating from some given reference point O) and are called vectors. In this chapter we study the properties of such vectors in some detail.
We (i)
begin by considering the following operations on vectors.
The resultant u + v of two vectors u obtained by the so-called parallelogram law, i.e. u + V is the diagonal of the parallelogram formed by u and v as shown on the right. Addition:
and V
(ii)
is
Scalar multiplication: The product kn of a real number fc by a vector u is obtained by multiplying the magnitude of u by A; and retaining the same direction if or the opposite direction if k<0, as shown on the right.
k^O
Now we assume the reader is
familiar with the representation of the points in the plane If the origin of the axes is chosen at the reference point above, then every vector is uniquely determined by the coordinates of its endpoint. The relationship between the above operations and endpoints follows.
by ordered
(i)
pairs of real numbers.
Addition:
If
(a, &)
and
will be the endpoint of
(c,
d) are the
endpoints of the vectors u and in Fig. (a) below.
v,
then (a +
c,
b
+ d)
u + v, as shown (a
+ c, b + d)
(ka, kb)
Fig. (a)
(ii)
Fig. (6)
Scalar multiplication: If (a, b) is the endpoint of the vector u, then {ka, kb) will be the endpoint of the vector kn, as shown in Fig. (6) above.
VECTORS IN
2
B"
AND
[CHAP.
C»
1
Mathematically, we identify a vector with its endpoint; that is, we call the ordered pair of real numbers a vector. In fact, we shall generalize this notion and call an «-tuple We shall again generalize and permit the coa«) of real numbers a vector. {ai, C2, ordinates of the «-tuple to be complex numbers and not just real numbers. Furthermore, in Chapter 4, we shall abstract properties of these %-tuples and formally define the mathematical system called a vector space. (a, 6)
.
.
.
,
We field
assume the reader is familiar with the elementary properties of the which we denote by R.
VECTORS IN The
real
number
R"
set of all w-tuples of real
numbers, denoted by R",
«-tuple in R", say
U —
Uz,
(til,
.
n-space.
A
particular
Un)
.,
.
is called
a point or vector; the real numbers im are called the components (or: coordinates) of the vector u. Moreover, when discussing the space R" we use the term scalar for the elements of R, i.e. for the real numbers. is called
Example
1.1:
Consider the following vectors: (1,-3),
(0,1),
(1, 2,
VS,
(-5,
4),
-1,
0,ff)
The first two vectors have two components and so are points in B^; the last two vectors have four components and so are points in B*.
Two ponents, vectors
u and v are eqtial, written u = v, if they have the same number of combelong to the same space, and if corresponding components are equal. The
vectors i.e.
(1, 2, 3)
Example
and
(2, 3, 1)
1.2:
are not equal, since corresponding elements are not equal.
Suppose (x-y, x
+ y, z-1) =
Then, by definition of equality of vectors,
(4, 2, 3).
X
—
y
x
+
y
= =
4:
2
2-1 =
3
x
Solving the above system of equations gives
=
3,
y
=
—1, and z
—
A.
VECTOR ADDITION AND SCALAR MULTIPLICATION Let u and v be vectors in R":
u =
(Ml, U2,
.
.
.
,
Un)
and
=
v
{Vi, Vz,
.,
Vn)
The sum of u and v, written u + v,is the vector obtained by adding corresponding components:
U + V
=
iUi
+ Vi,U2 + V2,
.
.,Un
number fc by the vector u, written ku, plying each component of u by k: kun) ku — (kui, ku2, The product
of a real
.
Observe that u + v and ku are also vectors in R".
-u = -1m The sum
of vectors with different
and
.
.
+ Vn)
is
the vector obtained by multi-
,
We
u- v —
also define
m+
numbers of components
{-v) is
not defined.
CHAP.
VECTORS IN
1]
Example
Let u
1.3:
=
+
5w
-
2m Example
1.4:
The vector
30)
(0, 0,
to the scalar
.
•
•
•
Then
•
denoted by 0, any vector u =
0) in P.",
.,
=
(3,5,-1,-2).
3
(1
in that, for
+
U
.
=
C"
+ 3, -3 + 5, 2 - 1, 4 - 2) - (4, 2, 1, 2) ^ (5, -15, 10, 20) (5 1, 5 (-3), 5 2, 5 4) = (-7, -21, (2, -6, 4, 8) + (-9, -15, 3, 6)
= =
V
AND
and v
(1,-3,2,4)
u
K«
(Ml
+ 0, M2 + 0,
• ,
•
M„
7,
called the zero vector.
is
(ztj,
%,
.
+ 0) =
.
.
,
14) It is similar
u„),
uj
(Ml, 2*2
= «
Basic properties of the vectors in R" under the operations of vector addition and scalar multiplication are described in the following theorem.
Theorem
For any vectors u,v,w G R" and any scalars
1.1:
+ v) + w = u + — u u+ u + {-u) = u +v — V +u (u
(i)
(ii)
(iii)
(iv)
Remark:
{v
+ w)
(v) (vi) (vii) (viii)
— kv
for some nonzero scalar said to be in the same direction as v if fe > 0, and in the op-
Suppose u and v are vectors in R" for which u A;
e
R.
Then u
is
G R: k(u + v) - ku + kv (ft + k')u = ku + k'u (kk')u = k{k'u) Vu, — u
k, k'
posite direction if
k <0.
DOT PRODUCT Let u and v be vectors in R":
u =
(ui, Ui,
.
.
.
,
and
t(„)
=
v
(vi, Vz,
.,
.
.
Vn)
The dot or inner product of u and v, denoted hy wv, is the scalar obtained by multiplying corresponding components and adding the resulting products:
U'V = The vectors u and v are zero: m v = 0.
+
UiVi
U^Vi,
+
•
•
+
•
UnVn
said to be orthogonal (or: perpendicular) if their dot product is
•
Example 15:
Let m
=
(1,
-2,
3,
-4),
u-v = 1-6 + M'W = 1'5 + Thus u and
w
=
-y
(-2)'7
(6, 7, 1,
+
(-2) •(-4)
-2)
and
w=
(5,
+ 3-5 +
-4,
Then
5, 7).
= 6-14 + 3 + 8 = = 5 + 8 + 15-28 =
3-1 + (-4)'(-2) (-4) -7
3
are orthogonal.
Basic properties of the dot product in R" follow.
Theorem
1.2:
For any vectors u.v.w G R" and any scalar (i)
(ii)
Remark:
+ v)"W {ku) V =
{u
'
=^
u-w +
k{u • v)
vw
(iii)
(iv)
fc
€ R:
= vu u-u^O, and
wv
The space R" with the above operations of vector addition, and dot product is usually called Euclidean n-space.
NORM AND DISTANCE
iff
u-d
scalar multiplication
IN R»
Let u and v be vectors in R": u = (uuUz, .. .,Vm) and v tance between the points m and v, written d{u, v), is defined by d(U,V)
wm =
=
\/(«l
- '^i? + {U2-V2)^+
=
(vi,V2,
+(Un- Vn)'''
.
.
.,Vn).
The
dis-
VECTORS IN The norm
length) of the vector u, written
(or:
root ot U'u:
By Theorem
1.2,
wu^O
=
= yul + ul+
y/u'U
Example
—
=
d(u,v)
V(l \\v\\
we
consider two points, say p
=
||p|l
Vo^TF
—
\\u
m = (1,-2, 4,1) and v =
Let
1.6:
•
•
Observe that
-
—
-5,
V32
+
Then
0).
- 1)2 +
(-2
(a, b)
and
v\\
(3, 1,
- 3)2 +
+ul
•
and so the square root exists. d{u,v)
if
1
defined to be the nonnegative square
is
||m||,
[CHAP.
C«
.
\\u\\
Now
AND
K"
12
+
(4
(-5)2
and q
=
5)2
+
+
02
= V35
(c,
(1
in the plane R2, then
d)
= V(a - c)" +
d{p,q)
- 0)2 = ^95
+
(&
-
corresponds to the usual Euclidean length of the arrow from the origin to the point p, and d{p, q) corresponds to the usual Euclidean distance between the points p and q, as shown below:
That
is, ||p||
P
=
1~
(a, b)
(a, 6)
i-d|
1
=
9
(c,
d)
1«
-
I
H
A similar result holds for points on the line R and in space R*. A vector e is called a unit vector if its norm is 1: Remark: any nonzero vector m
G
R", the vector eu
—
u/\\u\\
e\
—
•\
= 1. Observe that, for a unit vector in the same
||e||
is
direction as u.
We now Theorem
1.3
state a
fundamental relationship known as the Cauchy-Schwarz inequality.
Using the above by
vectors u,v
For any vectors u,v G
(Cauchy-Schwarz): inequality,
GW
we can now 6
-
(or:
6
cos
Note that
if
u-v =
0,
then
9
90°
\u-v\^
R",
define the angle 6
^ U
=
\\u\\ \\v\\.
between any two nonzero
,
V
This then agrees with our previous
ir/2).
definition of orthogonality.
COMPLEX NUMBERS complex numbers is denoted by C. Formally, a complex number is an ordered pair (a, b) of real numbers; equality, addition and multiplication of complex num-
The
set of
bers are defined as follows: (a, b)
{a,b)
= +
(c,
d)
=
a-
c
and
+ c,b + d) = {ac - bd, ad + be)
{c,d)
(a, b)(c, d)
iff
{a
b
=
d
CHAP.
We
VECTORS IN
1]
number a with
identify the real
<->
C"
number
the complex
a
AND
K«
(a, 0):
(a, 0)
This is possible since the operations of addition and multiplication of real numbers are preserved under the correspondence: (a, 0)
Thus we view
E
+
=
(b, 0)
C and
as a subset of
The complex number
0)
and
replace
(a, 0)
denoted by
(0, 1),
= a =
4-2
+ b,
(a
=
(0, 1)(0, 1)
=
(a, 0)(6, 0)
{ab, 0)
by o whenever convenient and
possible.
has the important property that
i,
= -1
(-1, 0)
or
= \/=l
i
Furthermore, using the fact (a, 6)
=
we have
+
(a, 0)
=
(a, 6)
and
(0,b)
+
(a, 0)
(0,6)
=
(6, 0)(0, 1)
=
(b, 0)(0, 1)
+
a
bi
The notation a + bi is more convenient than (a, b). For example, the sum and product of complex numbers can be obtained by simply using the commutative and distributive laws 9.71(1
7,^
^
1*
+ bi) +
{a (a
(c
+ di) -
+ bi){c + di) =
The conjugate
ac
+
a
bci
+
z are given
=
a^
+ bK)
di
=
a
—
«
#
0,
in addition,
{a
-
bdi^
(a,b)
z
-
+ c) +
{ac
w GC. We
=
a
z_
zz
b^
= a + bi
is
1.7:
+
{be
+ ad)i
denoted and defined by
bi
then the inverse z-^ of z and division by
—b a'
+ d)i
w = wz
,
.
and
b^
also define
— z = —Iz Example
+ .
+
a^
{b
- bd) +
by Z-1
where
If,
+
bi
=
number
of the complex
+
+ adi +
z
(Notice that zz
c
Suppose
z
+ Si
= 2 z
=
z
=
w=
and
+w =
zw
w-z = w+
and
(2
(2
+ 3i) +
(5
+ Si = 5 — 2i 2 + 3t
2
Then
- 2t) =
+ 3i)(5 - 2i) =
2
w
5-2i.
3i
(-z)
10
+
2 15t
and
+
5
-
+
4i
w =
5
-
2t
6t2
=
3i
-
2t
(5 - 2i)(2 - 3i) _ 4-19t _ 13 (2 + 3i)(2-3t)
=
7
i
+ Hi
16
=
5
i.
_
13
+
+
2i
31 13^ •
Just as the real numbers can be represented by the points on a line, the complex numbers can be represented by the points in the plane. Specifically, we let the point (a, b) in the plane represent the complex number z = a + bi,
The i.e. whose real part is a and whose imaginary part is b. absolute value of z, written |z|, is defined as the distance from z to the origin: \z\
Note that
|z|
is
Example
=
equal to the 1.8:
V^T&^ norm
Suppose
of the vector
z-2 + 3i 1^1
and
= V4 +
9
(a, 6).
Also,
\z
w = 12 - 5i. Then = v'iS and |wl =
ZZ.
V144 + 25
=
13
VECTORS IN
6
R«
AND
[CHAP.
C"
1
In Appendix B we define the algebraic structure called a field. We emphasize that the set C of complex numbers with the above operations of addition and multiplication is a field.
Remark:
VECTORS IN
C"
The set of all n-tuples of complex numbers, denoted by C", is called complex n-space. Just as in the real case, the elements of C" are called points or vectors, the elements of C are called scalars, and vector addition in C" and scalar multiplication on C" are given by (Zl, Z2,
.
.,
.
+
Zn)
(Wl, W2,
Z(2l, 22,
where
Zi,
wi, z
Example
Now
let
G
Wi
is real.
Wn)
=
.,Zn)
=
+ Wi, Z2 + Wi,
(^^l
{ZZi, 222,
.
.
+ 3i, 4-i, 3) + (3 -2i, 5i 4 - 6i) = = (-6 + 4i, 2 + 8i, 6i) 2i(2 + 3i, 4 - i, 3)
u and v be arbitrary
. ,
.
.,
.
ZZn)
(5
(2i, 22,
.
.
.
,
+
t,
4
+ 4i,
7
=
V
Zn),
product of u and v
The norm
Observe that
of
u
1.10:
Let
defined
is
= yJU'U =
wu and so
Example
is
{Wi, Wi,
.
.
.
,
Wn),
G C
Wi
2;,
defined as follows:
ZiWl
+ Z%W% +
•
•
•
+
ZnWn Wi
\/ZiZi
||«||
m =
+
2222
+
•
•
•
+
2„2„
are real and positive (2
+ 3i,
=
V'|2ip
+
|22p
+
•
•
•
+
Wi
when
|2„|2
when u =
when u¥=0, and
0.
= {S -2i, 5, 4- 61). Then u-v = (2 + 3i)(3 - 2i) + (4 - iXS) + (2i)(4 - 6i) = (2 + 3i)(3 + 2t) + (4 - 1)(5) + (2i)(4 + 6t) = 13i + 20 - 5t - 12 + 8i = 8 + 16i 4-i,
and
2i)
v
+ 3t)(2 + 3i) + (4 - i)(4 -i) + = (2 + 3i)(2 - 3i) + (4 - i)(4 + i) + = 13 + 17 + 4 = 34
||m||
(2
- Vu'u =
The space C" with the above operations is called
=
by
u'u =
Remark:
- 6t)
this definition reduces to the previous one in the real case, since
||m||
product,
+ W„)
Z„
vectors in C":
U'V = Note that
.,
(2
1.9:
dot, or inner,
.
C.
U — The
.
.
.
(2i)(2t)
(2i)(-2i)
\/34
of vector addition, scalar multiplication and dot
complex Euclidean n-space.
If
wv
be
real.
were defined by u-v = ziWi + ••• + ZnWn, then U'U-0 even though u¥-0, e.g. if u={l,i,0). In fact,
it
is
w%
possible for
may
not even
CHAP.
VECTORS IN
1]
AND
R«
C»
7
Solved Problems
VECTORS IN 1.1.
1.2.
R"
+
Compute: (i) (3,-4,5) (iv) -(-6,7,-8).
(1,1,-2);
(i)
Add
(ii)
The
(iii)
Multiply each component by the scalar:
(iv)
Multiply each component by —1:
corresponding components:
-4,
(1,2,-3)
+
-2)
=
+
5)
(1, 1,
stim is not defined since the vectors have different
u^ (2, -7, 1),
Let
(3,
(ii)
v
=
(-3, 0,
— (—6, 7, —8)
w=
4),
=
(6,
+ 1,-4 + 1,5-2) =
(4,
-3,
3).
(—12, 15, 18).
—7,
8).
Find
-8).
(0, 5,
(3
-3(4,-5,-6);
(iii)
numbers of components.
=
—3(4, —5, —6)
(4,-5);
3% - 4v,
(i)
(ii)
2u + Zv- 5w.
First perform the scalar multiplication and then the vector addition.
3u-4v = 3(2, -7, 1) - 4(-3, 0, 4) = (6, -21, 3) + (12, 0, -16) = 2u + 3v-5w = 2(2, -7, 1) + 3(-3, 0, 4) - 5(0, 5, -8) = (4, -14, 2) + (-9, 0, 12) + (0, -25, 40) = (4 - 9 + 0, -14 + - 25, 2 + 12 + 40) = (-5, -39,
(i)
(ii)
1.3.
Find x and y
if
{x, 3)
=
{2,x
(18,
-21, -13)
54)
+ y).
Since the two vectors are equal, the corresponding components are equal to each other:
=
X Substitute x
1.4.
=
Find x and y
=
3
2,
if
(4, y)
=
+
X
2 into the second equation to obtain y
=
y
Thus x
1.
(4,
=
y)
x{2, 3)
=
Solve the linear equations for x and
x,
y and
z if
(2,
y
=
and
j/
=
1.
(2x, Zx).
Set the corresponding components equal to each other:
Find
2
x(2, 3).
Multiply by the scalar x to obtain
1.5.
—
-3, 4)
=
y:
x
x{l, 1, 1)
=
2
and y
=
4
=
2x,
3a;.
6.
+ y{l, 1, 0) + z(l, 0, 0).
First multiply by the scalars x, y and z and then add:
= = =
(2,-3,4)
Now
set the
{X, X, x)
(x-^y
+
j/
+
z
=
2,
solve the system of equations, substitute 3/
—7. Then substitute into the
Prove Theorem (i)
(ii)
(iii)
(iv)
Let
0)
+ z,x-\-y,x)
Wj, Vj
1.1:
first
X
-\-
y
For any vectors u,v,w
+ v) + w — u-\— u u+ u + {-u) u+V = V +u {u
and Wj be the
ith
=
—3,
a;
=
4
k = 4 into the second equation to obtain 4 + j/ = equation to find z = 5. Thus x — 4, y = —7, z = 5.
To or
=
+ j/(l, 1, 0) + «(1, 0, + (y, y, 0) + (z, 0, 0)
corresponding components equal to each other: a;
1.6.
a;(l, 1, 1)
{v
+ w)
GW
and any scalars
(v) (vi) (vii) (viii)
fc,
fc'SR,
+ v) — ku + kv (fc + k')u — ku + k'u (kk')u - k{k'u) lu = u k{u
components of u, v and w, respectively.
—
VECTORS IN
8
AND
R"
[CHAP.
C»
1
definition, Mj + Vi is the ith component oi u + v and so (itj + Vj) + Wj is the tth component + (Vj + Wj) of (u + v) + w. On the other hand, Vi + Wj is the ith component oi v + w and so But Mj, Vj and Wj are real numbers for which the asis the tth component of u + (v + w). sociative law holds, that is,
By
(i)
%
+ Vi) + Wi - Mj-l-(Di + Wj) + tf = u+ {v-^w) since their
Accordingly,
=
Here,
(ii)
(m
+
(0, 0,
.
.
M
.,
= = .
M + (—m)
(Ml, M2.
•
(Ml + 0,
Mg
=
•.
•
+
Mn)
+ 0,
.
.
(0, 0,
M„
. ,
— M2, = (Ml, Mg, M„) + = (Ml - Ml, M2 - M2, .
.
,
m„)
(— Mi, .
.
By
.
.
(Ml, M2,
.
.
= M
M„)
,
.
— m„),
. ,
M„
. ,
.
0)
.,
.
.
+ 0) =
(—Ml,
. ,
.
(iv)
corresponding components are equal.
hence
0);
+
— m = — 1(mi, M2,
Since
(iii)
1))
i-\,...,n
for
(Ui
— M2, — M„) - M„) = (0, 0, .
. ,
.
.
=
0)
. ,
.
+ v^ is the ith component of u + v, and Vj + Mj is the ith. component of v are real numbers for which the commutative law holds, that is,
definition, n^
But
and
Mj
ijj
Wi
Hence m
+v =
1;
+m
+
ft
=
"(
+
since their corresponding
=
i
Mi,
1,
.
.
.
,
+ u.
w
components are equal.
+ Vj is the ith component of u + v, k(Ui + Vi) is the ith component of k(u + v). Since and kvi are the ith components of ku and kv respectively, Ajm; + fcvj is the ith component of ku + kv. But k, Mj and v^ are real numbers; hence n i = 1, fc(Mj + Vi) = ftMj + fc^j, Since Mj
(v)
fcMj
.
.
=
Thus k{u + v)
ku
+
.
,
kv, as corresponding components are equal.
first plus sign refers to the addition of the two scalars k and k' whereas the second plus sign refers to the vector addition of the two vectors ku and k'u. By definition, (fe + fc')Mj is the ith component of the vector (k + k')u. Since fcMj and Aj'Mj is the ith component of ku + k'u. are the ith components of ku and k'u respectively, kUf +
Observe that the
(vi)
k%
But
k, k'
and
numbers; hence
Mj are real
(fc
Thus
(k
Since
(vii)
ith
+ k')u =
fc'Mj is
ku
+
+ k')Ui —
k'u,
1.7.
•
=
i
k'Ui,
1,
.
.
,
.
n
the ith component of k'u, k(k'u^ is the ith component of (kk')u and, since fc, k' and Mj are real numbers,
fe(fe'M).
But
(fcfc')Mi is
the
component of
Hence 1
+
as corresponding components are equal.
(fcfc')Mj
(viii)
kUi
M
(kk')u
=
=
1(mi, M2,
Show that Ou =
.
,
.
i=l,
fc(fc'Mj),
...,n
as corresponding components are equal.
k(k'u),
.
=
-
M„)
=
(1mi, lu^, .... 1m„)
(mi, M2,
.
.
for any vector u, where clearly the
,
.
=
m„)
first
u.
a scalar and the second
is
a vector.
=
Method
1:
Om
Method
2:
By Theorem
0(mi, M2,
Adding
— Om
.
.
.
,
m„)
=
(Omi, OM2,
Om
1.1,
=
.
.
.
(0
,
Om„)
=
+ 0)m =
(0, 0,
Om
+
=
...,0)
Om
to both sides gives us the required result.
DOT PRODUCT 1.8.
=
(8, 2,
Multiply corresponding components and add:
wv
Compute u v where: •
(iii) (i)
tt
=
(3,-5,2,l),
(i)
u=
v-
(4,
-3, 6), v 1,-2, 5). (2,
=
2
The dot product
(iii)
Multiply corresponding components and add:
u-v =
u=
(ii)
•
8
+
not defined between vectors with different
(ii)
is
-3);
3
•
4
(—3)
(1,
•
-8,
+
2
6
•
0, 5),
(—3)
v
=
=
(3, 6, 4);
—8.
numbers of components.
+
(—5)
•
1
+
2
•
(—2)
+
1
•
5
=
8.
CHAP.
1.9.
VECTORS IN
1]
9
C"
Determine k so that the vectors u and v are orthogonal where u ^ (1, k, -3) and v = (2, -5, 4) (i) (ii)
u =
-4,
(2, 3fc,
1, 5)
In each case, compute u (i)
(ii)
1.10.
AND
R"
•
and v
=
v, set it
equal to
-1,
(6>
3, 7, 2fc)
and solve for
0,
-
-
= 1 2 + 2 5fe (-5) + 1*7 = 2-6 (-4)'3 + 3fc'(-l) + + + U'V = 12 - 3fe - 12 + 7 + lO/c = 0, u'v
•
fc
Prove Theorem
12
vw
wv
=
i;
O.
(Mi,M2
ku
Since
=
(ku)
U'V =
•
(ku^, ku^,
V
MjDi
+
•"
•
=
ku^Vi
M2''^2
.,
.
.
+
+ •
u'U =
•
fcM2'y2
+
iff
DISTANCE AND NORM IN
(iv)
M'M
-
0,
= -2
kGK,
=
and u-u •
•
iff
u =
•
•
.
•
•
•
•
Mj
=
•
•
•
=
+
=
ku^V^
'"l"!
+
''2'*2
HU^V^
+
'
"
+ M2'y2
+
'
d(u,v)
=
V(l
- 6)2 +
(7
(ii)
d(u,v)
=
V(3
- 6)2 +
(-5
(iii)
d(u,v)
=
V(5
- 2)2 +
(3
•
'^n^n)
=
*=(«* '
and since the sum of nonnegative real numbers
i,
2
2
I
for each
that
i,
=-
J. ,.2
I
m
is, iff
=
Find the norm
d{u, v)
is
non-
n 0.
=
d(u, v)
VW -
+ 5)2 = V25 + - 2)2 +
+ 1)2 +
(-2
+ 0)2 +
(-4
=
\/l69
+ 1)2 = V9 +
(4
=
+ •••+(«„ - vj^
v{)^
=
144
?;
49
+
+ 7)2 +
v = (6, -5); (2,-1,0,-7,2).
u=
(i)
(1, 7),
.
13 25
(-1
=
a/83
- 2)2 =
\/47
=
6 where
fc
||m||
")
= V'U
1>n^n
« = (2, fc, 1, -4) and v = (3, -1, 6, -3). (d(u, i;))2 = (2 - 3)2 + (fe + 1)2 + (1 - 6)2 + (-4 + 3)2 = fe2 + 2fe + 28 = 2, -4. + 2fc + 28 = 62 to obtain
Find k such that
fc2
•
R»
(i)
solve
1-
-I
Find the distance d{u, v) between the vectors u and v where: (iii) m = (5,3,-2,-4,-1), t; = (6,2,-1); (ii) «=(3,-5,4),
Now
1.13.
+
Mn''^n
In each case use the formula
1.12.
fc
ku^),
Since wf is nonnegative for each negative,
Furthermore,
1.11.
0,
tt
•
•
(iv)
==
= -l
k
(iii)
•
(iii)
10
5'2fc
•
(ii)
-5k -
0,
W„). = (yi.'y2. •'"n). W = (^1,^2. u + v = (mi + Vi, M2 + "2. •••.**„ + I'm). + (U„ + Vn)Wn (u + v)'W = (Ml + Vi)Wi + (% + '"2)^2 + = UiWi + ViWi + U2W2 + 1'2M'2 + + M„W„ + V„W„ = (MiWi + M2W2 + M„w„) + y„w„) + (viWi + V2W2 + + = U'W + VW
M =
Since
(i)
k.
=
For any vectors u,v,w G R" and any scalar
1.2:
+ v)'W = U'W + (fcM)-^ = k{u'v)
(ii)
Let
•
{u
(i)
=
(-3) -4
of the vector
In each case use the formula
u
||m|1
if
(i)
=
y/u^
(2,
+
4.
m2
-7), .
. .
+
(ii)
^2
u=
,
= ^53
(i)
IHI
=
V22
+
(-7)2
= V4 +
49
(ii)
11^11
=
V32
+
(-12)2
+
= V9 +
(_4)2
u=
144
+
16
= V169 =
13
(3,
-12, -4).
+
,
VECTORS IN
10
1.14.
Determine & such that
Now
Show
solve
that
fc2
||m||
By Theorem
1.16.
+
30
^ 0,
=
1.3
For any vectors u
+
12
and
=
||m||
3,
u=
ifi
=
C»
[CHAP.
1
{l,k,-2,5).
+
(-2)2
=
fc
and u'u
O,
+
fc2
and obtain
39
wu —
1.2,
Prove Theorem
=
AND
R"
= VS^ where u =
||tt||
I|m|I2
1.15.
.
=
52
+
A;2
30
-3.
0.
=
m
iff
Since
0.
=
||m||
yjii-u,
the result follows.
(Cauchy-Schwarz):
=
{u\,
.
.
and v —
m„)
.
(vi,
.
\u-v\
in B",
.,Vn)
.
^
]\u\\ \\v\\
n
We
shall prove the following stronger statement:
If M = or V = 0, then the inequality reduces to need only consider the case in which m # and v Furthermore,
-
\U'V\
+
IMjI)!
•
•
•
Thus we need only prove the second
Now
for any real
numbers
w,
+
^
M„V„|
=
and y
|mj|/||m||
=
3/
G
—
R,
(x
^
— j/)2 =
But, by definition of the {2)
norm i
—
2xy
+
y^
2
any
i,
^^'
IWP
It'iP
IMP
IMP
2 is,
ki^il ,1
II
,1
-
11
or, equivalently,
(1)
IMP
2kP
IMI IHI that
x^
2
^
2 kiVil
=
\u„vj
of a vector, ||m|| = 2m,^ = kiP and and using \u(Vi\ = ImjI |i;j|, we have
2M
HvH.
and is therefore true. Hence we where ||m|| # and ||i;|| # 0.
+
•••
I|m||
3/2
in (1) to obtain, for
Ifil/HvH
with respect to
+
a;2
IHI IHI
summing
+
\UiVi\
— i.e.
0,
^
|Mt'"tl
inequality.
2xy Set X
— j^
2
—
\u'v\
= S^f =
||i;||
IMI^
IMP
IMP
IblP
2|v,-|2.
Thus
1
IMI IHI Multiplying both sides by
1.17.
||m||
we
H'wH,
obtain the required inequality.
Prove Minkowski's inequality:
For any vectors u-{ui,...,Un) and v = If
IIm
Now
+ vjI = JMj
+ V(| —
jwjl
\\u
+
+
|i)j|
v\\2
=
2(«i +
=
2 ki 2 ki +
=
But by the Cauchy-Schwarz inequality
2M+fj|KI ^ Thus Dividing by
||M
+ f||2 ^
||m-|- v||,
we
numbers
for any real
(see
+
i;||
IHI
.
in R",
.,Vn)
mj, Vj
+
G
||m
+ vH
=^
||tt||
i'iP
2
^
2
Hence
R.
2k +
+ Vjj
\ui
Vil \ui\
ki +
ki
Vil
+ Vil
(kil
+ M)
Ivjj
preceding problem),
and
Ik+^IIIMI |Im
=
i'<)='
vil
.
Thus we need only consider the case
the inequality clearly holds.
0,
{vi,
+
||m
+ H|
obtain the required inequality.
2k + ||i;|l
=
'yilkl ||tt
+
i;||
^
Ik
+ HIIHI
(IMI
+
lbll)
||m
+ +
||v||.
i;1|
#0.
CHAP.
1.18.
VECTORS IN
1]
Prove that the norm
in
R"
For any vector w,
[Nz]:
For any vector M and any scalar For any vectors u and
[Na]:
was proved
[Ni]
v,
Problem
in
=
and |H|
||m||^0;
\\u
1.15,
+
11
^
\\u\\
[Ng] in
u=
iff
\\ku\\
A;,
v\\
and
C«
following laws:
satisfies the
[Ni]:
AND
R»
= +
|/cl
0.
||m||.
||t;||.
Problem
Hence we need only prove that
1.17.
[Ni] holds.
Suppose u
=
(ui, ii2,
=
||fcMl|2
The square
.
.
and so ku
u„)
.,
(fcMi)2
+
(kui)^
+
=
•••
(kui, ku^, ....
+
=
(fcM„)2
ku^. Then khi\
+
khi\
+
•••
+
khil
root of both sides of the equality gives us the required result.
COMPLEX NUMBERS 1.19.
Simplify:
+ 3i)(2-7i);
(4-3*^;
(ii)
+ 3t)(2 - 7i) = 10 + 6i - 35i - 21i? = (4 - 3t)2 = 16 - 24t + 9t2 = 7 - 24t (5
(i)
(ii)
(v) (vi)
(i)
-
/
.
(vu)
|^;
(i^)
31
-
(v)
»',
i*.
29i
(3 + 4t) _ 3 + 4i ^ A + Aj - 4i)(3 + 4t) 25 25 25 2-7i _ (2 7t)(5 3t) _ -11 41i _ _11_41. 34 34 34 5 + 3t ^ (5 + 3i)(5-3t) ts = i^'i = (-l)t = -i; P^ = (i*)7't^ = i* = v^"P = 1; - 12 - 8i = -11 - 2i (1 + 2i)8 = 1 + 6i + 121* + 8i^ = 1 + 6i
3
^'^'
Let
g"^''
("i)
_
1
r"\ ^*"^
1.20.
(5
+ 2i)''; (vii)(2^)'
(l
(vi)
(i)
4i
1
(3
Y
\^2-3iJ
1 _ _ ~ -5-12i ~
(-5
1^
•
(-t)
(-5 + 12t) _ -5 + 12i _ __5_ - 12t)(-5 + 12t) 169 169
= 2 - 3i and w = 4 + 5i Find: z + w and zw; (ii) z/w; (iii) « and w;
,
'
=
-i
12
169*
2!
(iv)
\z\
and
\w\.
+ w = 2 - 3i + 4 + 5i = 6 + 2i zw = (2 - 3i)(4 + 5t) = 8 - 12i + lOi - 15t2 = 23 - 2i - 5i) _ -7 - 22i ^ _ 1. _ 22 f\ £. - 2-3t _ (2 3i)(4 - 5t) 41 5i)(4 41 41 + 6i (4 4 + w « = 2 - 3t = 2 + 3t; w = 4 + 5i = 4 - 5t. (iii) Use a+U = a-bi: z
(i)
.
Use |a+6i|
(iv)
1.21.
(i)
\z\
=
|2
- 3t| = V4 + 9 =
Vl3;
For any complex numbers z,w G.C,
Prove: (i)
= V^^TP:
zw = zw, (iii) z = z. Suppose z = a+bi and w = e + di where a, b,c,d& R. z + w = (a+bi) + (e + di} = {a+c) + {b + d)i = {a + e) — {b + d)i = a + c — bi — di = (a — bi) + {e — di) = z + w
z
+ w = z + w,
(ii)
zw
(iii)
z
(ii)
= (a+ bi)(c + di) = (ae — bd) + (ad+ bc)i = {ac — bd) — (ad + bc)i = (a-bi)(c — di) - zw
— a+bi —
a
—
bi
= a—
(—b)i
= a+bi =
«
\w\
=
[4
+ 5i|
=:
Vl6 + 25= Vil.
,-31. *"
-
VECTORS IN
12
1.22.
= a + bi and w = c + di where |«|2 = o2 + 62, |w|2 = c2 + d2, z
Thus
Suppose
=
(c,
=
z
+
\z
=
w\
By Minkowski's
= Va2 +
+ c) +
\{a
1.24.
(i)
u+
(3
bd)
+
{ad
|2|2
+
bc)i
|^|2
+ d)i\ = V(« + c)2 +
(b
+
-
w\
\\u
+
||m
v\\
(6
+
w\
—
€
R.
Consider the vectors
d
=
jwl
||m||,
v,
+ 6i) and v = 4m, (iii) (1 + i)v,
+ d)2 =
+ v|| —
||m||
^
+
||m||
+
\z\
+
Vc2
\\v\\
||t;||
=
\w\.
=
rf2
||(a+
+
+ i,2-3i, 5). (iv) (1 - 2i)u +
2i, 4i, 1 (il)
(ii)
Multiply each component of u by the scalar
(iii)
Multiply each component of v by the scalar 1
6
and
+
=
d)||
\\u
+
=
+ t)i;
- 2i)u +
Find u-v and (3 - 2i,
so
+
lw|
+ 6i + i2,
(5
(8
— t,
2
4t:
4m =
+ i)v.
(3
+ i, 6 + 6t). +
(8
12i,
+ 4t).
—16, —24
i
+ 2 1 3i2, 6 + 5t) = i:
(4
+ 6i,
5 -
i,
5
+ 5i)
4i, 1
+ 6i),
vu v
(3
(5
(-1
+
(14
u = {l — 2i,S + i), v = + i,2-Si,l + 2i).
where:
=
- 8i 8 + 4i, 13 + 4i) + (13, 17 - 3i, 28 + 9t)
+ i)v = =
8i,
(4
(i)
9
- Ii,
+ 2i,
5
15
+ 5i)
— 6i);
Recall that the conjugates of the second vector appear in the dot product: .
(2l,
- 2t)(4 + 2i) + = (1 2i)(4 - 2i) +
M v = •
vu (ii)
v\\
First perform the scalar multiplication and then the vector addition: (1
(i)
and
(a, 6)
|j^||
c,
|z|
u=
Find:
{5
Add corresponding components: u + v —
(1
1.25.
=
62
\z
a, 6, c,
(i)
(iv)
-
{ac
C"
=
Let M
where
inequality (Problem 1.17), \z
VECTORS IN
w = c + di
a + bi and Note that \z\
and
Then
R.
of both sides gives us the desired result.
in R2.
d)
= =
(1
+ 2t)(l - 2i) + (4 + 2t)(l + 2i) + (4
u-v =
(3
=
(3
ij •
tt
1
\w\.
\z\
zw =
and
For any complex numbers z,w€lC,
Prove:
=
\zw\
b,c,dG
a,
[CHAP.
C"
= (ac - 6d)2 + (ad + 6c)2 = a2c2 - 2abcd + b^d? + aU^ + 2abcd + 62c2 = a^C^ + d2) + 62(c2 + d2) = ((j2 + 62)(c2 + (£2) =
|zw|2
The square root
V
AND
For any complex numbers z,w GC,
Prove:
Suppose
1.23.
R«
= =
In both
- 2i)(5 + i) + - 2i)(5 - i) +
.
.,
Z„)
•
(Wi,
...,WJ = «iWi
+ t)(5 - 6t) (3 + j)(5 + 6t) =
-lOi
- 6i)(3 + 1) (5 - 6z)(3 - i) =
lOi
+
•
+
•
2„W„
(3
+
9
9
-
+
23i
=
9
+
13i
(5
(4i)(2 (4i)(2
- 3i) + + 3i) +
(1 (1
+
23i
+ 6i)(7 + 2i) + 6i)(7 - 2i) =
=
20
9
+
-
13i
35i
+ i)(3 - 2i) + (2 - 3t)(4i) + (7 + 2i)(l + 6t) + i)(3 + 2i) + (2 - 3i)(-4t) + (7 + 2i)(l - 6i) = 20 - 35i (5 examples, vu — wv. This holds true in general, as seen (5
in
Problem
1.27.
(ii)
u
CHAP.
1.26.
VECTORS IN
1]
Find
where:
||tt||
||m|P
1.27.
||m1P
=
(3)2
(ii)
||m||2
=
42
+ 4i,
(3
- 2i,
5
- 3i);
1
C»
•
(4)2
+
(5)2
(-1)2
+
22
+
+
+
+
+
(-2)2
+
32
+
12
=
(-3)2
=
(i)
{zi, «2,
=
zu
Since
(zu) (iii)
.
=
and v
«„)
,
.
(wj, W2,
•
,
•
Using the properties of the conjugate established
VU (ii)
.
Method
=
V
•
WiZ^ .
Using
2.
+
=
W2Z2
+ +
•
•
•
ZZ2W2
+
•
•
•
(zwi, zwg,
.
.
.
W„Z„
+
ZZ„Wn
ZiZWl
and
(i)
u
W„Z„
2\/l5
wv = vu,
(ii)
Problem
+ +
WiZi Z^Wi
1.21,
+ +
W2Z2 Z2W2
•
•
•
•
•
•
+ +
W„Zn z„w„
=
WV
+ «2M'2 +
Z(ZiWi
"
"
+ ^nWn) =
"
«^(m
•
•
•
•
•
+ Z„ZW^ = ZiZWi + + Z^W^ = Z(U V)
Z2ZW2
•
+
•
•
•
+
«„2W„
•
(ii),
=
(zv)
•
=
(zv)
= z(vu) = z(vu) =
u
•
z(u'v)
MISCELLANEOUS PROBLEMS 1.28.
Let u (i)
1.29.
=
u+
v;
-
u + v
(ii)
4m
(iii)
2m
(iv)
u-v =
(v)
1|m|1
(vi)
d(u,v)
(i)
=
-
(3
=
-4,
=
4
V(3
- 2i,
(7
Sv;
=
=
u + v
(ii)
2m =
(14i
(iii)
(3-i)v
=
(iv)
u-v =
(7
=
(7
=
V72
(7
1
2
(-21, -3,
40
16
=
VSO,
(-2
- 1)2 +
(1
=
(1
+ 5i) and (iii)
- 2i + 1 +
v
{S-i)v; i,
- 4i2, 4t + 10t2) = (3
(4
9,
=
-18)
+
(-2)2
+
(2 (2
22
||v||
=
+ 3)2 +
V'49 (4
52
+
1
(iv)
u-v; (8
+
+
9
36
- 6)2 = V45 =
+ i, -3 - 6i). (v)
-
i,
=
\/95
3\/5
Find:
and
||mI|
-1
||v||.
- i)
+ 14i -10 + 4t)
+ 5i)(-3 - 6t) + 5i)(-3 + 6i) =
+
d{u,v).
(vi)
||i;||;
(-15, -7, 11, -10)
+ 3i - i - 12, -9 - ISi + 3i + 6P) -
- 2t)(TTt) + - 2t)(l - 1) +
and
||m||
(10,-1,-2,10)
+ 5i - 3 - 6z) =
2
(v)
(12,-8,4,16)
=
+
Find:
6).
u-v;
(iv)
24
- 7)2 +
2m;
(ii)
+
+
2, 8)
21-2-3 +
= V9 +
u + v;
IMI
(iii)
-3,
(7, 1,
+ 7,-2 + 1,1-3,4 + 6) =
(6,
(i)
(v)
2u —
=
(4'3, 4'(-2), 4-1, 4-4) 3i;
=
and v
1, 4)
4u;
(ii)
(i)
Let M
-2,
(3,
(zu)
w„). in
= =
+ +
= ^60 =
zwj,
,
+ Z^Wl + «(«iWi + Z^2 +
= =
(zv)
W2Z2
•
8
zz^),
.,
.
ZZiWi
Since zv
U•
+ +
WiZi
(zz^, zz2,
1.
Method
— =
•
==
||m||
Prove: For any vectors u,v EC" and any scalar z GC, (i) z{u'v), (iii) u-{zv) = z{u'v). (Compare with Theorem 1.2.) Suppose u
z„)
(z^, Z2
||m||
or
60,
+ 2t, 1 - hi).
3
2i,
i,
=
s
or
64,
=
(-5)2
u={4-
where
•
+
(1)2
+
22
•
13
(ii)
— w''+ 6^ when z = a+ hi. Use - u-u = ZiZi + Z2Z2 + + z„Zn
Recall that zz
(i)
m=
(i)
AND
R»
= V^,
5
\\v\\
-
(4
9i
=
+ 2t, -15 - 15t) -
36
Vl'^
-
+
3i
1^
+
= -31 (-3)2
+
12i
(-6)2
= V47
* '")
•
v
VECTORS IN
14
1.30.
C"
[CHAP.
1
pair of points P = {ou) and Q fines the directed line segment from
= (bi) in R» deP to Q, written vector v = Q-P:
Any
We identify PQ with the PQ = V = {bi — ai, 62 - a2,
PQ.
Find the vector v
1.31.
AND
R«
(i)
P=
(ii)
P=(l,-2,4), Q = (6,0,-3)
(i)
t)
(ii)
V
The Xi,
= =
.
Q-P Q-P
= (-3-2, 4-5) = (-5,-1) = (6 - 1, + 2, -3 - 4) = (6,
.,Xn of the
=
tion of
H.
(ci,
.
.,
.
-7)
R" which are solutions of a linear equation in n unknowns
in
^
c„)
+
+
C2X2
•
•
+
=
CnXn
&
(*)
in R", is called a
(We frequently
GH
kyperplane of R", and (*) is called an equawith (*).) Show that the directed line segment
H
identify
PQ of any pair of points P,Q u
2,
form CiXl
with u
where:
(-3.4)
H of elements
set
.
Q=
(2,5),
.,bn- a„)
.
PQ
with
identified
.
orthogonal to the coefficient vector u; the vector
is
said to be normal to the hyperplane H.
is
P=
Suppose
(«!,
.
Q =
and
.,aj
.
(fej,
.
.
.,6„).
Then the
and the
Oj
64
are solutions of the
given equation: Cjai
+
C2a2
+
•
+
•
Q-P^
v
m
• t;
•
•
Hence
1.32.
that
v,
is,
PQ,
c^bi
b,
= PQ = = Ci(&i - aj) + 62(63 - Og) + = C161 - citti + C262 - C2a2 + = (Ci6i + C262 + + c„bj -
Let
Then
=
c^an
is
•
•
orthogonal to
+
C262
+
•
•
•
+
c„&„
=
6
{b,~ai,b2-a^, ...,b„-a„)
- aj + c„fe„ - c„a„ (CjOi + 02(12 + + c„o„) = 6-6 =
•
+
•
•
•
c„(6„
•
•
•
•
u.
Find an equation of the hyperplane H in R* if: (i) H passes through P = (3, -2, 1, -4) and is normal to m = (2,5,-6,-2); (ii) passes through P = (1,-2, 3, 5) and is parallel to the hyperplane H' determined by 4:X — 5y + 2z + w = 11.
H
(i)
An
P (ii)
1,33.
equation of
H
is
of the form
into this equation to obtain
k
+
2x
=
5y
—2.
—
Qz
— 2w =
k since
Thus an equation of
H
H
normal to u. Substitute 2x + 5y — 6z — 2w = —2.
it is
is
and H' are parallel iff corresponding normal vectors are in the same or opposite direction. Hence an equation of H is of the form 4x — 5y + 2z + w = k. Substituting P into this equation, we find k = 25. Thus an equation of H is 4:X — 5y + 2z + w = 25.
The
line
and
in the direction of
points points
I
in
R" passing through the point
u=
X — P + tu, GK, that X= obtained from t
P=
(a,)
consists of the
(Ui) ¥= is,
consists
of the
(Xi)
Xi
ai
-I-
Uit
-f-
U2t
a;2
=
^2
n
—
an
(*)
where
t
takes on
all
called a parameter,
resentation of
I.
]
Unt
real values.
and
(*) is called
The
variable
t
is
a parametric rep-
CHAP.
VECTORS IN
1]
AND
B,"
Find a parametric representation of the tt where: (a)P = (2,5) and m
(i)
C"
15
P
passing through
line
and
in the direc-
and u
=
Find a parametric representation of the line passing through the points P and where: (a)P = (7,-2) and Q = (9,3); (6) P = (5, 4, -3) and Q = (l,-3,2).
Q
tion of
=
(-3,4);
(6)
P=
(4,
-2, 3,
1)
(2,5,-7,11). (ii)
In each case use the formula
(i)
(*).
= 4 + 2t y = -2+ 5t z = 3 - 7t w = 1 + nt X
= =
'x
\y
(In
K2 we usually eliminate Ax + Zy = 23.)
2
-
3t
5
+
4t
from the two equations and represent the
t
line
by a single
equation:
PQ = Q —
u =
First compute
(ii)
Q-P
u =
(a)
Then use the formula
P.
(x
= = 1+ = -2 +
Q-P
=
u
(b)
(2,5)
(*).
2t
X
5«
V z
we
(Note that in each case
u —
could also write
= (-4,-7,5)
= 5 -At = 4 -It = -3 + 5t
QP — P —
Q.)
Supplementary Problems VECTORS IN R» 1.34. Let u = (1,-2,5), and 1.35.
1.36.
Let
w=
(iii)
—u +
i;
=
-1, 0, -3), i; 2v — 2w, (iv)
(2,
= (1, -1, -i; 3), w = u'v,u'W and vw;
t;
=
(5,
-3, -1,
Determine & so that the vectors M and A;, -4, 2), i; = (1, -3, 2, 2fe). (iii) m
(5,
1.38.
Determine
a;
and
2/
if:
1.39.
Determine
a;
and
J/
if :
1.40.
Determine
a;,
y and «
1.41.
(i)
(3,-1,2)
(ii)
(-1,3,3)
Let (i)
«!
M
=
=
=
+
6e2
(i)
v;
-6m;
(ii)
(iii)
2u
-
5v;
(iv)
u-v;
(v)
||m|1
2, 7).
(1,3,-2,2). Find: (i) 2u (v) d(u,v) and d(v,w).
Find:
(i)
u+
v;
(ii)
3u
-
-
3v;
2v;
(ii)
(iii)
5u
-
M'^y;
-
3v
Aw;
(iv)
||mI|
=
+ y) = =
are orthogonal,
1)
(1, 7,
(y
fe
+ 2, -2),
-2,6);
2(y,-l);
(ii)
v
=
x(l,2)
x(2,y)
(ii)
u = (3,k,—2),
(i)
=
t;
=
(6,
— 4, — 3).
(ii)
u —
(3,k,-S,k).
=
-4(y,3).
y(l,-2).
if:
a:(l, 1, 1)
62
+
(x,x
x(3,2)
(i)
a;(l, 1, 0)
(!, 0,0),
aei
+
u
(i)
d(u,v).
(v)
ll-ull;
Find:
(3,l,-2).
d(u,v).
M = (2,1,-3,0,4),
Let
and 1.37.
(vi)
||t;||;
=
ceg;
+ j/(l, -1, 0) + + j/(0, 0, -1) + eg
(0,1,0), (ii)
m•
ej
=
= a,
z(l, 0, 0) z(0, 1, 1)
(0,0,1).
m•
62
=
Show that m 63 = c.
6,
•
for
any vector
u=
(a,b,c)
in
&:
.
VECTORS IN
16
1.42.
.
AND
R«
Let
Generalize the result in the preceding problem as follows. elsewhere: ith coordinate and
=
ei
Show
u—
that for any vector
u =
(i)
e K" has
1.43.
Suppose M
1.44.
Using d(u,v)
aiCi
+
=
62
(1,0,0, ...,0,0),
(aj, ag,
+
(1262
.
ej
...,
0,0),
e R"
1
be the vector with 1 in the
e„
=
(0,
0,0,
=
Oj
for
i
.
.,0, 1)
.
.,«„),
.
•
(0,1,0
[CHAP.
C»
•
+
•
m•
the property that
m
a„e„,
(ii)
=
for every
1)
•
ej
v
S
R".
Show
=
1,
.
that
..,n
it
=
0.
= \\u-v\\ and the norm properties [ATj], [iVj] and [N3] in Problem 1.18, show that the distance function satisfies the following properties for any vectors u,v,w G R": d(u,v)^Q,
(i)
and
=
d{u, v)
u=
itl
v;
d(u,v)
(ii)
=
d(v,u);
d{u,w)
(iii)
^
+
d{u,v)
d(v,w).
COMPLEX NUMBERS - 7t)(9 + 2i);
1.45.
Simplify:
(i)
(4
1.46.
Simplify:
(i)
^;
1.47.
Let z
= 2-5i
1.48.
Let z
=
1.49.
Show
that:
«
1.50.
=
(z
Show
1.52.
1.53.
and
(i)
(ii)
1;
=
\z\
z/w;
(i)
(^3^ zw;
(ii)
(ii)
«,
w;
«/«;
(iii)
(iii)
real part of
(iii)
\z\;
.'
(iv)
+ w;
(l-i)^.
(v)
1«1,
=
«
(iv)
«,
w;
(v)
1«|,
lw|
|w|.
^(z
+
z);
(iv)
imaginary part of
z)/2i.
zw =
implies
+ 7i, 2 - 6i)
(iv)
wwandvw;
(1
= (3 - 7t, tfvandvM; M
{u
=
«
(i)
Find:
^^*;
(iv)|-3||;
=
z
w=
or
0.
C»
=
Prove:
w = 6 - 5t.
««-i
It
Let
Find:
j^^;
(iii)
^2^
i^s,
(iii)
w = 7 + 3i
and
Let
(i)
1.54.
—
+i
that
VECTORS IN 1.51.
2
f^;
(ii)
(3-5i)2;
(ii)
2i,
v-{5- 2i, 3 - 4t).
and
(v)
and
1|m1|
and and Hiill
1|m||
For any vectors u,v,w G
+ v)-w =
WW
+ vw;
Prove that the norm in C"
=
1;
- i,
+ 2i,
11
w
(u
+
=0
For any vector u and any complex number
[Ns]:
For any vectors u and
(Compare with Problem
v;
(ii)
(3
+ i)u;
(iii)
2m +
(4
- 7t)v;
- Si).
Find:
(i)
m
-
v;
(ii)
(3
+
i)v;
(iii)
0;
\\u
+
(Compare with Theorem
1.2.)
following laws:
and
[A^a]:
v,
8
wu + wv.
=
v)
[Ni]:
\\u\\
u+
C":
(ii)
satisfies the
^
(4
For any vector
u,
(i)
||t)|l.
-1 + i)
(iv)
Find:
\\u\\
v\\
^
iff
+
=
1|zm||
z,
|H|
m
0.
=
\z\
\\u\\.
||i'l|.
1.18.)
MISCELLANEOUS PROBLEMS 1.55.
Find an equation of the hyperplane (i)
1.56.
passes through
(2,
—7,
1)
and
and
R3 which: to
-2,
2), (0, 1, 3)
(1,
-5,
2)
3x
(1,
(0, 2,
contains contains
parallel to
(ii)
is
in
normal
-1);
(iii)
and
is
-
Determine the value of k such that 2x - ky 8. (Two hyperplanes are perpendicular
2w =
(3, 1,
+
ly
+
iz
=
5.
- 5w = 11 is perpendicular to 7x + 2y -z corresponding normal vectors are orthogonal.)
4z
iff
—11);
+
CHAP.
1.57.
1.58.
VECTORS IN
1]
Find a parametric representation of the passes through
(ii)
passes through (1,9,-4,5) and (2,-3,0,4)
(iii)
passes through (4,-1,9) and
Q and R
Let P,
(7,
=
1.35.
u+v ||m||
||i;||
2u-Zv = (1, 1, 3, -15); (iv) wv — —6, M w = —7,
=
M+r
(7,
-2, -4, 2,
||i;||=2V22; 1.37.
(i)
k
=
6;
1.38.
(i)
»
=
2,
1.39.
(i)
x
=
-l, y
1.40.
(i)
x
=
2,
1.43.
We
1.45.
(i)
50-55i;
1.46.
(i)
-^i;
1.47.
(i)
z
=
J/
=
4;
-
=
y
fc
-3/2;
=
S, g
(5
(i)
z/w
1.50.
If
zw
1.51.
(i)
=
=
(7
0,
k
=
=
-6,
j/
(ii)
x
=
(ii)
=
-
30i;
=
y
=
(iii)
(iii)
j/
-i,
1.56.
k-0
1.57.
(i)
(x
=
T
+
18
t
that if
<
tj
(2
<
then
Ht
(-13, -9, 20);
(iv)
m
• -y
=
=
11,
\/38
-32); d(v,
,
-2,
1, »
=
4
=
0.
-m + 2i; - 2w =
(iii)
=
w)
z
(ii)
=
\w\
\z\
= -
- ei, + i,
-11,
31
-12;
=
u-v =
(iii)
(-2,-7,
2, 5);
(iv)
38;
||m1|
= VSO
-
29i;
(4
-2 -
(v)
2i.
+ 3i)/50.
=
z/w
(iii)
2
-
|0|
i,
=
w = 0.
6
+
Hence
+ 17i, 27-i)
(ii)
,
= -4
2/
(H-3i)/2;
(iv)
(iv)
-9 + 4t)
(ii)
-9;
3'\/2
(-1
- 41t)/58;
z
(iv)
=
2
+
5i,
w = 7 - 3f,
V5^-
\zw\
+ y-llz =
-14,
=
-1;
(3
Sx
- 5i; =
2m
w
i,
(ii)
(i)
—
a^-\- u„t
Show
t.
-7, -4, -2);
+ 7t)/65;
u-v =
(13
,
9,
t)
(-1
.
^or
*3
(-4,
(4
(i)
1.55.
(3,
or
0;
-1,
zw = 29
(ii)
+ 16t)/61;
+ t)i; =
z
3/2
(iii)
(ii)
—
+
3/2
0,
a;
.
.
^^^
(iii)
d(u, v)
(v)
6;
= (6 + 5i, 5 - lOi) - 16t) (3 + t)u = (-4 + 22t, 12 41i, -4 - 33i) 2iu + (4 7i)v = (-8 M+
x^
2y
= ^62
d{u,v)
which implies that m
+ 27t)/58;
then
=
Mgt,
(-6, 12, -30);
3u-2v =
(ii)
(iii)
-2;
-16
(ii)
(ii)
—
Supplementary Problems
vw = V^
a;
a;2
5u-3v-4w =
(ii)
=
3;
(ii)
M
1.48.
1.52.
(ii)
+ w = 9 - 2t; = = V29,
|z|
^2
to
11);
d(M,i>)
(v)
have that m • m
(v)
+
;
(i)
(i)
02 ti,
Mjt,
= (4, -1, 3); (ii) -6m = = V30, = a/U (vi)
•
1.36.
=
to the values
+
Oi
Answers (v)
Sx
perpendicular to the plane
is
be the points on the line determined by *i
(i)
—5)
8) in the direction of (1, 3,
which correspond respectively d(P,Q) ¥ d(Q,R) = d(P,R).
1.34.
17
line which:
(i)
—1,
C
AND
R»
13x
+
4j/
+
z
(iii)
or
u-v -
(v)
||m|1
(iv)
||m||
=
7;
=
(iii)
3»
(iii)
21
+
JHI
-
|m;|
,
+ ,
||i;||
vu
=
v/215
+
4z
so
vu
27i,
2i,
7i/
= Vei
=0; and
= 3VlO
12 8,
y/5
lw|
(iv)
u-v -
=
\z\
=
\z\
(iii)
= 1+ t y = 9 - 12* z = -4 + 4t w = 5—t x
5t;
=
=
=
z
=
= 21
3a/6
12
46.
-
2i
or
-
27i
w=
0.
chapter 2 Linear Equations INTRODUCTION The theory of linear equations plays an important and motivating role in the subject of linear algebra. In fact, many problems in linear algebra are equivalent to studying a system of linear equations, e.g. finding the kernel of a linear mapping and characterizing the subspace spanned by a set of vectors. Thus the techniques introduced in this chapter will be applicable to the more abstract treatment given later. On the other hand, some of
new
the results of the abstract treatment will give us crete" systems of linear equations.
insights into the structure of "con-
For simplicity, we assume that all equations in this chapter are over the real field R. We emphasize that the results and techniques also hold for equations over the complex field C or over any arbitrary field K.
LINEAR EQUATION By
a linear equation over the real aiXi
+
aix^
we mean an + anXn =
R,
field
+
•
•
•
expression of the form h
{!)
The
ai, & G R and the Xi are indeterminants (or: unknowns are called the coefficients of the Xi respectively, and b is called the constant term or simply constant of the equation. A set of values for the unknowns, say
or variables).
where the
scalars
ttt
=
Xl is
a solution of
(i) if
k\,
Xi=
ki,
.
.
Xn=
.,
kn
h
the statement obtained by substituting ttifci
+
a2k2
+
•
•
+
•
=
a„fc„
for
Xi,
b
then said to satisfy the equation. If there is no ambiguity about the position of the unknowns in the equation, then we denote this solution by simply thew-tuple , ,, , is
true.
This set of values
is
U =
Example
2.1
Consider the equation
:
The
4-tuple
u3
a true statement. equation since
is
1 is
x
,
(fcl,
fe,
.
.
.
,
kn)
+ 2y — 4z + w = is
(3, 2, 1, 0)
S.
a solution of the equation since
+ 2«2-4'l +
=
However, the 4-tuple
+ 2.2-4-4 +
5
v
=
3
or
=
(1, 2, 4, 5)
3
3
=
3
not a solution of the
is
-6 =
or
3
not a true statement.
Solutions of the equation
(1)
can be easily described and obtained.
There are three
cases:
Case
(i):
One
of the coefficients in (1) is not zero, say ai
¥- 0.
Then we can rewrite the
equation as follows axx-i
= b-
a2X2
-
•
•
•
—
anX„
or
Xi
18
=
a^^b
-
a^^aiXi
-
•
—
aj"
a„x„
CHAP.
LINEAR EQUATIONS
2]
19
By arbitrarily assigning values to the unknowns X2, .,x„, we obtain a value for Xi; these values form a solution of the equation. Furthermore, every solution of the equation can be obtained in this way. Note in particular that the linear equation in one unknown, .
=
ax has the unique solution x Example
—
We
with a¥'Q
h,
a^^b.
Consider the equation 2x
2.2:
.
—
+
4y
—
z
8.
rewrite the equation as
=
2a;
8
+
—
4j/
x
or
2
=
A
+
2y
—
^z
Any value
for y and z will yield a value for x, and the three values will be a solution of the equation. For example, let 2/ = 3 and z = 2; then x = 4 + 2'3 — ^-2 = 9. In other words, the 3-tuple u = (9, 3, 2) is a solution of the equation.
Case equation
the
is,
form
of the
Ojci
Then the equation has no Case
That
All the coefficients in (1) are zero, but the constant is not zero.
(ii): is
+
0*2
+
•
•
•
+
=
Oa;„
6,
solution.
All the coefficients in (1) are zero, the equation is of the form (ill):
Oxi
Then every n-tuple
#
with 6
of scalars in
R
+
is
Oa;2
+
•
•
+
•
and the constant Oa;„
is
also zero.
That
is,
=
a solution of the equation.
SYSTEM OF LINEAR EQUATIONS We now consider a system of m linear
equations in the
ttiiXi
+
ai2a;2
+
•
•
a2ia;2
+
a22X2
+
•
•
•
n unknowns
+
ainXn
=
+
a2nXn
=62
Xi,
.
.
.,
x„:
&i
(*)
OmlXl
where the
+ am2X2 +
•
"
+
•
OmnXn
=
&m
belong to the real field R. The system is said to be homogeneous if the con.,kn) of real numbers is a solution (or: 6m are all 0. An ?i-tuple u = (fci, a particular solution) if it satisfies each of the equations; the set of all such solutions is termed the solution set or the general solution. stants
61,
Oij, bi
.
.
. ,
.
The system of
.
linear equations
anXi
+
ai2a;2 ai2X2
+
aziXi
4+
n^aVa 4- ... • a22X2
•
+
••
•
•
+
ai„Xn ai„a;„
+
nn-fa2nXn
-I-
= ^ -
n (**)
amlXi
+ 0^2X2 +
•
•
the homogeneous system associated with = (0, 0, namely the zero w-tuple 0)
is called
tion,
•
other solution,
.
if it exists, is called
The fundamental
.
.
,
+
OmnXn
(*).
-
The above system always has a
called the zero or trivial solution.
a nonzero or nontrivial solution.
relationship between the systems (*)
and
(**) follows.
solu-
Any
LINEAR EQUATIONS
20
Theorem
Suppose m suppose
2.1:
W
is is
a particular solution of the nonhomogeneous system (*) and the general solution of the associated homogeneous system (**).
W
u + is
[CHAP. 2
=
{u
+ w: w G W)
the general solution of the nonhomogeneous system
(*).
We
emphasize that the above theorem is of theoretical interest and does not help us to obtain explicit solutions of the system (*). This is done by the usual method of elimination described in the next section.
SOLUTION OF A SYSTEM OF LINEAR EQUATIONS Consider the above system (*) of linear equations. We reduce
it
to a simpler
system as
follows:
Step
Step
1.
2.
unknown
Interchange equations so that the
first
cient in the first equation, that
so that
For each i>
1,
is,
an
xi
has a nonzero
coeffi-
¥-0.
apply the operation
+
—anLi
-*
Li
That
is,
ttiiLi
replace the ith linear equation Li by the equation obtained by mulby -an, multiplying the ith equation L, by
tiplying the first equation Li
an, and then adding.
We the
then obtain the following system which (Problem 2.13)
same
solution set as
equivalent to
is
(*), i.e.
has
(*):
anaii
+
ai2X2
+
a'laXs
+
•
•
•
+
a'mXn
=
&i
ayja^jg
+
+
0,2nXn
=
&2
amJ2^i2
"I"
^"
dmnXn
=
Om
¥= 0. Here Xj^ denotes the first unknown with a nonzero coefficient in an equation other than the first; by Step 2, Xj^ ¥^ Xi. This process which eliminates an unknown from succeeding equations is known as (Gauss) elimination.
where an
Example
2.3:
Consider the following system of linear equations:
2x 3x
4x
+ + +
iy
-
6y
+ +
8y
2v
z
+ —
z
+
5v
z
V
+ 2w - 1 + iw = —7 — w = 3
eliminate the unknown x from the second and third equations by applying the following operations:
We
L2
We
^ -3Li + -SLj:
compute
21,2:
-3Li + and
I/g:
-2Li
12y
+
L3:
5z
-
8v
+ 2w = -17
2z
—
4v
+
z
+
5v
3z
+
V
+
Sy
Lg
2v
6v
-4* — 8y + 4x
+
2z
+
+
-2Li
- 6w = -3 + Sw - -14
3z
ex
->
-
-6x - 12y +
2L2:
—21,1:
L3
and
2L2
- 'iw = -2 — w = 3 - 5w =
1
CHAP.
LINEAR EQUATIONS
2]
Thus the
21
original system has been reduced to the following equivalent system:
+
2x
iy
-
z
+
2v
+ 2w =
5z
-
8v
+ 2w = -17
32
+
V
1
— 5w =
1
Here
Observe that y has also been eliminated from the second and third equations. the
unknown
unknown
z plays the role of the
Xj
above.
We note that the above equations, excluding the first, form a subsystem which has fewer equations and fewer unknowns than the original system (*). We also note that: if
(i)
an equation Oa;i + + and has no solution; •
•
=
Oa;„
•
b ¥=0
5,
occurs, then the system is incon-
sistent
(ii)
+ Oa;„ an equation Oaji + without affecting the solution.
if
•
•
•
=
occurs, then the equation can be deleted
Continuing the above process with each new "smaller" subsystem, we obtain by induction that the system (*) is either inconsistent or is reducible to an equivalent system in the following form aiiXi
+
ai2X2
+
ttisccs
Cl'2uXj,
+
+ a2.j,+ lXi„ +
Ciri^Xj^
where
1
< ^2 <
•
•
•
< ^r and where an
1
+
+ar,j,+ ia;j^+i
+
•
•
•
+
(iinX„
=
bi
+
a2nXn
=
&2
+
arnX„
=
br
,
^
the leading coefficients are not zero:
¥ 0,
a2i2
^
^'
•
>
^^^r
^
^
(For notational convenience we use the same symbols an, bk in the system (***) as (*), but clearly they may denote different scalars.)
we
used
in the system
Definition:
(***) is said to be in echelon form; the unknowns Xi which are termed do not appear at the beginning of any equation {iy^lyjz, ., jr)
The above system
.
.
free variables.
The following theorem
Theorem
2.2:
applies.
The solution of the system two cases: (i)
(ii)
(***) in echelon
form
is
as follows.
r — n. That is, there are as many equations as unknowns. system has a unique solution.
There are
Then the
there are fewer equations than unknowns. Then we can arbitrarily assign values to the n — r free variables and obtain a solution of the system.
r
That
is,
Note in particular that the above theorem implies that the system (***) and any equivalent systems are consistent. Thus if the system (*) is consistent and reduces to case (ii) above, then we can assign many different values to the free variables and so obtain many solutions of the system. The following diagram illustrates this situation.
LINEAR EQUATIONS
22
fCHAP.
2
System of linear equations
Consistent
Inconsistent
No
Unique
solution
solution
More than one solution
In view of Theorem 2.1, the unique solution above can only occur when the associated homogeneous system has only the zero solution. Example
2.4:
We
reduce the following system by applying the operations L2 — 3Li + 2L3, and then the operation L3 -* —SL^ + Lgi
L3
-»
2x
+ + +
Sx 3a;
y 2y 32/
+
2z z
3z
+ 3w = + 2w = - 3w = =
The equation system
Example
2.5:
y 2/
5
Zy that
+ +
4
-8,
is inconsistent,
+
2x
1
4z 12z
and so has no
2x
1
+
5
—ZL^ + 2L2 and
y
-
2z
y
+
Az
7
+ Oi/ + Oz + Ow =
Oa;
is,
+ 3w = — 5w = - 15w =
22
-*
-8,
+ 3w = 1 — 5w = 5 = -8
shows that the original
solution.
—Lj + L2, L2 — Ls
We
L^ -> reduce the following system by applying the operations -* — — Lg 2Li + L4, and then the operations L3 2Lt + Lg and I/4 and L4 -» -21/2 + L^:
-»
-»
X X
2x 2x
+ + + +
2y Sy
5y ey
+ — +
Sz
4
z
11
4z 2z
=
+
a;
-
2]/
y
13
y
22
2y a;
+
2?/ 2/
+ + +
32
+
Az 2z
= = =
4z 2z
7
+
2y 2/
+
Sz 4z 2z
5
14
8z 3z
X
4
= = =
= = = =
4 7
2
4 7 2
Observe first that the system is consistent since there is no equation of the form = 6, with 6 # 0. Furthermore, since in echelon form there are three equations By the third equation, in the three unknowns, the system has a unique solution. Substitut2 = 1. Substituting z = 1 into the second equation, we obtain y = Z. Thus a; = 1, y = Z ing z = 1 and y — Z into the first equation, we find x = 1. and z = 1 or, in other words, the 3-tuple (1, 3, 1), is the unique solution of the system.
Example
2.6:
We
reduce th© "following system by applying the operations L2 —5Li + L3, and then the operation L3 -^ — 2L2 + L^:
L3
-»
X
+ + +
2x 5x
2y
2z
42/
3z
lOy
8z
+ Zw = + 4w = + llw =
X
2
+
2y
-
2z
5
z
12
2z
x
+
2y
+ Zw = - 2w = — 4w =
-2z + Zw = z - 2w =
2 1
2 1
2
-2Li
2y
-
2z z
+
L2 and
+ Zw = - 2w = =
2 1
CHAP.
LINEAR EQUATIONS
2]
23
The system is consistent, and since there are more unknowns than equations in echelon form, the system has an infinite number of solutions. In fact, there are two free variables, y and w, and so a particular solution can be obtained by giving y and w any values. For example, let w = 1 and y = —2. Substituting w = 1 into the second equation, we obtain z = 3. Putting w = X, z = 3 and 2/ = — into the first equation, we find a; = 9. Thus a; = 9, y — —2, z = 3 and w = 1 or, in other words, the 4-tuple (9, -2, 3, 1) is a particular solution of the system.
We
Remark:
find the general solution of the system in the above example as follows. Let the free variables be assigned arbitrary values; say, y = a and w = b. Substituting into the second equation, we obtain 2 = 1 + 26. Putting y = a, z = l + 2b and w = h into the first equation, we find a; = 4 - 2a + &. Thus the general solution of the system is
w-h a;
— 2a + b, = a, 2 = 1+ 26, w — h (4 — 2a + 6, a, 1 + 26, 6), where a and 6 are arbitrary num-
=
4
i/
other words, Frequently, the general solution and w (instead of a and 6) as follows:
or, in
bers.
X
We
=
A
— 2y +
w,
z
=
1
in
is left
+ 2w
or
terms of the free variables y
— 2y + w,y,l + 2w, w)
(4
will investigate further the representation of the general solution of a
system of linear equations in a later chapter. Example
2.7:
Consider two equations in two unknowns: a^x
+
6jj/
=
Cj
OgX
+
622/
=
C2
According to our theory, exactly one of the following three cases must occur: (i)
The system
is
inconsistent.
(ii)
The system
is
equivalent to two equations in echelon form.
(iii)
The system
is
equivalent to one equation in echelon form.
When
linear equations in two unknowns with real coefficients can be represented as lines in the plane R^, the above cases can be interpreted geometrically as follows:
The two
lines are parallel.
(ii)
The two
lines intersect in a unique point.
(iii)
The two
lines are coincident.
(i)
SOLUTION OF A HOMOGENEOUS SYSTEM OF LINEAR EQUATIONS we
begin with a homogeneous system of linear equations, then the system is clearly = (0, 0, ., 0). Thus it can always it has the zero solution be reduced to an equivalent homogeneous system in echelon form: If
consistent since, for example,
aiiXi
+
.
ai2X2
+
a2i2Xj2
+
a2.J2+lXj2+l
drj^Xj^
Hence we have the two
+ amXn = + a2nXn =
+
aisXs
1
+
({r.i^+lXj^+l
.
+
•
•
•
+
drnXn
^
possibilities:
(i)
r
= w.
Then the system has only the zero
(ii)
r
Then the system has a nonzero
solution.
solution.
If we begin with fewer equations than unknowns then, in echelon form, hence the system has a nonzero solution. That is,
r
and
[CHAP.
LINEAR EQUATIONS
24
Theorem
A
2.3:
2
homogeneous system of linear equations with more unknowns than
equations has a nonzero solution. Example
The homogeneous system
2.8:
- 3z+ w = x-Sy + z -2w = 2x + y - Zz + 5w = +
X
2y
has a nonzero solution since there are four unknowns but only three equations.
Example
We
2.9:
reduce the following system to echelon form: X
+
—
y
= -
z
2x-3y + x-4v +
z
2z
The system has a nonzero
Example
2.10:
We
—
z
-5j/
+
3z
= =
z
-5y +
3z
= =
-5y +
3z
=
we
obtained only two equations in the let z = 5; then y = S and x — 2.
y
solution, since
three unknowns in echelon form. In other words, the 3-tuple (2, 3,
y
—
+
X
For example, 5) is
+
X
a particular nonzero solution.
reduce the following system to echelon form:
2x
+ +
4y
-
Sx
+
2y
+
X
y
= = =
z z
2z
X
y
-
z
2y
+
z
-y +
5z
+
=
y
-
z
2y
+
z
= =
llz
=
+
x
= =
Since in echelon form there are three equations in three unknowns, the system has only the zero solution (0, 0, 0).
Solved Problems SOLUTION OF LINEAR EQUATIONS 2x-Sy + 6z + 2v 2.1.
y
Solve the system:
-
^z
+
5w =
= Zw -
v
3 1
.
2
Since the equations begin with the unknowns x,y and v respectively, the other unknowns, z and w, are the free variables. into the third equation, find the general solution, let, say, z = a and w = 6. Substituting
The system
is
in echelon form.
To
-
•y
=
36
2
or
-y
=
2
y
=
+
36
Substituting into the second equation, J/
-
4a
+
2
+
36
=
1
01
4a
-
36
-
1
Substituting into the first equation,
2x
-
Thus the general a;
=
3(4a
- 36 - 1) +
6a
+
2(2
+ 36) -
56
=
3a
3
-
56
solution of the system is
3a
-
56
-
2,
j/
=
4a
-
36
-
1,
z
-
u
=
2
+
36,
w =
6
real numbers. Some texts or (3a-56-2, 4a-36-l, a, 2 + 36, 6), where a and 6 are arbitrary a and 6 as follows: of instead w and z variables leave the general solution in terms of the free
CHAP.
LINEAR EQUATIONS
2]
X
=
Sz
y
=
4z
— 5w — — 3w —
V
-
2
+ 3w
25
2 or
1
(3«
— 5w — 2,
4z
- 3w -
2
1, z,
+
3w, w)
After finding the general solution, we can find a particular solution by substituting into the For example, let a = 2 and 6 = 1; then
general solution.
X is
=
-1,
Solve the system:
3a;
5x
1,2
=
z
4,
=
v
2,
w=
5,
or
1
(-1,
4, 2, 5, 1)
a particular solution of the given system.
X 2.2.
=
J/
+ 2y-3z = -1 — y+ + 3y -
2z
=
7
4z
=
2
.
Reduce to echelon form. Eliminate x from the second and third equations by the operations L2 and Lg -* —5Li + L3:
- — 3Li +
3x-
L2:
-3Li
+
-6y+9z=
-3x
-3Li:
Lg:
3
-SLj: L3:
y
+
2z
=
7
-7j/
+
llz
=
10
-5Li
+
-5x - lOy +
15z
3^/
-
4z
= =
2
-7j/
+
llz
=
7
5a:
+
L3:
5
Thus we obtain the equivalent system X
—
3z
-7y +
llz
=
10
+
llz
=
7
+
2y
-7y
The second and third equations show that the system Ox
2.3.
+
Oy
+
=
Oz
S
Solve the system:
Reduce L2
-*
-3Li
=
or
21,2
2x+ y-2z =
10
3x
+ 2y +
2z
=
1
5a;
+
+
32
=
4
-31,1: 21,2:
-3Li +
and
42/
-5Li +
-6x -3y +
+
we
obtain
.
= -30 =
+
4z
J/
+
lOz
= -28
y
-
2z
J/
+
lOz
= -28
3y
+
16z
= -42
=
-SLj:
-lOx
-
5y
2L3:
10a;
+
2
-5Li
Thus we obtain the following system from which operation L^ -» — 3L2 + Lgi
+
subtract
2L3:
6z
4j/
2L2:
2x
we
inconsistent, for if
Eliminate x from the second and third equations by the operations
"*
^3
6x
is
3.
to echelon form.
+
= -1
we
to
2L3:
6z
= -50 = 8
3y
+
16z
= -42
eliminate y from the third equation by the
2x
10
+
lOz
83/
+ +
+
y
-
2z
y
+
lOz
-142
-
10
= -28 =
42
In echelon form there are three equations in the three unknowns; hence the system has a unique By the third equation, z = —3. Substituting into the second equation, we find j/ = 2. Substituting into the first equation, we obtain a; = 1. Thus x = l, y = 2 and z = -3, i.e. the 3-tuple (1, 2, —3), is the unique solution of the system. solution.
LINEAR EQUATIONS
26
+ 2y-3z =
6
2x
—
=
2
4x
+ 3y-2z =
14
x 2.4.
Solve the system:
+
y
iz
[CHAP. 2
.
Reduce the system to echelon form. Eliminate x from the second and third equations by the Lz -* —2Li + L^ and L3 -» — 4In + L3:
operations
L2-
-
+
6«
= -12
2x- y+
4:Z
=
-2x
-2Li.
4j/
-4Li:
2
-ix - 8y +
12z
-
2«
4*
L3:
+
Thus the system
is
=
14
-5y + lOz = -10
+ 10« = -10 2 y - 2z -
-5j/
or
32/
= -24
y
or
-
2z
=
2
equivalent to
x
+
-Sz =
6
—
=
2
y-2z =
2
2y
y
+
X 2z
2y
-
3z
=
6
y
-
2z
=
2
or simply
we can
(Since the second and third equations are identical,
disregard one of them.)
In echelon form there are only two equations in the three unknowns; hence the system has an infinite number of solutions and, in particular, 3 — 2 = 1 free variable which is z. solution let, say, z - a. Substitute into the second equation to obtain or a; = 2 — o. Substitute into the first equation to obtain a + 2(2 + 2a) - 3o = 6 the general solution is
To obtain the general J/
= 2 + 2a.
Thus
a;
where a
is
The value,
2.5.
=
2
-
a,
y
=
2
+
=
z
2a,
a
or
(2
- a, 2 + 2o, o)
any real number. say,
o
=
1
yields the particular solution
x-Sy + 4:Z — 2w = 2y + 5z + w =
Solve the system:
a;
=
1,
3/
=
4, «
=
1
or
(1,4, 1).
5
2
.
=4
y-Sz
The system is not in echelon form since, for example, y appears as the first unknown in both unknown, the second and third equations. However, if we rewrite the system so that w is the second then
we
obtain the following system which is in echelon form: a;
— 2w —
w+
32/
2y 2/
+
+ -
4z
=
5
Bz
=
2
3z
=
4
Now if a 4-tuple (a, 6, c, d) is given as a solution, it is not clear if 6 should be substituted for Of course this or for y; hence for theoretical reasons we consider the two systems to be distinct. system. original the solution of obtain the to system new the using from us does not prohibit + Za. Substituting into the Let z = a. Substituting into the third equation, we find -6 - 11a. Substituting into the first second equation, we obtain w + 2(4 + 3a) + 5a = 2 or w = equation, = 5 - 17o or a; ^ _ ^^_^ _ ^^^^ _ 3(4 + Sa) + 4a = 5
w
y-A
Thus the general
solution of the original system is a;
where o
is
any
=
5
real number.
-
17o,
J/
=
4
+
3a,
z
=
a,
w = -6 -
11a
CHAP.
2.6.
UNEAR EQUATIONS
2]
27
Determine the values of a so that the following system in unknowns (i) no solution, (ii) more than one solution, (iii) a unique solution: X
+
—
y
+ Zy + + ay +
2x X
z
—
1
az
=
3
Sz
=
2
x,
y and z has:
Reduce the system to echelon form. Eliminate x from the second and third equations by the Lj ^ — 2Li + L^ and I/3 - — I/j + Lg:
operations
-2Li.
-2x -2y +
2z
+ 3y+
az
2x
y
Thus the equivalent system
+
(a
= -2 - 3
+ 2)z =
1
+
y
y
+
X
ay
+ +
Sz
= -1 = 2
(a-l)y
+
4z
=
—(a — V^L^
+
L^,
J/
—
=
z
+
(a
(a-l)y+
+ 2)z = 4z =
eliminate y from the third equation by the operation
-(a
-(a -
- 1)1,2:
+ (a-l)y +
La:
1
1)2/
(2
L3
1
1 1 -^
- a - a2)z = 14z = 1
- a - a^)z - 2(3 + a)(2 - a)z = 2 (6
or to obtain the equivalent
z
is
X
Now
—X—
a
a a
system
X
-V
y J/
—
=
z
1
+ 2)z = (3 + a)(2 - a)z =
+
(a
1
2
-
a
which has a unique solution if the coefficient of z in the third equation is not zero, that is, if a # 2 and a ¥= —3. In case a = 2, the third equation is = and the system has more than one solution. In case a = —3, the third equation is = 5 and the system has no solution. Summarizing, we have:
2.7.
Which X,
(i)
a
=
—3,
must be placed on has a solution?
condition
y and
z
(ii)
a,
b
o
=
and
(iii)
2,
c so that the
+ 2y — Sz — 2x + ey-llz X — 2y + Iz — X
Reduce — 2Li
1/2 -^
to echelon form.
+
L2 and
I/3 -»
a¥' 2 and o
¥=
—3.
following system in unknowns
a b c
Eliminating x from the second and third equation by the operations L3, we obtain the equivalent system
—Lj +
X
2y
— —
5z
— 4j/
4-
lOz
+
2y
Sz
= = =
a
— 2a —a
b c
Eliminating y from the third equation by the operation equivalent system X + 2y — 3z — a
2y
—
5z
= =
b
—
2a
c
+
26
Lg
-
-*
5o
2L2
+
L3,
we
finally obtain the
[CHAP.
LINEAR EQUATIONS
28 The system
will
+
5a
if
c
—
2b
=
have no solution if the third equation is of the form Thus the system will have at least one solution
+
-
26
=
5a
=
5a
or
Note, in this case, that the system will have more than cannot have a unique solution.
26
+
^
that
0;
is,
if
¥= 0.
c
k
with
fe,
2
c
In other words, the system
one solution.
HOMOGENEOUS SYSTEMS OF LINEAR EQUATIONS 2.8.
Determine whether each system has a nonzero
+ Sy +
5z
,
(i)
The system must have a nonzero
(ii)
Reduce to echelon form:
+ 2y-Zz = 2x + 5y + 2z = 3x- y-4z =
solution since there are
x
x
to
,
-
n
("i)
(ii)
(i)
_
,
2x
x
+ 2w =
o
+ 5y + 2z = x + Ay + 7z = x+3y + Sz =
+ 2y-3z = 2x + 5y + 2z = Sx- y-4z =
x-2y + Zz-2w = 3x-ly-2z + 4kw ^ Ax
solution:
more unknowns than equations.
+ 2y-Sz = y + 8z = -7y + 5z =
x
+
to
2y
-
3z
y
+
8z
61z
= = =
hence the system has In echelon form there are exactly three equations in the three unknowns; solution. zero the solution, unique a (iii)
Reduce to echelon form: X
2x a;
+ + + +
-
2y
+ + +
5y
%
z
2z 7z
= = = =
X
+
2y y
2y
+ + +
z
4z 8z 4z
= = = =
x
+
2y
-
z
y
+
Az
= =
x 3y the system has a In echelon form there are only two equations in the three unknowns; hence nonzero solution.
2.9.
The vectors Ui,...,Um
Sz
j/
say,
in,
R" are said to be linearly dependent, or simply
ki,...,km, not all of them zero, Otherwise they are said to be independent. kmiim = 0. kiui are dependent or independent where: whether the vectors u, v and
dependent,
there
if
+ • +
scalars
exist
such
that
Determme
w
(i)
(ii)
(iii)
w = (8, -7, 1) u = (1, -2, -3), V = (2, 3, -1), w = (3, 2, 1) u = (ai, a2), v = (bi, 62), w = (ci, C2) u=
-1), V
(1, 1,
=
(2,
-3,
1),
In each case:
+
yv
+ zw =
where
y and
z are
unknown
(a) let
XU
(6) find
the equivalent homogeneous system of equations;
(c)
(i)
x,
scalars;
then the vectors are determine whether the system has a nonzero solution. If the system does, independent. are they then not, does system dependent; if the
Let XU
+
yv
+ zw =
0:
x{l, 1,
or or
(a;, a;,
(x
-1)
-x)
+
+
j/(2,
(2j/,
-3,
-3y,
1)
y)
+ +
z{8,
(8«,
-7,
1)
=
(0, 0, 0)
-7z,
z)
=
(0, 0, 0)
+ 2y + 8z, x-Sy -Iz, -x + y + z)
=
(0, 0, 0)
CHAP.
LINEAR EQUATIONS
2]
29
Set corresponding components equal to each other and reduce the system to echelon form:
+
+ 8z = x-Zy -7z = X
2y
-X + y +
X
=
z
+ ~5y -
15z
= =
+
9z
=
+
2y
3y
Q
Sz
x
+
2y
+
Sz
y
+
3z
y
+
Sz
= = =
x
+
+ +
2y y
8z
-
3z
=
In echelon form there are only two equations in the three unknowns; hence the system has a nonzero solution. Accordingly, the vectors are dependent.
We need
Remark:
know
to
not solve the system to determine dependence or independence; a nonzero solution exists. x(l,
(ii)
(x,
(x
-2, -3)
+
-1)
3/(2, 3,
=
z{3, 2, 1)
only need
(0, 0, 0)
=
+ (2y, 3y, -y) + + 2y + 3z, -2x + 3y + 2z, -3x - y + z) = 3z
=
-2x + 3y +
2z
-3x - y +
z
= =
2y
+
-2x, -3x)
+
+
x
we
if
(3z, 2z, z)
x
+
2y
+
3z
-
7y
+ +
Sz
= =
5y
lOz
(0, 0, 0) (0, 0, 0)
x
Q
+
2y 7y
3z
=
8z
=
30z
=
+ +
In echelon form there are exactly three equations in the three unknowns; hence the system has only the zero solution. Accordingly, the vectors are independent. (iii)
x(,ai, 02) {ttix,
(dja;
+
+
a2x)
+
y{bi, 62)
z(ci, C2)
+
{byy, h^y)
+ CiZ,
61J/
+ +
a2X
=
(0, 0)
=
(c^z, c^z)
b^y
"''*'
and so
(0, 0)
02*
+ C2Z) =
+
62J/
~ =
"'^
'^
+
C2Z
(0, 0)
The system has a nonzero solution by Theorem 2.3, i.e. because there are more unknowns than equations; hence the vectors are dependent. In other words, we have proven that any three vectors in R2 are dependent.
2.10.
Suppose in a homogeneous system of linear equations the coefficients of one of the unknowns are all zero. Show that the system has a nonzero solution. Suppose are
all zero.
«!, ...,«„ are the unknowns of the system, and Then each equation of the system is of the form
«i^i
+
•
•
•
+
+
ttj-i^j-i
Then for example (0, .,0, 1, 0, .,0), where 1 equation and hence of the system. .
.
.
+
Oajj
.
is
aj + i^j +
i
Xj is the
+
•
+
•
•
the ;th component,
unknown whose
o„a;„ is
coefficients
=
a nonzero solution of each
MISCELLANEOUS PROBLEMS 2.11.
Prove Theorem (*) and suppose
2.1:
W
is
Suppose m is a particular solution of the homogeneous system the general solution of the associated homogeneous system (**).
Then is
W
u +
=
{u
+ w: w G W}
the general solution of the nonhomogeneous system Let
u = (%
U
(*).
denote the general solution of the nonhomogeneous system (*). Un). Since m is a solution of (*), we have for t = 1, , m,
Now
suppose
(**),
we have
.
w6 for
i
W =
and that 1,
.
.
w=
(w^,
.
.
.,
Since
w„).
w
is
+
aj2W2
+
•
•
+
fflin^^n
Suppose
uG U
and that
.
a solution of the homogeneous system
.,m, OiiWi
.
=
LINEAR EQUATIONS
30
Therefore, for
=
i
M+w
is,
+ Wi) +
Oi2(M2
=
OjiMi
=
(OilMl
=
6i
+
+ W2) +
+
OjiWi
+ ai2M2 + + = 6j
a solution of
is
•
•
+
•
aj2M2 •
Now
=
1,
.
.
=
V
suppose
(vi,
.
.
.
Observe that v
vj
+
"
'
+
ai2-U2
W
+
•
-
6i
+
•
*
=
ai„v„
+
Cmaml)Xl
+
+ «in«'n)
solution of
i.e.
•
•
.
m,
,
.
•
•
•
+ «{„«„)
u-mST^.
(*), i.e.
+
•
+
(Cittm
•
that
•
=
(fci,
.
.
,
.
fcj
ffliifci
To show that (Ciaii
But
this
tt
+
is,
+
•
u+W
the general solution of the
is
Multiplying the ith equation
18).
=
CmOmn)*™
+
Ci5i
by
which
2.13.
•
•
+
•
+
•
Cmd^Ofcl
•
•
+
0,lnK) Ci6i
(2),
is
(1),
•
•
+
•
we must
+
•
•
•
•
•
•
in (*).
+
Cmbm
(1)
Show
that
Then
(*).
=
«{«*:„
t
6i,
=
1,
. .
.,m
(2)
verify the equation
+
+
("ifflln
+
Cm(aml
•
+
•
•
•
•
=
emO'mn)K
«!&!
+
"
"
"
+
C^fe^
can be rearranged into Ci(anfel
or,
+ aah +
a solution of
is
a solution of
is
and hence
W
u +
U - u + W;
vGu+W,
Then
Such an equation is termed a linear combination of the equations any solution of () is also a solution of the linear combination (1). Suppose M
Then, for
(*).
bj
•
Consider the system (*) of linear equations (page by Ci, and adding, we obtain the equation •
'
'
a solution of the homogeneous system
•
"
=
6i
Both inclusion relations give us nonhomogeneous system (**).
•
'
U
c
•
U Q
+
«inWn
+ «i2W2 +
("il^l
.
=
(CiOn
+
linMn
= u+(v — u). We claim that v-uGW. For i = 1, + »m(^n ~ "n) ail(i;i — Ml) + ai2(t'2 — M2) + = (OjlVl + aj2'y2 + + ftin^n) ~ («il"l + «t2"2 + •
- M is
+
'
any arbitrary element of U,
is
"
2.12.
«t2"'2
+
.,w, ttji^i
Thus V
+ W„)
Thus u + w e^U, and hence
(*).
,
ai„{Un
+ «tnO +
•
•
u +
t
2
l, ...,nt,
0,i(Mi
That
[CHAP.
+
+
•
•
•
•
•
•
+
cj)„^
+ =
•
+
Cjbi
=
amn'«n)
+
•
•
•
+
Ci^l
+
'
"
"
+
C^fem
c^b^
clearly a true statement.
of linear equations, suppose an ¥= 0. Let (#) be the system ob^ -anLi + auU, i^l. Show that (*) and (#) tained from (*) by the operation solution set. same are equivalent systems, i.e. have the
In the system
(*)
U
equations In view of the above operation on (*), each equation in (#) is a linear combination of solution of also a (#). is of solution any (*) problem in (*); hence by the preceding
On
the other hand, applying the operation
!/{ -*
—
(-Oii^^i
+
i-O
to (#),
we
obtain the origi-
each equation in (*) is a linear combination of equations in (#); hence each
nal system (*). That is, solution of (#) is also a solution of
Both conditions show that
(*)
(*).
and (#) have the same solution
set.
CHAP.
2.14.
LINEAR EQUATIONS
2]
Prove Theorem
31
2.2:
Consider a system in echelon form:
aiia;i
+
+ aizXz + a2J2^J2 + a2,J2+l«J2 + +
ai^Xi
l
+
O'ri^^ir
1
where
= n.
r
(i)
< jr and where an ^
•
There are two
follows.
•
^0,
0, a2J2
Then the system has a unique solution. Then we can arbitrarily assign values
The proof is by induction on the the single linear equation aiXi
The free variables are
=
—
xs
Iti,
k^,
+
=
+
a^Xi
.
•
.
ainXn
= =
+
arnaJn
=
.
ari,
,
bi
62
for
^ 0,
The
solution is as
number r of equations
+
a^x^
•
•
+
•
=
a„a;„
n—r
to the
in the system.
where
6,
free variables and
If
r
=
then
1,
we have
Oj #•
Let us arbitrarily assign values to the free variables; say, Substituting into the equation and solving for Xi,
x^,
...,«„
•
ainXn
cases:
r
(ii)
«2
+
<*r,]V+ia;j^+l
+ +
., a;„.
fe„.
— (6 -
=
Xi
-
fflzfca
-
-
asks
o„fc„)
"1
These values constitute a solution of the equation;
—
"i
which
is
(* "" 02*^2
—
•
•
—
•
+
a„k„)
ajt^
+
we
on substituting,
for,
•
•
+
•
=
a„fc„
obtain or
6
6
=
6
a true statement.
Furthermore tion since
a(b/a)
r
if
=
5
=%=
we have ax =
then
1,
Moreover
is true.
where a
# 0.
a solution,
i.e.
b,
Note that x = b/a is a soluak = b, then k = b/a. Thus
=
k
is
true for a system of r
x
if
is
the equation has a unique solution as claimed.
Now
assume r
>
and that the theorem
1
—1
equations.
r — 1 equations
+
'*2J2*J2
'*2,J2+1*J2 +
as a system in the unknowns Xj^
we can
to obtain
a solution
(say,
values for the additional
Xj^ J2
—
fcj^,
—2
.
(n
—
x„
. ,
.
"^ "2na;n
1
Note that the system
«„.
arbitrarily assign values to the
"^
+
J2
—
—
1)
of the first equation with
—
=
Xi
(6j
-
012^2
view the
*2
echelon form.
By
induction
—
1) free variables in the reduced system in case r = 1, these values and arbitrary
(r
As
&„).
free variables (say,
is in
=
We
a;2
-
= •
•
^2,
•
-
•
•,
a^j,-!
=
'fja-i)'
yield
a solution
ai„k„)
Oil
(Note that there are (n — J2 + 1) — (r — 1) + (jg — 2) = n — r free variables.) Furthermore, these values for Xi, .,x„ also satisfy the other equations since, in these equations, the coefficients of «!,..., »j„-i are zero. .
.
Now if r = n, then 32 = 2. Thus by induction we obtain a unique solution of the subsystem and then a unique solution of the entire system. Accordingly, the theorem is proven. 2.15.
A
system
(*)
of linear equations
of its equations
is
Oa;i
Show
defined to be consistent
is
if
no linear combination
the equation
that the system
+
0*2
(*) is
+
•
•
•
+
consistent
Oa;„ if
=
b,
and only
where b if it is
¥-
(I)
reducible to echelon form.
reducible to echelon form. Then it has a solution which, by Problem 2.12, is a solution of every linear combination of its equations. Since (1) has no solution, it cannot be a linear combination of the equations in (*). That is, (*) is consistent.
Suppose
On it
must
(*) is
the other hand, suppose yield
Accordingly
(*) is
an equation of the form (*) is
not consistent,
not reducible to echelon form. Then, in the reduction process, That is, (J) is a linear combination of the equations in (*).
(1).
i.e. (*) is
inconsistent.
LINEAR EQUATIONS
32
[CHAP. 2
Supplementary Problems SOLUTION OF LINEAR EQUATIONS
+ +
2x 2.16.
Solve:
(i)
5x
2.17.
7y
+ y - Sz = -2y + 2z = 5x -Sy - z =
(i)
(ii)
3«
3
2x
5
Sx
5
+
3y
X- 2y 3x
+
2y
= =
4y 6y
= =
-2y =
5
-6x + 3y =
1
Ax
10 (iii)
15
+ 3y -2z = - 21/ + 3« = 4a; — + 4z =
2x
5
«
2
(ii)
16
1/
3
x
5
2x
(ii)
3x
7
X 3x Solve:
(i)
2x X
2.20.
+ +
2x
1
X 2x
(iii)
1
3a;
+ 2y + + 3y+ + 2j/ +
3« 8«
17?
= = =
3
4 1
Solve:
2x
2.19.
= =
Solve:
(i)
2.18.
Sy
+ 2y + -2y - 5y + + Ay +
+ 2y-3z + 2w + 5y - 8z + Gw + Ay - 5z + 2w = = 2 = 5 = -A —
2z z
3z 6z
X
2 5
3x
4
5y
3a;
+ -
2x
+
2y
X (ii)
2x
(iii)
Determine the values of fc such that the system in unknowns no solution, (iii) more than one solution:
y
x,
+ + +
-
+ + +
2y Ay 6j/
+ -
= + 5w = - Aw =
Az
2z 3z
13w
y and
z has:
+ +
= =
z
4z 2
+ 3w = + 3w = + 8w =
3 9
10
3
2 1
(i)
a unique solution,
(i)
a unique solution,
(ii)
(o)
1 l
X
2.21.
X (6) 2a;
(a)
3x
2x
Determine the condition on
(i)
+ + +
a, b
Ay
+ +
3y
—
y
and
-
kz
in
z
X (6)
3z 2z Bz
= =
2x X
1
c so that the
X + 2y — 3x- y + X — 5y +
unknowns
2
2z-k =
+ +
2y fc«
kz 8«
1
3
1
Determine the values of k such that the system (iii) more than one solution: (ii) no solution, X
2.22.
+y+z = + ky + z = + y + kz =
kx x
y and
-
+ ky + 2y +
(ii)
e
z has:
3z z
kz
system in unknowns
a b
x,
x,
= -3 = -2 - 1
y and
x — 2y + Az = 2x + Sy - z 3x + y + 2z =
z
has a solution:
a b e
HOMOGENEOUS SYSTEMS 2.23.
Determine whether each system has a nonzero
-2z = x-8y + 8z 3x-2y + Az = x
(i)
+
+ 2x 3x X
3y
(ii)
3y 3y 2y
solution:
-2z = + z = + 2z =
+ 2y — 2x - 3y + 4x — 7j/ + X
(iii)
+ Aw = 2z + 3w = z — 6w —
5z
CHAP.
2]
2.24.
Determine whether each system has a nonzero solution:
LINEAR EQUATIONS
X-
2y
2x+
y
+ -
2z
=
2z
=
(i)
3x+ 4y3x - lly +
2.25.
Determine whether the vectors
7z
9x
+
3y
+ -
2y
+ -
5y
12z
=
6x
u,
v and
w
(2, 0, 1),
w =
(1,
(ii)
u
=
(1, 1,
-1), V
=
(2, 1, 0),
w =
(-1,
(iii)
u =
(3, 2, 1,
-2),
4v
- 5w -
2z
+ ^
7v
+ w =
3z
+
v
+ 4z-
3v
+ 3w — 2w -
Q
are dependent or independent (see Problem 2.9) where:
=
=
+
5x
-1), V
V
4y
=
(1, 3,
3, 1),
-
6z
u =
-2,
2x
(ii)
(i)
(1,
33
-1,
1)
1, 2)
w =
-5, -4)
(1, 6,
MISCELLANEOUS PROBLEMS 2.26.
Consider two general linear equations in two unknowns x and y over the real
Show
+
by
=
e
ex
+
dy
=
f
K:
that:
it
(i)
ax
field
-¥'2,
i.e.
if
ad
-
6c ¥= 0,
then the system has the unique solution
x
=
_ af — ee ~ ad-bc' ,
^
2.27.
(ii)
i*
7 =
(iii)
ii
— = 2 =
J
'^
that if
~ —
^f he
7. tlien the system has no solution;
-f>
then the system has more than one solution.
Consider the system
Show
^^
ad
ax
+
by
=
ex
+
dy
=
1
ad-be¥'0, then the system has the imique solution x = d/(ad — be), y = —e/{ad if ad—be = 0,e¥'0 or d ^ 0, then the system has no solution.
be).
Also show that
=
2.28.
Show that an equation of the form Oki system without affecting the solution set.
2.29.
Consider a system of linear equations with the same number of equations as unknowns:
+
Oa;2
+
fflii*!
+
ai2«2
+
a^xi
+
a22«2
+
*
•
•
•
•
•
•
•
•
+
Oa;„
may
+
ai„x„
=
61
+
a2„x„
=
62
be added or deleted from a
(i)
Onl*! (i)
(ii)
+
01.2*2
+
•
•
+
«„„«„
=
6„
Suppose the associated homogeneous system has only the zero solution. unique solution for every choice of constants 6j. Suppose the associated homogeneous system has a nonzero solution. 64 for which {!) does not have a solution. Also show that if it has more than one.
constants
Show that
Show {1)
(i)
has a
that there are
has a solution, then
[CHAP.
LINEAR EQUATIONS
34
Answers =
=
(i)
X
2.17.
(i)
(1,-3,-2);
2.18.
(i)
a;
(ii)
(-a
(iii)
(7/2
2,
y
-1;
2.16.
(ii)
a;
=
Supplementary Problems
to
5
- 2a,
=
j/
a;
no solution
(iii)
rx
=
3,
no solution;
(ii)
{-1
(iii)
= -1
1/
ra;
+ 26, 1 + 2a - 26,
a, 6)
o*"
]
- 7a, 2 + 2a,
or
a)
= — z + 2w ^ + 2z — 2w = 7/2 or |^ ^ ^^^ + a;
2.19.
(i)
2.20.
(a)
2.21.
-
-1);
(2, 1,
-
56/2
(6)
(i)
never has
(a)
(i)
fc
fe
#
-2;
(ii)
# 3; ^2
and
+
=
A;
# -5;
(ii)
2.22.
(i)
2a
2.23.
(i)
yes;
(ii)
no;
(iii)
2.24.
(i)
yes;
(ii)
yes,
by Theorem
2.25.
(i)
dependent;
6
c
-2;
(ii)
(iii)
—
k
always has a solution;
(ii)
(i)
-
=
fc
a unique solution;
(6)
fc
6/2, 6)
5w/2
-
2j/
w/2
no solution
(ii)
k¥'l and
(i)
+
2a, a, 1/2
0.
(ii)
(ii)
Any
yes,
fc
=
(iii)
-5;
values for
by Theorem
(iii)
a, b
(iii)
1
(iii)
fe
fe
2.3.
dependent
fe
t^
4
=3 =2
and
2.3.
independent;
=
fe
4;
c yields
= -1 - 7z g + 2z
|^ ^
a solution.
2
chapter 3
Matrices INTRODUCTION In working with a system of linear equations, only the coefficients and their respective Also, in reducing the system to echelon form, it is essential to keep the equations carefully aligned. Thus these coefficients can be efficiently arranged in a rectangular array called a "matrix". Moreover, certain abstract objects introduced in later chapters, such as "change of basis", "linear operator" and "bilinear form", can also be represented by these rectangular arrays, i.e. matrices.
positions are important.
In this chapter, we will study these matrices and certain algebraic operations defined on The material introduced here is mainly computational. However, as with linear equations, the abstract treatment presented later on will give us new insight into the structure of these matrices.
them.
Unless otherwise stated, all the "entries" in our matrices shall come from some arbitrary, fixed, field K. (See Appendix B.) The elements of are called scalars. Nothing essential is lost if the reader assumes that is the real field R or the complex field C.
K
but
K
Lastly, we remark that the elements of R" or C" are conveniently represented by "row vectors" or "column vectors", which are special cases of matrices.
MATRICES Let
K be an arbitrary field. A
rectangular array of the form
\Q,ml
0,12
.
.
.
din
0,22
.
.
.
0,2n
Om2
...
fflr
where the Odi are scalars in K, is called a matrix over K, or simply a matrix if K is implicit. The above matrix is also denoted by (ohj), i = l, .,m, j = 1, .,n, or simply by (a«). The m horizontal «-tuples .
(ail, ai2,
.
.
.
,
ttln),
(tt21, 0^22,
.
.
.
,
.
a2n),
.
.
.
.,
.
{ami, am2,
.
.
.
,
Omn)
are the rows of the matrix, and the n vertical w-tuples lai2X a22 ,
•
•
.
f
\am2l
are its columns. Note that the element ay, called the ij-entry or ij-component, appears in the ith row and the yth column. A matrix with rows and n columns is called an by « matrix, or x n matrix; the pair of numbers (m, n) is called its size or shape.
m
m
35
w
MATRICES
36
Example
3.1:
The following Its
/I -3
a 2 X 3 matrix:
is
rows are
(1,
—3,
(
and
4)
[CHAP.
(0, 5,
3
4\
r
_c, )
—2);
its
•
columns are
(
«
)
.
(
r
)
and
I
j
.
and the elements of the Matrices will usually be denoted by capital letters A,B, ., Two matrices A and B are equal, written A = B, if field by lower case letters a,b, they have the same shape and if corresponding elements are equal. Thus the equality of matrices is equivalent to a system of mn equalities, one for each pair of elements. two .
K
.
.
.
.
.
mxn
Example
3.2:
The statement ..
.
^
'")=(,
"
z-wj
\x-y
(
equivalent to the following system
is
.)
4/
VI
of equations:
= = y +w = —w = + —
x
X 2z z
The
solution of the system is
A matrix with one
Remark:
row
as a column vector. a 1 X 1 matrix.
is also
x
=
=
y
2,
y
z
1,
I
5
4
=
a21
ai2 022
...
ain
vector, field
Oml
...
ami
and B, written
.
CLin
.
1
\
^^2
...
bin
^Hi
^22
...
ban
&ml
6m2
ai2
+
a22
&2I
+
+ +
Omi
bml
&12
...
am +
bin
622
...
a2n
+
?>2n
+
The product of a scalar k by the matrix A, written by multiplying each entry of A by k:
fc
&m2
.
A
•
kaml
.
•
Omn
.
.
&mti
+
or simply kA,
kai2
Ckaii fca2i
feain
ka22
.
fcOm2
.
.
•
.
.
kazn
kOmn I
A+B and kA are also mxn matrices. We -A = -1-A
The sum
.
the matrix obtained by adding corresponding entries:
an + &n
ami
^u
I
/
is
a21
_
g
.
ffimn
.
A + J?,
A + B„ =
Observe that
and with one column K can be viewed as
same number of rows and of
/
\
I
A
—1.
mxn matrices: (an
The sum of
w=
an element in the
MATRIX ADDITION AND SCALAR MULTIPLICATION Let A and B be two matrices with the same size, i.e. the columns, say,
3,
row
referred to as a
In particular,
3
of matrices with different sizes
and is
also define
A-B ^ A+
not defined.
{-B)
is
the matrix obtained
CHAP.
MATRICES
3]
Example
A =
Let
3.3:
(]
J\
^
2A-SB Example
°
^).
Then
1 + 3-2 + 3 + 2 4-7 5 + 1-6 + 8
3*1
=
3
•
-'
-c
(-2)
3
3-5
3 '4
r '8
=
mXn matrix
The
3.4:
fj
B =
and
A + B 3A
37
r
') +
-2 6
3-6
3
3 '(-6)
-12/
10
•
4
-3
12 ° "' -3 -24
V2I
whose entries are
all zero,
10
...
9
-18
15
-7
-4
29
7
-36
...
...
,0 is
called the zero matrix
any
that, for
mXn
and matrix
will be denoted
A=
0,
by
It is similar to the scalar
0.
A+ =
(a^),
(a^
+ 0) =
(Oy)
=
in
A.
Basic properties of matrices under the operations of matrix addition and scalar multiplication follow.
Theorem
F be the set of all m x n matrices over a field K. Then for any matrices
Let
3.1 :
A,B,C
GV
(i)
(ii) (iii)
(iv)
Using
Remark:
(vi)
and
(viii)
and any scalars
(A+B) + C = A + A+ = A A + (-A) = A+B = B + A
above,
Suppose vectors say,
we
in
u —
also
ki, kz
€ K,
{B + C)
(vi) (vii) (viii)
have that
A + A = 2A,A + A + A = ZA,
R" are represented by row vectors (ai, Oi,
.
.
. ,
ttn)
and
v
-
{ai
+ bi,a2 + b2,...,an + b„)
=
v
Then viewed as matrices, the sum u + v and the
u+
+ B) = kiA + kiB {ki + fe)^ = kiA + k^A (kiki)A = kiik^A) 1- A = A and OA = k,{A
(v)
by column
(or
(bi, 62,
.
.
.
,
scalar product
=
ku
and
...
.
vectors);
b„)
ku are as
(fcai,
follows:
kaz, ..., A;a„)
But
this corresponds precisely to the sum and scalar product as defined in Chapter 1. In other words, the above operations on matrices may be viewed as a generalization of the corresponding operations defined in Chapter 1.
MATRIX MULTIPLICATION The product of matrices A reason, (i)
we
and B, written AB, is include the following introductory remarks.
somewhat complicated.
For
this
Let A = (Oi) and B = (bi) belong to R", and A represented by a row vector and B by a column vector. Then their dot product A B may be found by combining the matrices •
as follows:
A-B
=
lbl\ (tti,
02,
.
.
.,a„)
(
M
=
aibi
+
a2b2
+
•
+
ttnbn
Wl Accordingly, above.
we
define the
matrix product of a row vector
A
by a column vector
B
as
MATRICES
38
(ii)
[CHAP.
bnXi
+
biiXi
+
feisics
=
y\
h2lXl
+
b22X2
+
b23X3
=
1/2
Consider the equations
This system
where vector
is
6n
b.
b.s\h\
&21
&22
&23/U3/
(x,)
and
_
(
""
"'-
fyA
^
Y—
""!(
we combine
if
(yi),
_
1
where Bi and B2 are the rows of B. vector yields another column vector.
Now
^^ ^ ^
^^^.^pjy
V^V
/feiiaJi
\b2iXl
(iii)
(1)
equivalent to the matrix equation
B — (&«), X = X as follows: Dv-
3
+ +
B
the matrix
&i2a;2
b22X2
+ +
fBi'X \B2-X
_
bisa^sN
b2SXs
and the column
J
Note that the product of a matrix and a column
auVi
+
ai22/2
=
zi
a2iyi
+
(i22y2
=
Z2
consider the equations
(2)
which we can represent, as above, by the matrix equation ^aii
,
^21
where of
or,
On
A=
(i) into
(Cij),
Y=
( Zx
ai2\/yi\ ,
,
,
022/^2/2/
{yi)
Z=
as above, and
the equations of
we
{2),
AY = Z
or simply
,
y22
Substituting the values of y\ and
(z^.
1/2
obtain
aii(&iia;i
+
6i2a;2
+
b\%x%)
+
ai2(62ia;i
+
622332
+
&23a:;3)
=
«i
a2i(&iia;i
+
&i2a;2
+
bisXs)
+
a22(&2ia;i
+
&22a;2
+
btzx^)
=
22
on rearranging terms, (ttubii
+ ai2&2i)a;i +
(aii6i2
+
ai2&22)a;2
+
(an&is
+
a\2b23)Xz
=
Zi
(azi&u
+
+
(«2i&i2
+
a22&22)a;2
+
(021613
+
022623)033
=
22
022&2i)a;i
the other hand, using the matrix equation we obtain the expression
BX = Y
(3)
and substituting for
Y
into
AY = Z,
ABX = Z This will represent the system
(3) if
ftii
ffli2\/6n 612 bisX
021
022/1621
622
623/
Ai-B'
ArB^ Ai-B^
A2-jB'
Aa-B^ A2'B^
we
_
define the product of
+ 012621 YO21611 + 022621
/aii6ii
A
011612 021612
and
B
+ 012622 + 022622
as follows: 011613 021613
+ 012623 + O22623
and J?S B^ and B^ are the columns of B. We emphasize that if these computations are done in general, then the main requirement is that the number of yi in (1) and (2) must be the same. This will then correspond to the fact that the number of columns of the matrix A must equal the number of rows of the matrix B.
where Ai and A2 are the rows of
A
CHAP.
MATRICES
3]
With the above
we now formally
introduction,
39
define matrix multiplication.
A = (a«) and B = (&«) are matrices such that the number of columns equal to the number of rows of B; say, A is an x p matrix and B is a matrix. Then the product is the matrix whose y-entry is
Suppose
Definition:
of
A
m
is
pxn
row A,
obtained by multiplying the ith
A
of
by the yth column B' of B:
A2-S1
Ai-B2 A2-B2
,A„-Bi
Am-B^
Ai-fii
AB = That
mxn
AB
Ai-5" A2-B"
. .
.
.
.
.
Am'B"!
is,
jail
Cm
...
/Cii
Cii
dml
...
\Cml
Opn
Ci
P
where
=
cy
+
aiiftij
ai2&23
+
•
•
2 1=
+ avpbp. =
•
fc
We emphasize that the qxn matrix, where p ^ q. Example
product
r
s
t
u
3.5:
Example
1
2
3
4
AB
not defined
is
<»i
(H
"3
raj
6i
62
^3
toi
1
1
3.6:
1
2
1
1
2
2
3
4
mxp
an
is
+ s6i + m6i
ra2 (02
+ 562 + M^2
1-1 + 2-0 3-1 + 4-0
I'l + 2-2 3«l + 4'2
1-1 + 1-3 0'l + 2-3
1-2 + 1'4 0*2 + 2*4
The above example shows that matrix multiplication
AB and BA
A
if
Cifc&fci-
l
is
matrix and
^'is
*"3
B
is
a
+ S63 + '^^s
not commutative,
1
5
3
11
4
6
6
8
i.e.
the products
of matrices need not be equal.
Matrix multiplication does, however, satisfy the following properties:
Theorem
3.2:
iAB)C = A{BC),
(i)
(ii)
(iii)
(iv)
(associative law)
A{B + C) = AB + AC, (left distributive law) (B + C)A = BA + CA, (right distributive law) k{AB) = {kA)B = A{kB), where is a scalar A;
We assume that the sums and products in the above theorem are We remark that OA = and BO =^ where is the zero matrix.
defined.
TRANSPOSE The transpose of a matrix A, written A*, A, in order, as columns: /ttii
0.12
.
.
.
Oln
0,21
ffi22
.
.
.
02n
Om2
.
^Oml
Observe that
if
A
is
an
.
.
m x « matrix,
is
the matrix obtained by writing the rows of
'
\ /
\
Omni
then A'
is
/ttli
0.21
.
.
.
aTOl\
O12
ffl22
.
.
.
Om2
\Oin
02n
.
.
OmnJ
an w x
m matrix.
MATRICES
40
Example
3.7:
The transpose operation on matrices
Theorem
3.3:
J
(J
(A+B)* = A* + B*
(ii)
(A')'
(iii)
(iv)
=
_IJ
/l
4^
(2
-5^
3
satisfies the following properties:
(i)
= A {kAy — kA\ {ABy = B«A«
[CHAP.
for k a scalar
MATRICES AND SYSTEMS OF LINEAR EQUATIONS The following system
of linear equations
a2iXi
+ ai2X2 + + a22X2 +
OmlXi
+
anXi
is
•
•
•
+
aina;n
=
•
•
•
+
annXn
=62
+
OmnXn
+
am2X2
•
•
•
&i
n\
equivalent to the matrix equation
/an
ai2
a2i
022
lOml
...
a2n
\IX2\
lb2
^
AX = B
or simply
\
(2)
fflm2
X=
B=
every solution of the system {1) is a solution of the matrix equation (2), and vice versa. Observe that the associated homogeneous = 0. system of (1) is then equivalent to the matrix equation
where
A=
(an),
{Xi)
and
That
(&i).
is,
AX
The above matrix
A
is called
its
/(111
O12
.
.
•
ttin
tt21
tt22
•
•
•
(^2n
^ttml
(lm2
•
.
.
Otnn
the augmented matrix of augmented matrix.
is called
by
the coefficient matrix of the system
Example
3.8:
The
coefficient
(1).
Observe that the system
(1) is
(1),
and the matrix
completely determined
matrix and the augmented matrix of the system
-
4z
x-2y-
5z
2a;
+
3j/
= -
7 3
are respectively the following matrices:
/2
(1 Observe that the system
3
_2 is
-4\ -5;
/2 \l
^*^
3-4 -2
-5
7 3
equivalent to the matrix equation
X\
'2
3
1
-2
,rj
In studying linear equations it is usually simpler to use the language and theory of matrices, as indicated by the following theorems.
CHAP.
MATRICES
3]
Theorem
are solutions of a homogeneous system of linear Then every linear combination of the m of the form kiUi + kiUz + + krOin where the fe are scalars, is also a solution of AX = 0. Thus, in particular, every multiple ku of any solution u of
Suppose
3.4:
Ui,U2,
equations
AX = We
.
.
.,tin
AX = 0. •
Proof.
41
•
•
AX = 0.
a solution of
is also
— 0, Au2 = 0, Aun = 0. Hence + knAun + fettn) = kiAui + kiAu^ + = fciO + ^20 + = fc„0 +
are given that Aui
A{kui
+
kui
+
•
•
•
.
.
. ,
•
•
Accordingly, kiUi
Theorem
+
•
•
•
+ k„iia
Suppose the
3.5:
is
•
•
•
•
a solution of the homogeneous system
field
K is
infinite (e.g. if
Then the system AX = an infinite number of solutions. field C).
B
AX = 0.
K is the real field R or the complex has no solution, a unique solution or
AX
= B has more than one solution, then Proof. It suffices to show that if = B; that is, Au = many. Suppose u and v are distinct solutions of — B. Then, for any k GK,
AX
infinitely
Av
Au + k{Au-Av) = B + k(B-B) = B solution of AX = B. Since each k e K, u + k(u-v) is (Problem 3,31), AX = B has an infinite number of
it
B
has
and
A{u + k{u-v)) = In other words, for are distinct
tions
a.
all
such solu-
solutions
as
claimed.
ECHELON MATRICES A matrix A = (an) is an echelon matrix, or is said to be
aiii, '^^h'
with the property that aij
We call
ttijj,
.
Example
.
.
, ttrj,.
3.9:
=
•
•'
"'^'r'
for
where
i^r,
j
< ji,
^i
< ^2 <
and for
number
in echelon form, if the
row by row
of zeros preceding the first nonzero entry of a row increases rows remain; that is, if there exist nonzero entries •
•
•
until only zero
< jr
i>r
the distinguished elements of the echelon matrix A.
The following are echelon matrices where the distinguished elements have been circled: /(i)
3
2
4
5
1-3
2
-6\ 2
0/
In particular, an echelon matrix tinguished elements are:
is called
a row reduced echelon matrix
(i)
the only nonzero entries in their respective columns;
(ii)
each equal to
if
the dis-
1.
third matrix above is an example of a row reduced echelon matrix, the other two are not. Note that the zero matrix 0, for any number of rows or of columns, is also a row reduced echelon matrix.
The
ROW EQUIVALENCE AND ELEMENTARY ROW OPERATIONS A matrix A is said to be row finite
equivalent to a matrix
sequence of the following operations called
B
if
B
can be obtained from
elementary row
operations:
A
by a
MATRICES
42
[CHAP. 3
[Et]:
Interchange the ith row and the yth row: Rt <^
[E2]:
Multiply the ith row by a nonzero scalar
[Es]:
Replace the ith row by k times the jth row plus the ith row: Ri
In actual practice
we apply
Replace the ith row by Ri -* k'Rj + kRi, k^-O.
[E]:
[£^2] fe'
and then
k:
[£"3]
Ri
Rj. -»
kR,,
fc
v^ 0.
in one step,
-*
kRj
+
R,.
the operation
i.e.
times the jth row plus k (nonzero) times the ith row:
The reader no doubt recognizes the
and those used In fact, two systems with row equivalent augmented matrices have the same solution set (Problem 3.71). The following algorithm is also similar to the one used with linear equations (page 20). similarity of the above operations
in solving systems of linear equations.
Algorithm which row reduces a matrix Step
1.
Suppose the ji column is the first column with a nonzero entry. Interchange the rows so that this nonzero entry appears in the first row, that is, so that
Step
2.
form:
to echelon
¥- 0.
ttijj
For each
i
>
1,
apply the operation Ri
-*
+
—ttij^Ri
aijjiJt
Repeat Steps 1 and 2 with the submatrix formed by all the rows excluding the Continue the process until the matrix is in echelon form.
The term row reduce
Remark:
Example
3.10:
shall
The following matrix R2 ^ -2Ri + ^2 and
mean
A
is
i?3
to
transform by elementary row operations.
row reduced
form by applying the operations and then the operation R3 -» -SKj + 4^3:
to echelon
^ -3fii + R3,
-3
2
1
aijj,
.
.
.
,
2
5
3
-3 4
2 2
a matrix in echelon form with distinguished elements Apply the operations
suppose Orj^.
A=
4
2
1
A=2 4-22to0042to Now
first.
is
(oij)
Rk
-^
-ak^Ri
+
Oii-Rk,
fc
=
1,
.
.
.,
i-
1
then i = 3, ...,i = r. Thus A is replaced by an echelon matrix whose distinguished elements are the only nonzero entries in their respective columns. Next, multiply Ri by a~^, i~r. Thus, in addition, the distinguished elements are each 1. In other words, the above process row reduces an echelon matrix to one in row reduced echelon form. for
i
=
2,
Example
3.11:
On the following echelon matrix A, apply the operation the operations fii - ^3 + Bi and R^ -> — 5K3 + 2i22: /2
A=0
3
4
5
3
2
1/6,
R2 by
9
7 3
2
1/6
and ^3 by 1/2
-^
— 4^2 + 3i2i
-2\
/6
5to0 2/
\0
2/
\0
Next multiply Ri by
/6
6\
5toO
R^
\0
to obtain the
9
and then
7 6
0^
4 2/
row reduced echelon
matrix /l
3/2
7/6
0\
12/3 \0
1/
The above remarks show that any arbitrary matrix A is row equivalent to at least one row reduced echelon matrix. In the next chapter we prove, Theorem 4.8, that A is row equivalent to only one such matrix; we call it the row canonical form of A.
CHAP.
MATRICES
3]
SQUARE MATRICES A matrix with the same number of rows matrix with n rows and n columns
The diagonal an,
a22,
.
.
. ,
(or:
main
43
as columns
a square matrix. A square an n-square matrix. consists of the elements (Oij)
is called
and
said to be of order n,
is
diagonal) of the n-square matrix
A=
is called
ftjin.
Example
The following
3.12:
is
/l
2
3^
4
5
6
\7
8
9,
a 3-square matrix:
Its diagonal elements are 1, 5
and
9.
An upper
triangular matrix or simply a triangular matrix below the main diagonal are all zero: entries
/an
ai2
.
.
O22
/ail
ain\
...
.
.
a square matrix whose
am a2n
or
ann/
whose entries above the main
Similarly, a lower triangular matrix is a square matrix
diagonal are
.
a22
ain
\0
ai2
is
all zero.
A diagonal matrix
is
a square matrix whose non-diagonal entries are
/a,
...
'ai
\
...
a2
\o
"-^
or
'
a„/
...
all zero:
an
In particular, the n-square matrix with I's on the diagonal and O's elsewhere, denoted by /« or simply /, is called the unit or identity matrix; e.g., /l
0^
10
h = \0 This matrix I
is
1,
similar to the scalar 1 in that, for any n-square matrix A,
AI = lA = A The matrix
a scalar k diagonal entries are each k. kl, for
G K,
is called
a scalar matrix;
it is
a diagonal matrix whose
ALGEBRA OF SQUARE MATRICES two matrices can be added or multiplied. However, if we only consider square matrices of some given order n, then this inconvenience disappears. Specifically, the operations of addition, multiplication, scalar multiplication, and transpose can be performed on any nxn matrices and the result is again an n x n matrix. Recall that not every
In particular,
if
A
is
any n-square matrix, we can form powers of A: A^
= AA, A^ =
We can also form polynomials
in the
f{x)
=
A^A,
.
and A"
..
=/
matrix A: for any polynomial
ao
+ ai* +
UiX^
+
•
•
•
+
ttnX"
MATRICES
44
where the
aj
we
are scalars,
is
Example
Let
3.13:
=
aiA
+
the zero matrix, then
A
aol
+
A = (J _l);
+ 3x- 10,
then
2a;2
If g{x)
=
x^
=
''^'
A
is
A^ =
then
(J
•
is called
then
=
+
a2A^
- 3a; + 5,
If f(x)
Thus
3
define f(A) to be the matrix
/(A)
In the case that f{A)
[CHAP.
^:) -
•
+ a„A"
a zero or root of the polynomial f{x).
(J
Ka
•
J)(J
-I) -
_^2)
<
= (_^ "^
^
:)
c
:
a zero of the polynomial g(x).
INVERTIBLE MATRICES A square matrix A is said
to be invertible if there exists a
matrix
B
with the property
that
AB = BA = I where / is the identity matrix. Such a matrix B is unique; for ABi - BiA = / and AB2 = B2A = I implies Bi = BJ - BiiABz) = iBiA)Bi = IB2 = B2 We call such a matrix B the inverse of A and denote it by A~*. Observe that the above relation is symmetric; that is, if B is the inverse of A, then A is the inverse of B. Example
2
5\/
3
-5\
1
3/1^-1
2)
-5\/2 2j\l
5\
s)
/
3
-5\
|
^
3
-1 ,'2
Thus
(
,
1
5\ „
)
and
„
3/"""\^-i
)
/6-5
_ ~
3.14:
-10 + 10 -5 + 6
1^3-3
_ "
/
6-5
V-2 +
1 1
15-15 -5 +
2
1
6
1
are invertible and are inverses of each other.
2
(Problem 3.37) that for square matrices, AB = I if and only if BA = /; hence it is necessary to test only one product to determine whether two given matrices are inverses, as in the next example.
We show
(2-1
3.15:
3)(-4
2 4
-44-4 + Thus the two matrices are
We now calculate X, y, z, w such that a
^\( ^
+ 0-2 2 + 0-2 + 0-3 4-1-3 8 + 0-8 8 + 1-8
+ + 12
l|=|-22 + 4 + 18
-11
Example
invertible
the inverse of a general
48
and are inverses of each
2x2
matrix
fa A —
y\ _ /l
0\
cd)\zwj~\0l)
^^
fax + bz \cx + dz
ay cy
b\
{
^
+ bw\ + dwj
other.
1
.
We
seek scalars
''
_ /l " \0
1
CHAP.
MATRICES
3]
which reduces
two systems of linear equations
to solving the following
+ bz = \cx + d2 = iax
If
we
let
and only
jay + bw =
1
\cy
+ dw =
1
then by Problem 2.27, page 33, the above systems have solutions such solutions are unique and are as follows:
\A\ ¥= 0;
d ad
two unknowns:
in
= ad — be,
|A| if
45
d
-be
—b
_
Accordingly,
_ ad
\A\'
,i^i
,
'^-c/|A|
— -
_ -
-b/\A\\
d/\A\
-
..
^
_
-be
ad
"
\A\'
i
—
^
"'
ad
\A\'
\ ( d i^ii \A\\^-c
0'l\A\J
—
z£.
be
if
-be
\A\
-b" a
The reader no doubt recognizes \A\ = ad — bc as the determinant of the matrix A; thus we see that a 2 x 2 matrix has an inverse if and only if its determinant
Remark:
not zero. This relationship, which holds true in general, will be further investigated in Chapter 9 on determinants.
is
BLOCK MATRICES Using a system of horizontal and vertical lines, we can partition a matrix A into smaller matrices called bloeks (or: eells) of A. The matrix A is then called a block matrix. Clearly, a given matrix may be divided into blocks in different ways; for example, 1
-2
2
3
5
7-2
3
1
4
5
The convenience
-2
3\
1
9/
3\
1 j
/I
-2
2
\3
'
3
3
5]7
-2
1
4
"1
1',
= \s
3
5
7 |-2
1
4
5
of the partition into blocks
is
=
9/
1
1
5
that the result of operations on block matrices if they were the
can be obtained by carrying out the computation with the blocks, just as actual elements of the matrices. This is illustrated below.
Suppose
A
is
partitioned into blocks; say
Ain Ain
Multiplying each block by a scalar
Now suppose
a matrix
B
is
k, multiplies
each element of
(kAii.
kAi2
feAai
iCAjnl
.
.
.
&A22
.
.
.
rCAjn2
.
.
.
partitioned into the
B =
A
by
k; thus
kAm kA2n iCAmn j
same number of blocks as A; say
Bn
B12
.
.
.
Bin
B21
B22
.
.
.
B2n
\^Bml
Bm2
...
j
9
B,
MATRICES
46
[CHAP. 3
Furthermore, suppose the corresponding blocks of A and B have the same size. Adding these corresponding blocks, adds the corresponding elements of A and B. Accordingly, /All A
I
—
I>
-^21
+ fill + -^21
Ai2
+ Bm\
Am2 + Bm2
A22
+ Bi2 + B22
The case of matrix multiplication
is less obvious but partitioned into blocks as follows
-Zl
^
JJ
U12
...
C/ip\
C/22
...
U2P
Vmi
...
Umpj
^^^
.
•
•
still
Aln
.
+
Bin
such that the number of columns of each block Uik block Vkj. Then
Amn + Bm true.
^
y
I
Wn
...
Wm
W-n
W22
...
Wzn
Ui2V2i
+
•
•
+
suppose matrices
is,
7i2
...
Vin\
V2I
V22
...
V2:
\Vj,l
F22
...
Fpn/
equal to the
is
That
/Fa I
'Wn
Wa = UnVn +
where
.
I
\Am\
U and V are
Am + Em
...
number
of rows of each
UipVpj
of the above formula for UV is straightforward, but detailed and lengthy. as a supplementary problem (Problem 3.68).
The proof is left
Solved Problems
MATRIX ADDITION AND SCALAR MULTIPLICATION 3.1.
Compute:
-{I ^^ (i)
2
-3
-5
1
[1 -1 -
Add
1)
/3 -5
4^\
-1
^
/3
-1\
-2
-3y
-
I2
J
6
5\ (iii)
-{I
2
-3
-5
6
corresponding entries:
n 2\Q -5
-5 I
-0
+
3
(0 +
2
/I
6
-
-2 -^)
(I
-
2-5
-3
+
(ii)
The sum
(iii)
Multiply each entry in the matrix by the scalar —3:
is
4-333
4-- l^ -1 --3y
6
1-2
-5 +
2
not defined since the matrices have different shapes.
0/1
2
-3N
^
-
-
'
'
'
_ -
/ '
-3 -12
-6
9
15-18
-5 -1 -4
It
CHAP.
««
3^.
MATRICES
3]
T
.
Let
-5
A = 72 .
-2 -3\
/I
„
1\
0-4)'^ = (0-1
(3
47
/O
-2\
1
= (l-l-lj-F^"
5J'^
First perform the scalar multiplication, and then the matrix addition:
/6 -15
„^ .„ „^ 3A + 4B - 2C ,
„ 3.3.
Fmda;,i/,zandwif
/4 -8 -12\ 2o}
3\
/
-12) + (0 -4
= (^
fxy\
3
w
z
^
X
6
—1
2w
/
=
\
I
-2
4\
2
2)
+ (-2
4
/
\
+
\z
X
+w
=
/lO -25
-5\
-2
lo)
( 7
+y 3
First write each side as a single matrix: /3a;
3y\
\3z
BwJ
_ ~
x
/
\z +
+
X
4
w-l
+
y
+
2w +
6
3
Set corresponding entries equal to each other to obtain the system of four equations,
= «+4 3y = X + y + 6 3z = z + w — 1 3w = 2w + 3 = 4, z = 1, w = 3.
= 4 = 6 + x 2y = 2z w—1 w = 3 2a;
3as
The
3.4.
solution
x
is:
=
Prove Theorem
2,
j/
Let
3.1(v):
A
and
B
or
be
mxn
matrices and k a scalar.
Then
kiA+B) = kA + kB. Suppose A — (Ojj) and B — is
Then Oy + 6jj is the y-entry of A + B, and so &(ajj + 6^) (bij). +B). On the other hand, ka^j and fcfty are the ij-entries of kA and kB respecfe6y is the ti-entry of kA + kB. But k, ay and &„• are scalars in a field; hence
the v-entry of k(A
and so
tively
fcay
+
k(aij
Thus k(A Remark:
+ B)
= kA +
+ 6jj) =
fcfflij
+
kbij,
for every
i,
j
kB, as corresponding entries are equal.
Observe the similarity of this proof and the proof of Theorem l.l(v) in Problem 1.6, page In fact, all other sections in the above theorem are proven in the same way as the corresponding sections of Theorem 1.1. 7.
MATRIX MULTIPLICATION 3.5.
Let if
(r
x
s)
denote a matrix with shape rxs.
the product
Find the shape of the following products
is defined:
(i)
(2x3)(3x4)
(iii)
(1
x
2)(3
x
1)
(v)
(3
x
4)(3
x
4)
(ii)
(4xl)(lx2)
(iv)
(5
x
2)(2
x
3)
(vi)
(2
x
2)(2
x
4)
matrix are multipliable only when p = q, and then X p matrix and a Recall that an X n matrix. Thus each of the above products is defined if the "inner" numbers the product is an are equal, and then the product will have the shape of the "outer" numbers in the given order.
m
qXn
m
(i)
The product
is
a 2 X 4 matrix,
(ii)
The product
is
a 4
(iii)
The product
is
not defined since the inner numbers 2 and 3 are not equal.
(iv)
The product
is
a 5
(v)
The product
is
not defined even though the matrices have the same shape.
(vi)
The product
is
a 2 X 4 matrix,
X
X
2 matrix.
3 matrix.
MATRICES
48
3.6.
Let
(i)
^ = Since
and
(2 _!)
A
2
is
X
and
2
B
is
row
entries in the first
^ ^ X
2
-2
(3
)
AB
the product
3,
AB, multiply
of
6
[CHAP. 3
^^^^
"
is
the first
defined and
row
a 2
is
A
of
3)
(1,
^^'
^^^
^^'
^"^
X
To obtain the /2\ /
3 matrix.
by the columns
x3y'V-2
_4S and
-
(
of B, respectively:
1
0-4\
S\/2
1
-1
2
-2
j(^ 3
/1-2
_ ~
6 y
entries in the second
3- (-2)
1
•
+
(-4)
3
•
6
V
+
2
To obtain the
+ 3-3 1-0 +
row
0-6
9
-4 + 18\
AB, multiply
of
-6
/ll
row
the second
(2,
14
—1) of
A
by the
columns of B, respectively: 1
3
2
-1
y2 A
-4
^ ~
N
ey
-2
3
(ii)
Note that
BA
3.7.
(i)
is
2
J5 is
X
2.
+
•
14
(-1)
•
2
(-2)
•
(-4)
+
\
(-1)
•
6/
2-14
( 1
Since the inner numbers 3 and 2 are not equal, the product
5-3)' *"^
/^
^^'
^^^
^^*
^"^
AB
Note that
BA
is
\
Since first
B
is
2
X
3
"!
® )
5
and
A
—8
/
is 1
X
= 2.
(2
•
1
+
1
.
4,
2
•
(-2)
+
1
•
5,
2
+
•
1
•
(-3))
=
(6, 1,
-3)
Since the inner numbers 3 and 1 are not equal, the product
not defined.
A =
Given
(i)
X
=
2
is defined and is a 1 X 3 matrix, i.e. a row Since A is 1 X 2 and B is 2 X 3, the product vector with 3 components. To obtain the components of AB, multiply the row of A by each column of B:
\ 4
3.8.
2
is
B =
and
(2,1)
AB = (%,!)( (ii)
A
and
3
(-l)-3
not defined.
A =
Given
V2-2 +
^^
Thus
-6
11
/
A 2
-3\
X 2 and B is AB, multiply
is 3
row
of
-1
B =
and
1
(
\
find
1
AB,
(ii)
BA.
AB
/2-3 -4-4
( 3
(i)
^
2 X 3, the product is defined and is a 3 X 3 matrix. the first row of A by each column of B, respectively:
,
"
^V
[^
"o )
To obtain the second row
0\
row
of
1-1
-8
-10^
= AB, multiply
of
+
-10
To obtain the
the second
A
by each column of B,
respectively: /
2
-1
\
1 \
-6
1(
,
1-2-5 ^ ^ ^
3 i
To obtain the
-1
/ ]
=
I
1
+
-8 -2 +
-10 -5 +
/-I
\ I
=
11
-8 -2
-10^
-5
4
/
third
row of AB, multiply the
third
row
of
A
by each column of B, respectively:
CHAP.
MATRICES
3]
49
-8 -2
-1
=
"(s -:-:)
Since
5
2 of
is
row
first
-3
+
12
6
+
16
X
3 and
i4 is
BA, multiply
-2
-5
'
"«
+
6
A =
o\
B =
and
2
15
B
0-20
Since
(ii)
Now
A
is
2
X
3
and
£
is
3
X
4,
by each column of A,
=
c,4
-
(2,-1,0)
C21
=
(1, 0,
0,
(i)
1
6
2
5
-7
(ii)
-3 first
factor is 2
a 2
X
=
1
•
1
+
•
3
2
+
+
(iii)
X
2
and the second
(iv)
is
2
X
Cu and
ith column of B.
2-1 + (-l)'(-l) + 0-0
•
c^a,
C21.
4 matrix.
=
5/(2 -1
-3
The
is
row and
in the ith
(-3)
•
(-3) -4
=
1
2
+
+
Hence:
+
+
1
6
=
0-12 =
-^(-l
(V)
6
=
+
6\/4
1
Compute:
=
-3)
AB
+
•
-3
1
(-2)
1
-21
/15 (lO
+
respectively:
-1
row of A by the
Cy is defined as the product of the ith
-21
-21 > "-1)
-2
the product
To obtain the
15
Determine the ahaite of AB. (ii) Let Ca denote the element column of the product matrix AJB, that is, AB = (co). Find:
(i)
(i)
22
i
-21
3
\4
3.10.
9
+
are defined, but they are not equal; in fact they
-4 -1
/I
-1
BA
1
(i)
-10 -5
-3 +
+
.10
Observe that in this case both AB and do not even have the same shape.
2
4
'15
BA
Thus
:;th
-8 -2
-1 +
15
15
)l
4
Let
1
15
To obtain the second row of BA, multiply the second row of
3.9.
1
-5
3 X 2, the product BA is defined and is a 2 X 2 matrix. the first row of B by each column of A, respectively:
2-2 +
Remark:
-10
AB
Thus
(ii)
1
-11
(2,-1)
-6
(3,2)
2,
so the product is defined
and
is
a 2
X
2 matrix:
MATRICES
50
1
The
(ii)
factor
first
0\
6Y4 5A2
^-3
X 2 and -3
Now
the first factor
product
distinct, the
is is
Here the first factor matrix:
(iv)
_
2\
2 X 1 and the second not defined. 2
X
=
=
3.11.
Prove Theorem Let
A =
B=
+ 5'(-7))
1,
[-2
-5^
and
a 2
is
1
matrix:
[-41^
so the product is defined
2,
X
Since the inner numbers 1 and 2 are
18
is
a 2X 2
12
so the product is defined
and
=
8
=
(-1). (-6))
and
2^
3
=
i:i)
+
(2-1
X
-6'
/-40'
_ ^
\
/ 16
(8)
is
a 1
X
1
matrix
(AB)C = A{BC).
3.2(i):
(oy),
X
2
is
6- (-7)
2X2.
is 1
il--l
The first factor is 1 X 2 and the second which we frequently write as a scalar. (2,-l)(^_g)
+
is
and the second
1
so the product is defined
1,
1-2
/
_ ~
(-3)-0
\(-3)'2
{>'' (v)
X
the second is 2
5A-7;
is
l-0 + 6'(-l) \ + 5-(-l)y
/
^V
1
(iii)
l'4 + 6-2 V(-3)-4 + 5.2
^
-1/
2
is
[CHAP. 3
C=
and
(bfl,)
Furthermore,
(e^).
AB = S =
let
and
(sj^)
BC = T =
(t,,).
Then
=
Sjfc
+
ajiftifc
+
at2b2k
•
•
•
+
2
=
at„6mfc
Oyftjj.
3=1 n
=
hi
Now
S by
multiplying
{AB)C
C,
+
i.e.
+
+
bj„c„i
=
(AB) by C, the element in the
ith
^ji'^ii
bjiCn
•
•
•
is
+
»ilCll
SJ2C21
+
On the other hand, multiplying of the matrix A{BC) is
•
•
A
+
•
by
2
=
Si„C„i
A
i.e.
+
«i2*2!
+
•
•
+
•
3.12.
Prove Theorem Let
and
A=
(tty),
F = AC =
n
m
fc=l
j=l
Ith
{"'ifijkiOkl
=22
»ij*jl
column of the matrix
in the tth
m
2
=
aim*ml
Since the above sums are equal, the theorem
row and
by BC, the element tn
»il*ll
fcjfcCfci
=22
StfcCfcl
k=l
T,
2 lc=l
row and
fth
column
n
««(6jfcCfci)
proven.
is
A{B + C) = AB + AC. B = (6jfc) and C = (Cj^). Furthermore,
3.2(ii):
(fik)-
D = B + C=
let
(dj^),
E = AB =
(ej^)
Then djk
=
6jfc
+
e*
=
aii6ik
+
ai2*'2fc
/ifc
=
Ojl^lfc
+
<»i2<'2fc
Cjfc
m
+
•
•
+
ajm^mk
+
"vm^mk
-
2 j=i
—
2 =
«ij6jic
m
+
•
•
3
Hence the element
in the ith
row and
feth
column of the matrix
m «ik
+
-
fik
2
ayftjfc
On
the other hand, the element in the ith
+
2
=
ttyCj-fc
3=1
row and
AB + AC
2
+
aisd^k
+
•••
+
otmdmk
=
fcth
2
}=l
Thus A{B + C)
— AB + AC
«i)(6jic
+ c^k)
j=l
column of the matrix
m Ojidifc
is
m
tn
i=l
"ijCjfc
1
AD = A(B + C)
m o-ad-jk
=
2
a.ij(bjk
i=l
since the corresponding elements are equal.
+ Cjk)
is
CHAP.
MATRICES
3]
51
TRANSPOSE 3.13.
Find the transpose A* of the matrix
Rewrite the rows of
3.14.
Let
A
A
A =
as the columns of A':
Under what
be an arbitrary matrix.
conditions is the product
m
Suppose A is an X n matrix; then A* is n X m. Observe that A*A is also defined. Here AA* is an X
Thus the product AA*
w m
3.15.
^ =
Let
(
3
_!
4)
Find
•
(i)
AA\
(ii)
matrix, whereas
A
A*, rewrite the rows of
-
-
-'
G
-
31
1-1 2
•
1
+
+
-I
Prove Theorem Let
AB
A=
(oy)
8.3(iv):
and
|
.
Then
B=
1-3 + 2'(-l) + 0'4 \ 3'3 + (-1) •(-!) + 4-4/
l'2 + 3'(-l) 2
•
2
+
(-1)
•
(bj^).
/5 \1
1
26
+ 3-4 \ + (-1) 4 +4 4
l'0
(-1)
2
•
0-2 + 4' (-1)
{AB)*
_ ~
:)
3-3
(-1) '3
0«l + 4'3
3.16.
—1
2
:)(: -:
-i (3
I
always defined. an n X n matrix.
is
is
^^
l'l + 2'2 + 0'0 3'1 + (-1)'2 + 4-0
A*A
=
A*
as columns:
3-
/I
defined?
A*A. '1
To obtain
A*A
AA*
•
•
•
j
= B*AK Then the element
in the ith
row and
jth column of the matrix
is
anbij
Thus
(1) is
On
+
ai^hzj
the element which appears in the jth
the other hand, the jth
row
+
•
•
•
row and
+
ai^h^j
ith
column of the transpose matrix (AB)*.
of B* consists of the elements
(6„
bzj
...
(1)
from the jth column of B:
6„j)
(2)
Furthermore, the tth column of A* consists of the elements from the ith row of A:
(3)
Consequently, the element appearing in the ;th row and ith column of the matrix B*A* product of (2) by (S) which gives (1). Thus (AB)* = B*A*.
is
the
[CHAP.
MATRICES
52
3
ECHELON MATRICES AND ELEMENTARY ROW OPERATIONS 3.17.
Which
Circle the distinguished elements in each of the following echelon matrices.
are
row reduced echelon matrices?
-3
2
/l
/O
l\
-5
7
1
2-4,
5
7
\0
1,
3/
0l\
[l)2-3
0/
\0
The distinguished elements are the
first
nonzero entries in the rows; hence
/0©7-5
0\
/0O5O2^
\0
0/
\0
00,00
0(1)2 -4, 3/
An
echelon matrix
is
0\
row reduced
20
4 7,
distinguished elements are each 1 and are the only nonzero Thus the second and third matrices are row reduced, but the
if its
entries in their respective columns. first is not.
2
-2 -1
\3
1
/I 3.18.
Given
A =
canonical form, (i)
i.e.
3 -1\
to
2
2
2
3/
(i)
.
Reduce
A
to echelon form,
Apply the operations ^2 -* -2i?i + Rz and R^ iJj -> -7B2 + 3B3 to reduce A to echelon form:
^
+
-ZRi
-2
3
7
^0
1.
^
Apply the operation B, 4/?3
+
7i?2
to the last
-»
2i?2
matrix in
+ (i)
to
I
3-4
4
by
1/21,
I
R^ by 1/21 and '1
-^3 +
7i?i
45^
fig
by 1/7
-10
to obtain the
J
row canonical form
of A:
15/7^
10
-4/7
1
-10/7 J
Method 2. In the last matrix in (i), multiply R^ by 1/3 and matrix where the distinguished elements are each 1:
'1-2 1
3
-4/3
^0
1
fig
by 1/7
Observe that one advantage of the until the very last step.
first
to obtain
an echelon
-1 4/3
-10/7/
apply the operation R^ -» 2^2 + Ru and then the operations {—1/3)R3 + jBi to obtain the above row canonical form of A.
Remark:
iSj -^
I
^0
jBj -^
4
-lOy
0-12
21
to
7
Now
row
-1^
ZRi, and then the operations to further reduce A: IZX
Finally, multiply Bj
to
and then the operation
^3.
Ato|0 3-4 4|to|0 3-4 Method and Ri
A
row reduced echelon form.
'1
ii)
Reduce
(ii)
method
is
R2, -»
(4/3)fi3
+ R^
and
that fractions did not appear
CHAP.
MATRICES
3]
53 /O
3.19.
A =
Determine the row canonical form of
1-4
2
3
\2 1
-4
is
already in echelon form.
1^
A
A = -4
1
-6
to echelon form,
row canonical form.
i.e.
to its
first
The computations are usually simpler and third rows:
A
if
the "pivotal" element
is 1.
Hence
first
already in echelon form.
is
that each of the following elementary
row operations has an
of the
same
m
Multiply the zth row by a nonzero scalar
inverse operation
type.
Interchange the ith row and the
jth.
row: Ri <^ Rj. k:
-»
Ri
kRi,
fc
^ 0.
Replace the ith row by k times the jth row plus the ith row: Ri (i)
Interchanging the same two rows twice, is its
(ii)
(iii)
own
interchange the
to
Note that the third matrix
Show
and then to row reduced echelon form,
2-5/
1
\
3.21.
1
I2
3 -4\
6
/
Reduce
3
2 -ll
to
Note that the third matrix
3.20.
Z\
3 -2\
1
we
obtain the original matrix; that
^
kRj
is,
this operation
+ Ru
inverse.
Multiplying the ith row by k and then by fc-i, or by fc-i and then by k, we obtain the original matrix. In other words, the operations iJj -» kRi and i?j -^ fe-iiJj are inverses. flj -» kRj + Ri and then the operation fij -^ —kRj + fij, or applying the operation fij -* -kRj + i?j and then the operation fij -» kRj + flj, we obtain the original matrix. In other words, the operations Ri -» kRj + fij and iJj -» —kRj + flj are
Applying the operation
inverses.
SQUARE MATRICES 3.22.
Let
A =
^ _g -3^ ^4
(
(i)
A^
)
Find
.
1
= AA =
4 /
(i)
A^
(ii)
A*,
(iii)
/(A),
where
fix)
2
-3 )il 1-1
4)
+ 2-4
V4-l + (-3)-4
1-2 + 2 -(-3) \ 4-2 + (-3) -(-3)/
^ ~
9
-4\
[-8
17/
/
=
2a^
-
4x
+
5.
MATRICES
54
« (iii)
To
=
-'
=
c
-:)(-: ») / l-9 + 2-(-8) (^4-9 + (-3) -(-8)
A
find /(A), first substitute
=
f(x)
2x9
/(A)
_
4a;
[CHAP. 3
+
+
l-(-4) 4 -(-4)
+
2-17
/-7 \eO
^
\
(-3) -17/
30\
-67 J
for x and 57 for the constant 5 in the given polynomial
5:
= 2A3-4A +
Then multiply each matrix by
2(-; -Z) -
=
5/
^
-s) + K'o
{\
l)
respective scalar:
its
/-14
60\
1^120
-134y
-4
-8\
/5
0\
V-16
12/
\0
5y
/
Lastly, add the corresponding elements in the matrices:
/ -14 -4 + 1^120-16 +
_ "
3.23.
Referring to Problem 3.22, show that
A
is
i.e. first
A
is
60-8 +
5
A
if the matrix g(A) is the zero matrix. for x and 11/ for the constant 11 in g(x)
= .....
„,.,
-n,
Then multiply each matrix by the
=
12
^
/
-IS
52 \
104
-117/
=
+ 2a! - 11.
+ 5/
=
x^
/(A),
+ ^x- 11:
-"GO
it:
-4X _4x
9
a;^
Compute g(A) as was done for
(-:.)-(! 4)
scalar preceding
g{A)
+
a zero of the polynomial g{x)
a zero of g(x)
substitute
_
\
I'
-134
17;
V-8
4\
/2
<
V8
-(6/
,
/-ll -11
V
Lastly, add the corresponding elements in the matrices:
Since g{A)
3.24.
Given
=
A -
First set
0,
A
is
[
/9 + 2-11 -4 + 4 + 0\ ^ ^_g^8^.Q i7_g_;^iy
-
g{A)
/O (^0
a zero of the polynomial g(x).
)
.
Find a nonzero column vector u
Au =
up the matrix equation
Write each side as a single matrix (column
)
Au -
such that
3m.
~
^\y
vector):
+ 3y\ \Ax-3yJ /
I
3u:
-3/U/
4
=
x
^
/Sx^ V3j/y
equations (and reduce to Set corresponding elements equal to each other to obtain the system of echelon form):
+ 3J, = Ax-Zy a;
3a;
Zy
__
Ax-
2x
3y 6y
= =
_
2x
-
Sy
=
0-0
^^
The system reduces to one homogeneous equation in two unknowns, and so has an To obtain a nonzero solution let, say, ?/ = 2; then » = 3. That is, a;
of solutions.
solution of the system.
Thus the vector w
=
(
is nonzero g j
2x
-
infinite
=
3,
=
Sy
i/
and has the property that
number
=
2
Au =
is
a
3m.
,
.
CHAP.
3.25.
MATRICES
3]
5^ „
/3
Find the inverse of
^
f
l2 Method
We
1.
3
seek scalars
,.,..»
" \0
r3a; <
or which satisfy
l2a;
The solution of the
+ +
system
first
We
2.
= =
5«
3« is
*""
a;
\2x +
1
=
rTT 1^1
A=Q
,h.„
ly
=
|A|
(
\2j/
=
—
—3, z
/-3 is
2y
GO
+ 3w)
+ 5w = + 3w =
1
and of the second system
2,
is
2/
=
5,
w=
—3.
5\
1
)
V-c
A-^
where
1
(
a'
MISCELLANEOUS PROBLEMS 3.26, Compute AB using block multiplication,
oi the
2X2
=
—
lAI
= -1 .„d A-. =
9-1(1
2
Sz
[31/
and
derived the general formula for the inverse
A^i Ttaslf
for which
l)
Thus the inverse of the given matrix Method
w
and
x, y, z
w/
3/\2
,2
55
ad
-l(4
matrix
A
6c
-^) =
(-J
4).
where
1\ I
4
3
I
\0
„
Hence
^1 GJ
I
(
)
TJ
\0
//9
ER ES + FT\ (^^^ ^^^+/^)
=
AB
GT
= ~
J
Suppose fined.
B=
Show
{Ri, R2,
that
.
.,
.
BA =
i?„),
where E, F, G, R,
12 15N
33/ jVl9 Vl9 26 33y \
3.27.
=
2,
S =
and
B
and
1
i.e.
(RiA, RzA,
/3N
are the given blocks.
/I
V^yVoyj \0 V^/
0)
(
S and T
=
ji9
26
33
7
(2)
that Ri is the ith row of B. Suppose BA is de.,RnA), i.e. that RiA is the ith row of BA. .
.
A"» denote the columns of A. By definition of matrix multiplication, the ith row of BA is {Ri •A\Ri'A\ But by matrix multiplication, BjA = (Bj • A^, i?i • A2, i2i • A"). Bj-A™). Thus the ith row of BA is ftjA.
Let
i4i,
A2,
.
.
.,
•
3.28.
Let Ci = (0, 1, where. Show that CjA
3.29.
Show:
(i)
.
,
.
be the row vector with 1 in the tth position and the ith row of A. Observe that Cj is the ith row of /, the identity matrix. By the preceding problem, the IA is BjA. But lA = A. Accordingly, CjA = JBj, the ith row of A. .
of
.
Let
.
. ,
.
.
. ,
.
else-
0)
= Ri,
tth
A has a zero row, then AB has a zero row. B has a zero column, then AB has a zero column.
(i)
If
(ii)
If
(iii)
Any matrix with
jBj
.
a zero row or a zero column
be the zero row of A, and B^,
.
.
{RrB\ Ri'B^
.,£" the columns of B. ...,
Ri-B^)
=
(0, 0,
is
not invertible.
Then the ..., 0)
ith
row of
AB
is
row
MATRICES
56
Let Cj be the zero column of B, and Aj,
(ii)
.
.
.,
A„
[CHAP. 3
the rows of A.
Then the
jth
column of
AB
is
/Ai-C/ A^'Cj
m'Cj (iii)
matrix A is invertible means that there exists a matrix A~^ such that AA"^ = A~^A —I. But the identity matrix / has no zero row or zero column; hence by (i) and (ii) A cannot have a zero row or a zero column. In other words, a matrix with a zero row or a zero column cannot
A
be invertible.
3.30.
Let
A
AB
is also
An"*
•
•
B
be invertible matrices (of the same order). Show that the product and (AB)-^ = B'^A'K Thus by induction, (AiA2- -An^^ = -Az^Ai^ where the Ai are invertible.
and
invertible
3.31.
= A(BB-i)A-i =
=
/
(B-iA-i)(AB)
= B-i(A-iA)B = B-^B = B^B =
I
Show
show that
if
+
u fcj
=
k^.
Suppose
(1)
holds.
kGK,
the vectors
u and v are
distinct,
ki{u
— v) =
M
+
k2(u
— v)
(1)
Then
ki(u — v) =
Since
that, for each scalar
are distinct.
It suffices to
then
i
= B-iA-i.
Let u and v be distinct vectors.
u + k{u — v)
= AA
A/A-i
(AB)(B-iA-i) and
Thus (AB)-i
•
k^iu
— v)
u — v¥'0. Hence
or fci
{ki
— fcg —
— k2)(u — v) = and
=
fci
/Cj.
ELEMENTARY MATRICES AND APPLICATIONS* 3.32. A matrix obtained from the identity matrix by a single
elementary row operation is Determine the 8-square elementary matrices corre-
an elementary matrix. sponding to the operations Ri <^ R2, Ra
called
^ —IRs and =
R2.
o\
/I
Apply the operations to the identity matrix /g
— 3i?i +
722 -*
to obtain
1
\o
1/
/I £^1
=
1
^2
,
=
3.33.
Eo
1
=
-7
\o
Prove: Let e be an elementary row operation and E the corresponding m-square elementary matrix, i.e. E-e(l-m). Then for any TO X % matrix A, e{A) = EA. That is, the result e(A) of applying the operation e on the matrix A can be obtained by multiplying A by the corresponding elementary matrix E.
B
is
Let iJj be the tth row of A; we denote this by writing A = (B^ a matrix for which AB is defined, then AB = (R^B, ..., R-^B). ej
=
(0,
...,0,1,0
*This section is rather detailed and may be omitted in a results in Chapter 9 on determinants.
0),
first
reading.
A =
R^).
We
By Problem
3.27, if
also let
i
It is
not needed except for certain
CHAP.
MATRICES
3]
=
Here a /
=
(cj,
.
Let
(i)
e„)
1 is the ith component. the identity matrix.
is
be the elementary row operation
e
E and
=
(iBj,
.
.
Thus
=
e(/)
^-A
=
(6i, ...,fcej, ...,
=
e(I)
(ftej
.,£^,
(fijA, ...,A;ejA,
+
ej)A
(ei, ...,fcej
=
fc(ej.A)
ej
EA = (M,
+
...,(fce^
+ ei)A,
kRj
=
+
.,
.
.,
.
and A
i
.
.
(ftj,
.
and
.
,
.
+
fefij,
a =
., fcfij,
.
.
.
.
,
«„)
Then, for
Kj.
e(A)
.
=:
(fij,
.
.
.
=
fij
., ffj
.
Then, for
=
e(A)
kRj
j,
A
., i?,-,
.
t^ 0.
fc
—
ej
/s (fii,
remark that
also
BJ
Rt
(fij,
^
JBj
,
i,
fcfij
.
.
= /\
e(A)
.,
BJ
e(A)
—
i,
+ Bj,
.
.
.
,
i2
J
we have
Rf,
...,e„A)
=
We
iJj.
/\
=
...,e^A)
+ ej, ...,6j BjA
.
and
row operation
-
.
ej,
jB^ -> fcjBj,
/\
Lastly, let e be the elementary
E =
Bj
.
be the elementary row operation
let e
Using
(ej
/\ ...,e^, ...,6iA, ...,e^A)
(fijA,
E =
(iii)
=
=
e^A
3.28,
Then, for a
Rj.
^
^A = Now
i^j «->
e(I)
e(A)
Thus
(ii)
By Problem
means that
t
.,
.
57
=
(R^,
.
.
.,
kRj + Ri,
.
.
.,
RJ =
e(A)
Thus we have proven the theorem.
A
row equivalent to B if and only if there exist elementary matrices E2E1A = B. By definition, A is row equivalent to B if there exist elementary row operations ej, ..e, for which es(---(e2(ei(A)))- ••) = B. But, by the preceding problem, the above holds if and only if Eg- -E^EiA = B where is the elementary matrix corresponding to e^.
3^. Show
that
is
El, ...,Es such that Es-
•
.
•
3^5.
.
JS7j
Show
that the elementary matrices are invertible and that their inverses are also elementary matrices. Let E be the elementary matrix corresponding to the elementary row operation e: e(I) = E. Let e' be the inverse operation of e (see Problem 3.21) and E' its corresponding elementary matrix. Then, by Problem 3.33, /
Therefore E'
3M. Prove (i)
(ii)
(iii)
is
=
=
e'(e(/))
e'E
= E'E
and
/
=
e(e'(I))
=
eE'
= EE'
the inverse of E.
that the following are equivalent:
A is invertible. A is row equivalent to the identity matrix /. A is a product of elementary matrices.
Suppose A is invertible and suppose A is row equivalent to the row reduced echelon matrix B. Then there exist elementary matrices Ei,E2, -yE^ such that Eg- E2E1A = B. Since A is invertible and each elementary matrix E^ is invertible, the product is invertible. But if B ^ I, then B has a zero row (Problem 3.47); hence B is not invertible (Problem 3.29). Thus B = I. In other •
words,
(i)
Now
implies if
(ii)
(ii).
holds, then there exist elementary matrices E^, E^,
E.-'-E^EiA
By
=
the preceding problem, the Ei
Now matrices
if
(iii)
holds
is invertible.
(A
=
and so
/,
A =
(E,-
-E^Ei)-^
are also elementary matrices.
EiE^-
-E^),
then
(i)
must follow
.
.
.,Eg such that
= E^^E^^-'-EJ^ Thus
(ii)
implies
(iii).
since the product of invertible
MATRICES
58
3.37.
[CHAP. 3
Show
A and B be square B = A-K Thus AB = I
Let
A
Suppose
is
matrices of the same order. if and only if BA = I. Then A is not row equivalent invertible. not
that
if
AB = I,
to the identity matrix
/,
then
and so
A
row equivalent to a matrix with a zero row. In other words, there exist elementary matrices E^ such that E^- -E^E^A has a zero row. Hence E^- -EJE^AB has a zero row. Accordingly, El AB is row equivalent to a matrix with a zero row and so is not row equivalent to /. But this contradicts the fact that AB = /. Thus A is invertible. Consequently, is
•
B = IB =
1.38.
= A-HAB) = A'^I = A"!
(A'-->^A)B
reducible to the identity matrix / by the (i) Show that this sequence of elemen., e„. sequence of elementary operations ei, Use this result to obtain the inverse (ii) tary row operations applied to / yields A-K
Suppose
A
is
invertible and, say,
row
it is
.
.
/I
of
(i)
A =
2\
2
-1
\4
1
3 8/
hypothesis and Let Ei be the elementary matrix corresponding to the operation ej. Then, by i = E^---EJEJ Problem 3.34, E„- -E^EiA = I. Thus (E„- EiEJ)A ^ I and hence A e« ej, operations row elementary -i the applying / by from can be obtained In other words, A •
(ii)
Form
the block matrix (A,
and row reduce
I)
2
=
(A, /)
[2-1
3
1
8
I
2
-1 -1
2
-1
to
1 I
to
/l
0\
1
I
row canonical form:
it to
I
I
1
I
-2 -6
I
1
1
I
to
final block
Observe that the
matrix
the form
is in
(/,
B).
Hence
A
is
invertible
and
B
is its
inverse:
A-i
Remark:
=
In case the final block matrix is not of the form row equivalent to I and so is not invertible.
(/,
Supplementary Problems MATRIX OPERATIONS In Problems 3.39-3.41, let
-(o1 3.39.
Find:
(i)
A+
3.40.
Find:
(i)
AB,
3.41.
Find:
(i)
A*,
4
-"i-l
!)• B,
(ii)
(ii)
(ii)
A+
AC,
A'C,
C,
(iii)
(iii)
(iii)
AZ), IJtA',
3A
0-3 2
-
3
4B.
(W) BC, (iv)
B«A,
(y)
BD,
(y)
(wi)
DW.
CD.
{wi)
DDK
B), then the given matrix is not
CHAP.
3.43.
MATRICES
3]
=
Let
Cj
(i)
Be*.
(ii)
If
(iii)
If
(0,
=
.
where
... 0, 1, 0, .... 0)
the ith column of B.
Cj,
= ejB = Ae\ Be\ e^A
for each
i,
then
for each
i,
then
59
(By Problem
=
ejA
3.28,
2-1 2 1-2 4 2-6 6
A =
(i)
2 \3
3.45.
Reduce
A
A
11
3
2-5
3
1
1
5/
\4
3.46.
3.47.
Describe
all
1
A =
(ii)
,
3.49.
5
2
\0
-5
6
Show
that
(ii)
(iii)
row equivalence
row equivalent
row equivalent
to
row equivalent
to
is
Ij
-3
5
4,
matrices which are in row reduced echelon form.
that every square echelon matrix
is
1^
^
Show
A A A
the following:
4
Suppose A is a square row reduced echelon matrix. Show that has a zero row.
(i)
ejA,
where
if
A 3.48.
(i)
4-13 ^=0021
,..^
(")
2X2
find
C4/
row canonical form, where /O 1 3 -2^
to its
2\
the possible
-1 \4 -5 3
5/
-1 -5
3
C3
Bj.)
3-2
/2
form and then
to echelon
/l (i)
l\ 3
h h],
C2
A = B. A = B.
ECHELON MATRICES AND ELEMENTARY ROW OPERATIONS 3.44. Reduce A to echelon form and then to its row canonical form, /l
&2
Show
component.
1 is the ith
61 C,
is
A;
B B
implies
and
B
relation:
row equivalent
B row
the identity matrix, then
upper triangular, but not vice versa.
an equivalence
to
A # 7,
to
C
equivalent to
A;
Implies
A
row equivalent
vfl
-
to C.
SQUARE MATRICES 3.50.
3.51.
g(x)
A = ( = a;2 - -
Let
B =
"Let
(iii)
3.52.
a;
f
\
.
(i)
Find A^ and A3,
].
B
(i)U
f(x)
are said to
(
3.53.
Let
1
A =
=
Zx^
-
2x
+
i,
=
2x2
-
u =
commute
+
Z,
(
]
such that
if
AB =
4x
find f{B).
(ii)
Bu =
BA. Find
If
g{x)
=
x^
-
)
.
find /(A),
(iii)
If
Find A".
4x
-
12,
find g(B).
6m.
all
matrices
(
Vz
.
VO
/(«)
find g(A).
8,
Find a nonzero column vector
Matrices A and /I 1\ , with
If
(ii)
^
)
w/
which commute
MATRICES
60
3.54.
A =
Let
Find:
3.56.
(i)
Let (i)
3.58.
3/
A+
B,
(
(ii)
AB,
(iii)
B
i.e.
mXn matrix A,
for any
D^A =
kA;
(ii)
that the sum, product and scalar multiple
(i)
upper triangular matrices
is
(ii)
lower triangular matrices
is
(iii)
diagonal matrices
(iv)
scalar matrices
3.62.
BA. Show that
Show
that:
BD^ =
kB.
of:
lower triangular;
diagonal;
is
is scalar.
(i)
(
Find the inverse of each matrix:
Show
AB =
i.e.
upper triangular;
Find the inverse of each matrix:
Find the inverse of
k.
nXm matrix B,
for any
-
-'
^
c
(i)
2
|
1
3
3
-1
l-l
5
|
-3\
'2 )
>
(")
2
-3\
(
1
i
3/
2 ,
,4-2
t.61.
/(A) for a polynomial f{x).
a scalar matrix.
is
-1 3.60.
(v)
commutes with every 2-square matrix A,
INVERTIBLE MATRICES 3.59.
A",
(iv)
be the m-square scalar matrix with diagonal elements
Dfc
Show
11
A^ and A3,
B
3
,,
for some scalar k,
)
,
kj
\0
„
(
\0
Suppose the 2-square matrix
B = 3.57.
\0
B =
and
„
„
(
[CHAP.
5/
-V
1
(ii)
\5
2
1
2
-3/
4\ 6 1/
that the operations of inverse and transpose commute; that A is invertible if and only if A* is invertible.
is,
(A«)-i
=
Thus,
(A-i)«.
m
particular,
...
(»! °.
3.64.
Show
that
A
is
A
is invertible if
row equivalent
to
B
."!
///.
.
.M
if
and only
if
the system
invertible,
and what
is its
inverse?
there exists an invertible matrix
if
P
such that
B = PA. 3.65.
Show
that
and only
AX =
has only the zero solution.
MISCELLANEOUS PROBLEMS 3.66.
Prove Theorem (Parts
3.67.
3.68.
(i)
and
3.2:
(ii)
(iii)
(B
+ QA = BA + CA;
were proven
Prove Theorem 3.3: (Part (iv) was proven
(i)
in
(A
+
in
B)*
Problem
(iv)
Problem 3.11 and
=
A*
+ BH
k(AB)
=
(kA)B
=
A(kB),
where
fc
is
a scalar.
3.12.)
(ii)
(A')«
=
A;
(iii)
(feA)'
=
kA*,
for
k a scalar.
3.16.)
defined and the number of Suppose A = (A^) and B = (B^,) are block matrices for which AB is Show that AB - (Gy) B^j. block each of rows of columns of each block Aj^ is equal to the number = Ag^B^j. where Cy
2
CHAP.
3.69.
MATRICES
3]
The following operations are
called elementary
column operations:
Interchange the tth column and the jth column.
[El]
Multiply the tth column by a nonzero scalar
Show
3.71.
that each of the operations has an inverse operation of the same type.
BUB
A
matrix A is said to be equivalent to a matrix can be obtained from A by a finite sequence of operations, each being an elementary row or column operation. Show that matrix equivalence is an equivalence relation.
Show that two consistent systems of linear equations have the same solution set if and only if their augmented matrices are row equivalent. (We assume that zero rows are added so that both augmented matrices have the same number of rows.)
Answers 3.39.
3.40.
«
-1
i-l Not
(i)
(")
1
-1
1)
Not
defined.
-13 -3
18 \
17
0/
f
(iii)
4
<"«
-2 4 -3 -12
( 11
(i)
(ii)
Supplementary Problems
to
defined.
/-5
3.41.
A;.
Replace the ith column by k times the jth column plus the ith column.
[^3]
3.70.
61
«')
C)
m
<">
":
1)
8
0\ 3
I
/ (ii)
Not
defined.
(iii)
-7
4
CI
(vi)
Not
4\
0-6-8
(iv)
(9, 9)
<^'
/ (v)
14
4
-2
(vi)
I
2
3.42.
(i)
\-3
4/
(ai, ag. 03. a*)
(ii)
(61, 62.
K K)
(iii)
'1 3.44.
(i)
3-6
(
1
|
and
/2
3-2 -11
10
5
-15
2
4/3^
1
3.45.
3
-1 -5
r''-° 11
(i)
^0
/o
(ii)
L
1
0/
\0
/I
/l
and
5
3 "
and
4/11
5/11
15/11
13/11
-5/11
3/11
0/
1
:
3
-2
-13
11
35 0/
o\
1
and
13/11
\
-5/11 /
4/11 1
-1/6^
-10/11
\0
ll
2
\
1
(
l\
6/
(Ci. Cg, Cg, C4)
^0
(ii)
12
1 1
0/
6
-2
1-3 -3
9y
'''
Co
^0 3.48.
[CHAP.
MATRICES
62
(
1
V
1
1
'^
^"^
'^
J)
I) ''Co
«'=^'^"
upper triangular but not an echelon matrix.
Is
)
-^^'^
D'Co
o)'(2
1/
iO
3.52.
Only matrices of the form
/I
2n\
3.53.
^" = (o
i)
3.54.
« ^+^
= (o
«
/9 14
- c =
(»)^^=(o
^J
- c =
:)
^"=U
('^^
33;
5
-2\
(^
3)
1-5
4
/
3.60.
(1)
,..,
(n)
/
1/3
1/3
f
^/9
2/g
-3\
10-7
(i)
8-6
-
3«, 3d,^
/3ci
3.59.
'" '<^'
.;)
2"
ON
/14
„
commute with (^
]
(
8
/ 11
6
5/
\
-1 -3^
-5
12
10
-1 -4y
/31/2 -17/2 -11^
-5/2
9/2
3.61.
\-7
3.62.
-3
4
Given
AA-i =
51
/.
Then
7
=
7'
=
(AA-i)'
= {A'^YAK
That
is,
/a-i 3.63.
A
is
invertible
iff
each
aj 9^ 0.
Then
A
^
-
.0 '
\0
a-i
...
(A^^)'
=
(A*)"!.
(T
;,;
3
:
chapter 4
Vector Spaces and Subspaces INTRODUCTION we studied the concrete structures B" and C" and derived various propercertain of these properties will play the role of axioms as we define abstract
In Chapter 1
Now
ties.
"vector spaces" or, as
sometimes
In particular, the conclu[Afi]-[M4] below. 1.1, 3, We will see that, in a certain sense, we get nothing new. In fact, we prove in Chapter 5 that every vector space over R which has "finite dimension" (defined there) can be identified with R" for some n. sions
(i)
through
(viii)
it is
of
called, "linear spaces".
Theorem
become axioms
page
[A]]-[A4],
The definition of a vector space involves an arbitrary field (see Appendix B) whose elements are called scalars. We adopt the following notation (unless otherwise stated or implied):
K or
a, &, c
u, V,
We
remark that nothing essential or the complex field C.
the field of scalars, the elements of K,
A;
V
the given vector space,
w
the elements of V.
is lost if
the reader assumes that
K
is
the real field
R
Lastly, we mention that the "dot product", and related notions such as orthogonality, not considered as part of the fundamental vector space structure, but as an additional structure which may or may not be introduced. Such spaces shall be investigated in the latter part of the text. is
Definition :
V
be a nonempty set with rules of addition and and to any u,v a sum u + v any uGV,kGK a product ku G V. Then V is called a vector space over K (and the elements of V are called vectors) if the following axioms hold: Let
iiT
be a given
field
and
let
[Ai]:
For any vectors u,v,w GV,
{u
+ v)
[A2]:
There is a vector in V, denoted by for any vector u GV.
[A3]
For each vector
[A4]:
For any vectors u,v GV,
[Ml]:
[M2]
:
[Ms]:
[Mi]:
uGV K
there
is
+w and
= u+
{v-i-
w).
called the zero vector, for
scalar k
G
which u
a vector in V, denoted by —u, for which u
u+v = v +
u.
+ v) = ku + (a + b)u = au + For any scalars a,b GK and any vector {ab)u = a{bu). For any scalars a,b G K and any vector For the unit scalar 1 G K, lu = u for any vector u GV. For any
GV
GV
scalar multiplication which assigns to
GV, u GV, u GV,
and any vectors u,v
63
k{u
kv. bu.
+
Q
—u
+ {—u) =
0.
VECTOR SPACES AND SUBSPACES
64
[CHAP.
4
The above axioms naturally split into two sets. The first four are only concerned with the additive structure of V and can be summarized by saying that 7 is a commutative group (see Appendix B) under addition. It follows that any sum of vectors of the form Vi
+
+
V2
•
•
•
+ Vm
requires no parenthesis and does not depend upon the order of the summands, the zero is unique, the negative —u of u is unique, and the cancellation law holds: vector
u + for any vectors
u,v,w G V.
w =
+w
V
u — v
implies
by
Also, subtraction is defined
u—V = u
+
{—v)
the the other hand, the remaining four axioms are concerned with the "action" of these Using splitting. this reflects axioms the of on V. Observe that the labelling field space. additional axioms we prove (Problem 4.1) the following simple properties of a vector
On
K
Theorem
4.1:
Let
7
(i)
For any scalar
(ii)
For
(iii)
(iv)
be a vector space over a
gK
kGK
field
K.
G
7,
and
=
fcO
0.
uGV, Ou = 0. = or m = 0. If ku ^ 0, where kGK and uGV, then For any scalar kGK and any vector uGV, {-k)u = k{-u) = and any vector
A;
EXAMPLES OF VECTOR SPACES We now list a number of important
The
examples of vector spaces.
-ku.
example
first
is
a
generalization of the space R". Examplje
4.1:
of be an arbitrary field. The set of all n-tuples of elements Let addition and scalar multiplication defined by
K
(«!,
+
a„)
a2
and
fc(ai. <»2.
where
«<, 64,
k&K,
is
=
(61,62, ...,6„)
•
• .
«n)
=
(01
•
•
.
^O
denote this space by X". The zero The proof that K" is a vector 0). which we may now regard as stating
we
a vector space over K;
=
•
with vector
a„+6„)
+ 61,02+62
C^^i- '««2.
K
(0, 0, ...
vector in K» is the w-tuple of zeros, space is identical to the proof of Theorem 1.1, that R" with the operations defined there is a vector space over R.
Example
4.2:
Let
y
X n matrices with entries from an arbitrary field K. Then be the set of all with respect to the operations of matrix addition and over space vector a
m
V
is
K
scalar multiplication, by
Example
4.3:
4.4:
Theorem
3.1.
+ a„t" with coefficienis oj polynomials Oo + a^t + Ogt^ + respect to the usual operations with over vector space a is from a of addition of polynomials and multiplication by a constant. Let
V be
the set of
field
Example
,
K
K.
•
all
Then
•
K
y
X
be any nonempty set. be an arbitrary field and let Let into K. The sum of any two functions functions from f
+ gGV
X
defined by {f
and the product of a scalar defined by
+ g){x) =
kEK
f(x)
+
=
,
„
^
kf(x)
eV
g(x)
and a function / e
(kf){x)
Consider the set V of all is the function
f,g
y
is
the function
kfeV
CHAP.
VECTOR SPACES AND SUBSPACES
4]
65
a vector space over K (Problem 4.5). The zero which maps each x G X into S K: 0{x) = x G X. Furthermore, for any function f G V, —f is that function in V for which (—/)(») = —f(x), for every x G X.
V
Then
with the above operations
vector in for every
V
is
is
the zero function
Suppose S is a field which contains a subfield K. Then E can be considered to be a vector space over K, taking the usual addition in to be the vector addition and defining the scalar product kv of and v S jF to be the product of k and v as element of the field E. Thus the complex field C is a vector space over the real field E, and the real field R is a vector space over the rational field Q.
Example 45:
E
kGK
SUBSPACES
W
Let TF be a subset of a vector space over a field K. is called a subspace of V if TF is a vector space over K with respect to the operations of vector addition and scalar multiplication on V. Simple criteria for identifying subspaces follow. itself
Theorem
W is& subspace of V and only W nonempty, W closed under vector addition: v,w G W implies v + w G W, W closed under scalar multiplication: v GW implies kv GW
4.2:
if
(i)
is
(ii)
is
kGK.
every
W
4.3:
ia
4.6:
V if and only
a subspace of
+ bw G
implies av
Example
Let
V be
W
4.7:
(i)
Let
V
Let
V
V
GW (or W # 0),
(i)
is
W
zero,
Then the
be the space of
all
—
{{a,b,0)
square
nX n
set :
(iy)
Let
V
be the space of set
W
(A function / € every x G X.)
4.8:
is
v,w
GW
a subspace of V.
=
for which
(oy)
4.2).
ay
Let V be the space of polynomials (see Example 4.3). Then the set of polynomials with degree — n, for a fixed n, is a subspace of V.
Then the
Example
GR},
matrices (see Example
consisting of those matrices A symmetric matrices, is a subspace of V. (iii)
(ii)
W consisting of those vectors whose
a,b
W
set
and
GK.
Then the set {0} consisting of the zero vector alone, and are subspaces of V.
be the vector space R^.
third component (ii)
if
for every a,b
any vector space.
also the entire space
Example
for
is
(iii)
Corollary
if
all
functions from a nonempty set all bounded functions in
consisting of
V
is
bounded
if
there exists
M GR
X V
=
Then the Ojj,
called
W consisting
into the real field R.
is a subspace of V. for such that |/(a;)| -
M
Consider any homogeneous system of linear equations in n unknowns with, say, real coefficients:
aiiXi
+
a-y^Xi
4-
•
•
•
+
ai„a;„
=
a2iXi
+
a^sx^
+
•
•
•
+
a2„a;„
—
any particular
may
be viewed as a point in R". a subspace of R" (Problem We comment that the solution set of a nonhomo4.16) called the solution space. geneous system of linear equations in n unknowns is not a subspace of R". Recall that
The
set
W of
all
solution of the system
solutions of the
homogeneous system
is
VECTOR SPACES AND SUBSPACES
66
Example
4.9:
[CHAP. 4
W
be subspaces of a vector space V. We show that the intersection are subsince U and G C/ and S a subspace of V. Clearly and u,v e UdW. Now suppose m,v e.Ur\W. Then u,v spaces; whence are subspaces, and, since U and
Let
V
Vr\W
and
W
i& also
&U
W
&W
W
aw for any scalars space of V.
The
result in the preceding
Theorem
4.4:
The
+
a,b€K.
6i)
G
aw
and
?7
Accordingly, au
+
bv
+
&
6v
e
UnW
W and so [7nTF
is
a sub-
example generalizes as follows.
any number of subspaces of a vector space
intersection of
7
is
a
subspace of V.
LINEAR COMBINATIONS, LINEAR SPANS Let F be a vector space over a field K and let form aiVi
where the OiGK,
+
a2V2 4-
•
•
vi,
•
Any
...,VmGV.
V
of the
+ amVm The following theorem
a linear combination of vi,...,Vm.
is called
vector in
applies.
Theorem
4.5:
be a nonempty subset of V. The set of all linear combinations of vectors in S, denoted by L{S), is a subspace of V containing S. Furtheris any other subspace of V containing S, then L{S) CW. more, if
Let
S
W
In other words, L{S) is the smallest subspace of V containing S; hence it subspace spanned or generated by S. For convenience, we define L{0) = {0}. Example
4.10:
is called
the
be the vector space R3. The linear span of any nonzero vector u consists multiples of u; geometrically, it is the line through the origin and the scalar of point u. The linear space of any two vectors u and v which are not multiples of each other is the plane through the origin and the points u and v.
Let
V
all
Example
4.11:
The vectors 6i = (1,0,0), eg = (0,1,0) and eg = (0,0,1) generate the vector space specifically, R3. For any vector (a, 6, c) G R^ is a linear combination of the ej; (a, b. e)
Example
4.12:
The polynomials (in*):
y=
powers of
t.
L(l,
= =
1, t, t^, t^, ... t,
t^
.
.
.).
+ 6(0, 1, + ftej + 063
a(l, 0, 0)
aej
0)
+
c(0, 0, 1)
generate the vector space V of all polynomials is a linear combination of 1 and
For any polynomial
CHAP.
VECTOR SPACES AND SUBSPACES
4]
Example
4.13:
67
Determine whether or not the vector v - (3, 9, -4, -2) the vectors u^ = (1, -2, 0, 3), U2 == (2, 3, 0, -1) and Wg to the space spanned by the Mj.
=
V
is
=
a linear combination of -1, 2, 1), i.e. belongs
(2,
Set r as a linear combination of the Mj using unknowns XU^ + J/M2 + ZM3: (3, 9,
= =
-4, -2)
!B(1,
-2,
0, 3)
4
i/(2, 3, 0,
-1)
+ 2y + 2z, -2x + 3y-z,
(x
2«,
+
x,
y and
«(2,
-1,
z;
that
is,
set
2, 1)
3a; + z) 3/
Form
the equivalent system of equations by setting corresponding components equal to each other, and then reduce to echelon form:
+
2y
-2x +
3j/
X
+ -
2z
=
3
2
=:=
9
22 3a;
-
y
+
z
X
+
2y 7y
+ +
= = = =
2z 3z
or
= -4 = -2
2z
-7y a;
+
+ +
2|/
or
-
7y
5z
x
3
+
15
+ +
2y 7y
2z 3z
or
-4
22
-11
-2z
- 3 - 15 = -4 = 4
= 3 - 15 = -4
2z Bz 22
Note that the above system is consistent and so has a solution; hence v is a linear combination of the Mj. Solving for the unknowns we obtain x = 1, y = 3, z — —2. Thus V — Ui + 3m2 — 2M3. Note that
if the system of linear equations were not consistent, i.e. had no soluthen the vector v would not be a linear combination of the Mj.
tion,
ROW SPACE OF A MATRIX Let A be an arbitrary mxn matrix over a field K:
\fflml
The rows
(Hi
...
ai„
(I22
.
.
.
a,2n
flm2
.
.
.
dmn/
\
of A,
Rl viewed as vectors in
=
.K",
(ttll, 0,21,
.
.,
am),
.
.
.
,
Rm =
(Oml, am2,
.
.
.
,
dmn)
span a subspace of K" called the row space of A.
row space
of
A —
L{Ri, R2,
.
.
.
,
That
Rm)
Analogously, the columns of A, viewed as vectors in K"", span a subspace of column space of A.
Now
suppose (i)
we apply an elementary row Ri <^ Rj,
(ii)
Ri
^
kRi,
is,
X"
called the
operation on A,
k¥'0,
or
(iii)
Ri
->
kRj
+ Ri
and obtain a matrix B. Then each row of B is clearly a row of A or a linear combination of rows of A. Hence the row space of B is contained in the row space of A. On the other hand, we can apply the inverse elementary row operation on B and obtain A; hence the row space of A is contained in the row space of B. Accordingly, A and B have the same row space.
This leads us to the following theorem.
Theorem
4.6:
Row
equivalent matrices have the
same row
space.
We shall prove (Problem 4.31), in particular, the following fundamental result concerning row reduced echelon matrices.
VECTOR SPACES AND SUBSPACES
68
Theorem
reduced echelon matrices have the same row space have the same nonzero rows.
Row
4.7:
Thus every matrix row canonical form.
We
[CHAP. 4
row equivalent
is
if
and only
row reduced echelon matrix
to a unique
if
they
called its
apply the above results in the next example.
Example
Show
4.14:
that the space
=
Ml
(1, 2,
V
and the space
U
-1,
M2
3),
=
(2, 4, 1,
and
-2),
wg
=
-7)
(3, 6, 3,
generated by the vectors vi
are equal; that
generated by the vectors
=
U=
is,
and
-4, 11)
(1, 2,
=
v^
-5, 14)
(2, 4,
V.
Show that each Mj is a linear combination of v^ and V2, and show that 1. each Vi is a linear combination of Mj, M2 and M3. Observe that we have to show that six systems of linear equations are consistent.
Method
Method
Form
2.
A
the matrix
whose rows are the
and row reduce
Mj,
A
to
row
canonical form: 2
1
=
A
2
4
1
-2
-1 3
to
I
I
6
3\
-8 -16/
3
3
-8
\o "^
-8/3
1
to
-1
to
1/3
2
1
2
1
/
Now
form the matrix
B
whose rows are Vi and
t>2,
and row reduce
B
to
row canonical
form:
B
_ = ~
2-4 4-5
,x /I (
\2
11\ )
14/
to
/I
-4 UN 2 2-4
(„ VO
„
o
3
_o) -8/
/I to
(j,
1/3
2 1
_8/3
Since the nonzero rows of the reduced matrices are identical, the row spaces of and B are equal and so U = V.
A
SUMS AND DIRECT SUMS Let
U
and
consists of all
Note that u'
+ w'
W be subspaces of a vector space V. The sum of U and W, written U + W, sums u + w where uGU and w &W: U + W = {u + w:uGU,wGW} = + eU + W, since OeU.OGW. Furthermore, suppose u + w and + W, with u,u' GU and w,w' e W. Then (u + w) + (u' + w') = {u + u') + {w + w') G U +
belong \joU
and, for any scalar
k{u
k,
+ w) = ku + kw G U +
W
W
Thus we have proven the following theorem.
Theorem
4.8:
Example
The sum 4.15:
U+
W
of the subspaces
U
and TF of
F
is also
a subspace of V.
U consist of those Let V be the vector space of 2 by 2 matrices over R. Let V consist of those matrices matrices in V whose second row is zero, and let whose second column is zero:
m
W
- =
{(::)^«'— }
-
=
{(::)«-'}
CHAP.
VECTOR SPACES AND SUBSPACES
4]
Now U
W are subspaces of V.
and
U+W
=
We
have:
VnW
*"d
"'^'"^A
{(" o)
69
=
[(I °)
U+
That is, W^ consists of those matrices whose lower right entry is 0, consists of those matrices whose second row and second column are zero.
u&V
The following theorem
{i)V
if:
Example
GF
said to be the direct
Ur\W
U
and W,
as v
= u+w
of its subspaces
w way
can be written in one and only one
applies.
^
V
is
U+W, U =
sum
the direct
and
In the vector space R^,
4.16:
sum
and
w gW.
and
The vector space
4.9:
is
V = V ®
every vector v
if
where
Theorem
V
The vector space denoted by
Definition:
aeR
:
(ii)
U
let
{{a, 6, 0)
let
W
=
6
S R}
and
W
if
and only
{0}.
be the xy plane and a,
:
U
of its subspaces
UnW =
and
W be the yz plane: {(0,
U+W
Then R^ = since every vector in R3 is the sum in W. However, R* is not the direct sum of U and
h,c& R}
b,c):
of a vector in
W
U
and a vector
sums are not
since such
unique; for example,
=
(3, 5, 7)
Example
In R3, let
4,17:
U
Now any
=:
vector
V
+
{(o, 6, 0):
=
(3, 5, 7)
W be the z axis: a,6GR} and W =
(3,
-4,
0)
+
{(0, 0, c)
G R}
c
:
G R^ can be written as the sum of a vector and only one way: {a, b, c)
is
the direct
sum
(0, 9, 7)
let
(a, b, c)
in one
Accordingly, R3
and also
(0, 4, 7)
be the xy plane and
U vector in
(3, 1, 0)
of
=
U
(a, 6, 0)
+
in
U
and a
(0, 0, c)
and W, that
is,
R^
®
= U
W.
Solved Problems
VECTOR SPACES 4.1.
Prove Theorem (i)
For any scalar
(ii)
For
(iii)
If
(iv)
For any
(i)
By axiom
(ii)
ku
Let
4.1:
F be
kGK
a vector space over a
and
GK
and any vector
—
where
0,
kGK [A^]
and any
fcO
=
K.
0.
uGV, Ou = 0. = or u = and uGV, then uGV, {-k)u = k{-u) = - ku. fc
0.
with m = 0, we have + = 0. Hence by axiom [Mi], — kO to both sides gives the desired result.
fee
+
By
a property of K,
fcO.
kGK
GV,
field
fcO
=
fc(0
+
0)
=
Adding
+
=
0.
Hence by axiom
to both sides yields the required result.
[Mg],
Om
=
(0
+ 0)m =
Qu
+
Ou.
Adding
- Om
VECTOR SPACES AND SUBSPACES
70
Suppose
(iii)
=
fcw
and k
exists a scalar fc^i such that
Then there
¥= 0.
u = lu = {k-^k)u = k-Hku) =
+
Using u
(iv)
—ku —
+
Using k
Show
=
=
kO
k{u
+
=
(-m))
few
=
1;
hence
=
fe-iQ
+
Adding -ku
k{-u).
(-fe)
=
we
0,
obtain
= Oit = + (-k))u = = k(—u) = —ku. (fe
Thus (—k)u
{—k)u.
u and
+
to both
-v) =
k(u
+ (-v)) = ku +
+
In the statement of axiom [Mz], (a sign represent?
=
b)u
=
k(-v)
au
+
Adding -ku
(-k)u.
ku-
k{u-v) =
v,
{u-v = u+ (-v))
of subtraction
definition
ku
+
ku
(-kv)
= ku -
both
to
kv.
and the result of Theorem
—kv), k(u
4.3.
=
obtain
that for any scalar k and any vectors
Using the {k(—v)
we
0,
fc~ifc
4
k(—u).
—ku =
sides yields
4.2.
=
{-u)
sides gives
[CHAP.
4.1(iv)
kv
which operation does each plus
bu,
addition of the two scalars a and 6; hence it represents the addithe other hand, the + in au+ bu denotes the addition of the two K. On the field in tion operation represents a vectors au and bu; hence it represents the operation of vector addition. Thus each
The
+
(a+b)u denotes the
in
+
different operation.
4.4.
In the statement of axiom [Ma], represent?
=
(ab)u
which operation does each product
a{bu),
In (ab)u the product ab of the scalars a and 6 denotes multiplication in the product of the scalar ab and the vector u denotes scalar multiplication.
field
K, whereas the
multiplication; In a{bu) the product bu of the scalar 6 and the vector u denotes scalar product of the scalar a and the vector bu denotes scalar multiplication.
4.5.
the
also,
V be the set of all functions from a nonempty set X into a field K. For any funcin V defined tions f.gGV and any scalar k G K, let f + g and kf be the functions
Let
as follows:
(The symbol
{f
+ 9){x) -
V
means "for
fix)
X is nonempty, V is
(f
+
g)
+
+ g) + h)(x) = (f+(g + h))(x) = ((f
f(x), g(x)
and
[AJ:
(f
+ g) +
+
=
+ g){x) +
f(x)
+
(g
h
=
f
+
g(x))
+
/,
+ h)(x) =
+ 0)(a;)
and
is
K
X
a vector space over K.
is
need to show that
all
(fix)
+
g{x))
+
h(x),
f(x)
+
(g(x)
+
h(x)),
where addition of scalars
=
h(x)
G
the axioms of a vector
= f + (g + h), it is necessary to show + h) both assign the same value to
=
h{x)
yfx
kf(x),
f(x)
+
(g(x)
+
Vo;
G
=
f(x)
+
0(a;)
0(a!)
= =
the zero vector in V.
0, Va;
f(x)
G X. Then
+
=
yfxGX is
associative; hence
h(x))
for any function
f(x),
that each
X
+(g + h).
denote the zero function:
Let
(/
Thus /
if
h(x) are scalars in the field (f(x)
Accordingly,
We now
that (f + g) + h and the function f + (g
h
7
Prove that
nonempty.
=
(kf){x)
f.g.hGV. To show
Let the function Now, a; e X. [All-
But
and
g{x)
every".)
also
Since space hold.
+
Vo;
G
X
/
G
V,
CHAP.
VECTOR SPACES AND SUBSPACES
4]
For any function / G V,
[A3]:
+
(/
+
Hence /
=
(-/)
+
Hence f
=
g
Let
(-/)(*)
-
f(x)
f(x)
(-/)(«)
=
=
Oix),
= - f(x). Then, yfx&X
Then
+
g
+
f(x)
=
gix)
(Note that
f.
g(x) are scalars in the field [Ml]:
the function defined by
=
0.
+ ffKx) =
(/
-/ be
let
+
f(x)
f.g^V.
Let
[AJ:
=
(-/))(«)
71
/(*)
K where
+
g(x)
+
f(x)
=
g(x)
+
g(x)
follows
f{x)
y/x&X
+ f)(x),
(g
from the fact that
/(«)
and
addition is commutative.)
f,g&V and k & K. Then + 9))i.x) = k((f + g)(x)) =
W
=
=
(kf)(x)
+
+
k(f(x)
=
(kg)(x)
(kf
=
g(x))
kf(x)
+ kg)(x),
+ kg(x) ^fxeX
Hence k(f + g) = kf + kg. (Note that k(f(x) + g{x)) = kf(x) + kg(x) follows from the fact that k, f(x) and g(x) are scalars in the field K where multiplication is distributive over addition.) Let
[M2]:
/ey
and
= =
((a+6)/)(a;)
Hence
(a
+
6)/
Let
[Mg]:
=
+
af
f&V
(ab)f
Since
(af+hf)(x),
=
G
a, 6
X.
(af)(x)
+
6/(a;)
Then,
=
(a6)/(x)
G
=
a(6/(a;))
o(6/)(a;)
leK,
Then, for the unit
y.
the axioms are satisfied,
all
=
+ hfi^x) VaseX
af(x)
=
(a(6/))(a;),
Va;
G
;f
a(6/).
Let /
[AfJ:
4.6.
=
=
(a+h)f(x)
bf.
and
({ab)f)(x)
Hence
6 X. Then
o, 6
y
is
(!/)(»)
=
l/(a;)
=
f{x),
V« G
X.
Hence 1/
=
/.
a vector space over K.
V be the set of ordered pairs of real numbers: V = {{a,b): a,bGR}. Show that V is not a vector space over R with respect to each of the following operations of addition in V and scalar multiplication on V:
Let
(i)
(a, b)
(ii)
(a, 6)
(iii) (a,
6)
+ (c, d) = + (c, d) = + (c, d) =
(a
+ c,b + d) and
(a, 6)
(o
+ c,
and 6
k{a, b)
+ d) and
k{a, b)
=
=
kb);
(fee,
{ka, b);
fe(a, 6) := (fc^a, fe^ft).
In each case show that one of the axioms of a vector space does not hold. (i)
Let r
=
=
l, 8
2,
=
v
(3, 4).
Then (r
Since
(ii)
Let
0)
(r
=
+ s)v
¥=
(1,2),
w=
rv
rv
+
+
sv,
sv
^
+ s)v =
1(3, 4)
+
(iii)
Let r
=
1,
s
w+
¥=
=
2,
i;
v,
=
Thus
{r
+ s)v
¥=
rv
+
+
(1, 2) (3, 4)
axiom [AJ does not
(3, 4).
SV sv,
+
(6, 4)
=
(9, 8)
hold.
=
+ +
(3, 4)
(1,2)
= =
(1, 2)
(3,4)
hold.
Then (r
rv
(9,4)
(3, 4)
Then
(3,4).
+w w+v = +w
=
2(3, 4)
axiom [M^] does not
v
Since v
=
3(3,4)
+ s)v =
1(3, 4)
+
3(3, 4)
2(3, 4)
=
=
(3,
4)
and so axiom [M2] does not
(27, 36)
+
(12, 16)
hold.
=
(15, 20)
VECTOR SPACES AND SUBSPACES
72
[CHAP. 4
SUBSPACES 4.7.
Wis + wGW, and
Prove Theorem 4.2: implies v
Suppose
W satisfies
a.
subspace of V if and only implies kv v and
(i), (ii)
By
(iii).
if (i)
GW
GW
(iii)
W
(i),
W
is
nonempty; and by
is
nonempty,
(ii)
(ii)
and
eW
v,w
kGK.
for every scalar
the operations
(iii),
Moreover, the axioms [A,], [AJ, of vector addition and scalar multiplication are well defined for W. belong to V. Hence we need only show since the vectors in [Ml], [Ma], [Mg] and [MJ hold in Then by (iii), Ou - S is nonempty, say that [A2] and [A3] also hold in W. By (i), G then (-l)v = -v £ v it Lastly, satisfies [Ag]. and v + = v for every v G W. Hence of V. subspace is a Thus = [A3]. satisfies hence 0; and V + (-v)
W
W
W
W
Conversely,
4.8.
TF
if
Prove Corollary implies av
a subspace of
is
4.3:
GW
+ bw
W satisfies
Suppose by (ii), v + w
by Theorem
=
lv
4.9.
W
if
(i)
w = {(a, b,0):
a,b
third component
=
W
(ii) tt
kGK
and
then,
V
W G
is
R},
then clearly
a subspace of
W
i.e.
is
(i)
V
and
(ii),
kv
=
kv
W
then,
+ Ove W. Thus
GW
= =
k'w
i.e.
0},
k(a, b, 0) (ka, kb, 0)
and so
where:
W
consists of those vectors each with the is zero.
For any vectors
W
is
+ k'(c, d, 0) + (fc'c, k'd,
0)
=
(ka
+ k'c,
kb
1;
a' +
=
+ k'd,
(a, 6, 0),
w =
0)
a subspace of V.
+ + = 0. Suppose v = (a, b, c), w = (a', b', e') + C = 0. Then for any scalars k and k', kv + k'w = k(a, b, c) + k'(a', b', c') = (ka, kb, kc) + {k'a', k'b', k'c') = (ka + k'a', kb + k'b', kc + k'c')
since
and
W.
the xy plane consisting of those vectors whose
of its components
sum
+ k'w e W,
= (0, 0,0) +6+c =
by
hold in
(ii)
since the third
+
kv
Thus kv
v&W
if
is 0. component of W, and any scalars (real numbers) k and k',
e
0,0)
d, 0) in
(c,
and
+b+c =
{{a,b,c): a
(0,
hold.
is 0;
property that the (i)
(iii)
v,wGW
a subspace of
that
W
and
(ii)
a subspace of V. is
y = R^ Show
(ii)
(i),
e TF and (ii) a subspace of V if and only if (i) for all scalars a,b GK. Then, by (i), W is nonempty. Furthermore, if v,w G (i) and (ii).
Let
=
then clearly
ia
+ lweW;
4.2, Tf^ is
Conversely,
W
W W
W
W
W
V
uGW.
belong to
W,
i.e.
6'
and furthermore, (ka
Thus kv
4.10.
Let (i)
+ k'a') +
+ k'w e W,
(kb
and so
V = R^ Show that W PF = {{a, b,c): a ^ 0},
+ k'b') +
W
is
(kc
(iii)
Pf
=
{(a, b, c):
not exceed
1;
W = {(a, 6,
c)
:
d'
+ c) + + fc'O =
k(a+ fcO
b
k'{a'
+
b'
+ e')
not a subspace of V where: consists of those vectors whose i.e.
is
W
+ b^ + c^^
a, b, c
= =
a subspace of V.
nonnegative; (ii)
+ k'c')
e Q},
1},
i.e.
i.e.
first
component
is
W consists of those vectors whose length does
W consists of those vectors whose components are
rational numbers. In each case, show that one of the properties (i)
,,
=
(1,2,3)
W since -5
GW
and
fc
is negative.
= -5 e Hence
R.
W
is
of, say.
=
Theorem
-5(1,2,3) fc. not a subspace of V.
But
=
4.2 does not hold.
(-5, -10,-15)
does not belong to
CHAP.
VECTOR SPACES AND SUBSPACES
4]
eW
V = (1, 0,0) belong to
(ii)
W
(iii)
v
=
4.11.
its
W
(ii)
consists of the
square
all
nxn
Is
+
(1, 0, 0)
=
(0, 1, 0)
does not
(1, 1, 0)
not a subspace of V.
=
2V2, 3\/2) does not belong to not a subspace of V.
(\/2,
W
is
matrices over a
symmetric matrices,
W consists of matrices W= {AgV: AT = TA}.
Show
K.
A=
matrices
all
i.e.
field
(otj)
that
W
for which
which commute with a given matrix T; that
all
OSW
(i)
W
and k = y/2GK. But fcr = \/2 (1,2,3) components are not rational numbers. Hence
Let V be the vector space of is a subspace of V where: (i)
+w =
But v Hence
+
1^
GW
(1,2,3)
W since
w = (0, 1, 0)eW. 1^ + 0^ = 2 > 1.
and
since
73
is,
are and hence equal. Now suppose A = (ay) and B = (6y) = ay and 5jj = 6y. For any scalars a, 6 G if aA + bB is the matrix aa^ + 66y. But aa^j + 66ji = aoy + 66y. Thus aA + 6B is also symmetric,
since all entries of
W,
belong to
i.e.
ftjj
,
whose ii-entry is and so TF is a subspace of V.
OeW
(ii)
or =
since
=
Now
TO.
A,BgW;
suppose
that
is,
AT - TA
and
BT =
+
b(TB)
TB. For
any scalars a,b G K,
+ bB)T = =
{aA
Thus aA
4.12.
+ 6B
(aA)T T(aA)
commutes with
Let V be the vector space of is not a subspace of V where:
(ii)
W consists of W consists of
(i)
(Recall that
(i)
to
W since
det (A
=
W
W; hence
=
A
is
(
»
0\ I
a subspace of V.
=
0.
But
field
Show
R.
that
W
= A.
The matrices
be.)
a{TA)
A =
A+B =
f
and
)
f
B =
f
W since
does not belong to
j
belong
j
not a subspace of V.
=
i
for which A^
and det(B)
W
is
2 x 2 matrices over the real
all
^ = ad — Hence
/2
r(aA
+ b(BT) = + 5B)
matrices with zero determinant;
But 2/
belongs to
a(AT)
matrices
The unit matrix
(ii)
T{hB)
= =
all
det(A) 1.
i.e.
{bB)T
all
det(
+ B) =
T,
+ +
,
)
=
belongs to
i'l
W since
X
n
W
since
does not belong to
(
=
/1 (' Vo
") 1
=
^
¥^
2/
4 4
Hence
4.13.
W is not a
subspace of V.
V be the vector space of a subspace of V where:
Let is (i)
(ii)
w = {f: W = {f:
/(3) /(7)
functions from the real
= 0}, i.e. W consists of = /(!)}, i.e. W consists
value to 7 and (iii)
all
field
R
into R.
those functions which
map
Show
that
W
3 into 0;
of those functions which assign the
same
1;
W consists of the odd functions,
i.e.
those functions / for which /(-«)
= - /(«)•
VECTOR SPACES AND SUBSPACES
74
Here (i)
denotes the zero function:
OeW
since
0(3)
numbers a and
=
(af
Hence af + bg (11)
OeW
since
& W,
=
any real numbers a and
(ill)
OeW
since
— g{x).
4.14.
& W,
=
+
R.
and
=
bg{3)
f.g^W,
real
(a/+6f?)(-a!)
=
a/(-a;)
Hence af + bg
€:
W, and
+
af(7)
W
+
aO
I.e.
=
/(7)
W={f:
(ii)
W
=
/(7)
2
=
Then for any
0.
real
=
60
and
/(I)
=
«r(7)
Then, for
flr(l).
W
is
=
6ff(l)
(a/+6fl-)(l)
f,g&W,
=
/(-x)
i.e.
and
-/(«)
g{-x)
=
b,
= - af(x) -
6flr(-a;)
so
+
a/(l)
Suppose
-0(a;).
numbers a and
+
=
bg(l)
= - (a/(x) +
bg{x)
= -(af + bg)(x)
6flr(a;))
a subspace of V.
Let V be the vector space of all functions from the real is not a subspace of V where: (i)
sr(3)
a subspace of V.
is
= -0 =
=
0(-a;)
/(3)
i.e.
&
a subspace of V.
is
=
and so
Then for any
x
(1/(3)
Suppose
0(1).
for every
4
b,
(af+bg){7)
Hence af + bg
=
bg)(3)
and so TF
=
0(7)
+
0,
f.gGW,
Suppose
0.
b,
=
0(a;)
[CHAP.
R
field
Show
into R.
that
W
+ /(!)};
consists of all nonnegative functions,
function / for which
all
i.e.
f{x)
^ 0,
yfxGR. (i)
f.geW, I.e. /(7) = 2 + /(l) and flr(7) = 2 + flr(l). Then = /(7) + fl-C?) = 2 + /(I) + 2 + flr(l) (f + g)i^) = 4 + /(I) + ir(l) = 4 + (/ + flr)(l) ^
Suppose
Hence f + g^W, and so (11)
Let
fc
V« e
= -2
and
But
R.
/
let
(fc/)(5)
=
fe/(5)
(/
+ sf)(l)
not a subspace of V.
Tl' is
GV
+
2
be defined by
=
(-2)(52)
=
/(a)
= -50 <
W
since Then / G Hence kf € W, and so
0.
x^.
W
/(«) Is
=
a;2 s= Q,
not a sub-
space of V.
4.15.
(i)
(ii)
(iii)
V be
W consists of W consists of W consists of
(1)
(ii)
+
a^t^
•
all all
polynomials with degree
all
polynomials &o
+
&it^
+
•
•
+
a„f"
V
a subspace of
is
polynomials with integral coefficients;
only even powers of
4.16.
+
the vector space of polynomials ao + ait ficients, i.e. Oi e R. Determine whether or not VF
Let
with real coefwhere:
— 3;
h^t^
+
•
•
•
+
bnt^",
i.e.
polynomials with
t.
do not always belong to W. For example, v = No, since scalar multiples of vectors in is "closed" under vector = (Observe that but ^i' 3 + 5t + 7(2 e f + |« + 1*^ ^ W. belong to W.) in elements addition, i.e. sums of
W
^
W
For, in each case, Yes. the scalar multiples of any element in
and
W
W
(iii).
is
nonempty, the sum of elements
W belong to
in
Consider a homogeneous system of linear equations in n unknowns field K: ^ + ainXn = n anXi + ai2X2 + ,
•
Show that the = (0, 0, .
.
solution set
. ,
0)
e
PF
aziXl
+
022*2
+
OmlXl
+
am2X2
+
W
is
•
•
•
•
+
a2nX„
=
•
+
dmnXn
=
a subspace of the vector space K".
since, clearly,
a«0
+
ajjO
+
•
•
•
+
ttinO
=
0,
W belong to W, and
W.
for
t
=
1,
. .
.,m
Xi,
.
..,Xn over a
CHAP.
VECTOR SPACES AND SUBSPACES
4]
Suppose M
=
(«!, M2,
•
•
.
,
=
and v
M„)
«il"l OjiVi
(vj, Vg,
+ +
.
.
+ +
ai2W2 ai2'y2
,
.
•
•
belong to W,
v„) •
•
•
+ +
75
i.e.
for
i
—
I,
.
.
.,Tn
= =
ttinMn
ai„v„
Let a and 6 be scalars in K. Then and, for
i
=
au 1,
.
.
.
,
m,
aji(aMi
+
=
bv
(ciMx
+
+ 6v
"
•
a solution of the system,
is
+
6^2.
•
•
>
<*w„
+
6v„)
+ bvj) + ai2(au2 + 6f 2) + + ain(aMn + bv^) = o(ajiMi + OJ2M2 + + «!„«*„) + 6(ajii;i + (ij2i;2 + = aO + 60 = •
Hence au
6^1, au2
•
•
•
belongs to W.
i.e.
Accordingly,
•
•
+
•
W
aini'n)
is
a subspace of K".
LINEAR COMBINATIONS 4.17.
= =
Write the vector v 62 = (1,2, 3) and 63
We we
(1,
—2,
(2,
-1,1).
wish to express v as v
=
xei
require (1,
Form
-2,
as a linear combination of the vectors
5)
+
= = =
5)
+
3/62
with
ze^,
x,
y and
z
as yet
unknown
ei
=
(1, 1, 1),
scalars.
Thus
+ j/(l, 2, 3) + z(2, -1, 1) (x, X, x) + (y, 2y, 3y) + (22, -z, z) + 2/ + 2z, + 2j/ — z, + 32/ + 2) x{l, 1, 1)
(a;
a;
a;
the equivalent system of equations by setting corresponding components equal to each other,
and then reduce to echelon form:
x+
y
+ +
2y
X x
3y
+ —
2z z
+
z
= 1 = —2 = 5
+
x
y
or
j/
2y
+ 2z = 1 - 3z = -3 - z =
x or
2z Sz
= =
1
S
4,
Solve for the
unknowns
to obtain
Write the vector v = (2, -5, 3) in R^ as a linear combination of the vectors (1,-3,2), 62 = (2, -4,-1) and 63 = (1,-5, 7). Set V as a linear combination of the (2,
Form
-5,
3)
= =
x{\,
{x
using the unknowns
Cj
-3,
+
2)
+ 2y + z,
+ -3x 2x
—
2y+z=2 4y
-
5z
y
+
Tz
= -5 = 3
For which value of k
=
the vectors v Set u = XV
(1,
+
x
z:
y{2,
v
=
xe^
+ j/eg + zeg.
-4, -1)
2y+z-2 -
2z
-5y +
5z
2y
or
(3, 0,
+
-2)
and w =
x
= 1 = -1
+
2y
2^
or
+ -
z
2z
= = =
2 1
3
Accordingly, v cannot be written as a linear com-
u=
will the vector (2,
(1, -2, k) in R" be a linear combination of -1, -5) ?
yw:
-2,
fe)
=
a(3, 0,
-2)
+
j/(2,
-1, -5)
=
(3a;
+ 2y,
-y, -2x
- 5y)
the equivalent system of equations:
+ 2y = 1, = x = —1,
-y =
3x
By
y and
=
+ z(l, -5, 7) -3x -4y-5z,2x-y + 7z)
The system is inconsistent and so has no solution. bination of the vectors Ci, e^ and 63.
Form
x,
Ci
the equivalent system of equations and reduce to echelon form:
x
4.19.
+ y -
y
5z=10
Note that the above system is consistent and so has a solution. X = —6, y = B, z — 2. Hence v = — 6ei + 862 + 263.
4.18.
+
the first two equations,
j/
2.
-2,
-2x - 5y = k
Substitute into the last equation to obtain
k
=
—8.
VECTOR SPACES AND SUBSPACES
76
4^0.
=
Write the polynomial v nomials
= t^-2t +
ei
+ 4t — 3 over R as = 2t^ - St and ca =
5,
Set D as a linear combination of the
+
t2
-
4t
= = -
3
+ —2x —
=
2y
+
Note that the system a;
4.21.
=
-3,
2/
=
2, z
=
=
v
z:
+
xe^
ye^
+
263.
2xt
equal to each other, and reduce the system to echelon form:
t
+
X
=
2y
2/+z=6
or
-IQy +
+
x
1
=
2y
or
2/
= -8
3z
1
z=6
+
13z
=
52
Solve for the unknowns to obtain
consistent and so has a solution. - -Ssj + ie^ + 463.
is
y and
x,
+ 5) + 3/(2«2-3t) + 2(f + 3) + 5a; + 22/t2 - s^/t + zt + 3z {x + 2y)fi + {-2x-3y + z)t + (5a; + 3z) -
a;t2
= -3
3z
S.
using the unknowns
ej
1
3j/+z=4
hx
+
t
a;(t2-2t
Set coefficients of the same powers of
x
a linear combination of the poly-
t^
62
[CHAP. 4
Thus v
4.
E —
Write the matrix
A =
as a linear combination of the matrices
I j
;j).-a?)--(:-x Set
Form
£
as a linear combination of
/I
3
IN
1
_iy
=
A,B,C
/O
1\
oy +
^'(i
iKN,/0
X
\y
0/
/O
2
^o
-1
0\
1) +
\i
X
E — xA + yB +
using the unknowns x,y,z:
0N/0 /O
/
X
VJ
\a;
+
22 2z\ -s \0-«/ \0
X
y
2/
+ 2z —z
the equivalent system of equations by setting corresponding entries equal to each other: a;
=
X
3,
+
y
—
+
x
1,
=
2z
—
y
1,
= —1
z
Substitute a; = 3 in the second and third equations to obtain y = —2 and z values also satisfy the last equation, they form a solution of the system. Hence
4.22.
zC.
Suppose m is a linear combination of the vectors linear combination of the vectors Wi, Wn.
u =
+
aiVi
+
a2V2
•
•
•
Vi,
.
.
=
—1.
E =
•
•
•
+ OmVm
and
Vi
=
+ baWi +
haWi
•
•
+
•
•
u
-
•
•
"S,
m
i=l
^in
=
/
2 i=l
^i
L{S) (ZL{T).
•
•
•
'
n
2 \3=1
(
•
•
•
n
\
h'^j
/
2 3=1
=
)
/
•
7n
•
+
frmn'"'™)
\
2 \i=l
(
•
•
cuba
)
Wj
/
LINEAR SPANS, GENERATORS 4.23.
Show
u=
that the vectors
We Set
(1, 2, 3),
v
=
need to show that an arbitrary vector (a, 5, c)
=
xu
(a, b, c)
+ =
yv
+
(0, 1, 2) (a, 6, c)
w=
and
S R3
is
(0, 0, 1)
generate
+
1/(0, 1, 2)
+
z(0, 0, 1)
W.
a linear combination of u, v and w.
zw:
x(l, 2, 3)
a
bi„w„
•
m or simply
Vi is
C.
. ,
.
•
•
— 2B —
.,Vm and suppose each
Show that u is also a linear combination of the wu Thus if ScL{T), then u = a^Vi + a^v^ + + a^v^ = ai(6iiWi + + b2nWn) + + a^(h.ml1»l + + 6i„W„) + 02(62l"'l + = (diftji + (12621 + + aJi^^Wn + am6mi)wi + + (ai6i„ + a^h^n + •
Since these
BA
=
{x,
2x
+ y,Sx + 2y + z)
CHAP.
VECTOR SPACES AND SUBSPACES
4]
Then form the system
77
of equations
=
X
2x Zx
+ +
a
+
2y
=
z
+
«
=6
y
+ 3a;=:c + 2x = h X = a
2i/
or
y
c
The above system is in echelon form and a solution. Thus u, v and w generate R^.
—
x
consistent; in fact
is
a,
—
y
b
— 2a,
z
= e — 2b + a
is
4.24.
Set
+
xu
yv
+
X
using unknowns
+y —y+ 2y -
x,
y and
(a, b, c)
z:
=
zw.
=
+
x(2, 1, 0)
=a Zz Az
— =
-1,
j/(l,
2x or
b
+
+
2)
z{0, 3,
= = =
y
-
3j/
2y
c
=
-4)
6z
iz
{2x
+ y,x-y + Sz, +
2x
-
=a
y
Zy
or
26
- 4z)
echelon form:
it to
a a
2y
-
= =
^z
c
a
- 2b - 46 -
2a
3c
belongs to the space generated by u, v and w if and only if the above system is consistent if and only if 2a - 46 - 3c = 0. Note, in particular, that u, v and do not generate the whole space R^.
The vector
(a, b, c)
consistent,
and
w
belongs to the space generated by
the equivalent system of linear equations and reduce
Form 2x
w
as a linear combination of u, v and
(a, 6, c)
{a, b, e)
4.25.
W
Find conditions on a, b and c so that (a, b, c) G u = (2, 1, 0), V = (1, -1, 2) and w = (0, 3, -4).
Show
it is
that the xy plane
(1, 2, 0)
and v
=
W = {(a,
(0, 1, 0);
(ii)
generated by -1, 0) and v = (1, 3, 0).
b, 0)} in
u=
(2,
In each case show that an arbitrary vector Set
(i)
(a, b, 0)
=
xu
+
R^
is
(a,
b,0)eW
u and v where:
(i)
u=
a linear combination of u and
is
v.
yv:
=
{a, b, 0)
x(l, 2, 0)
+
=
y(0, 1, 0)
(x,
+ y,
2x
0)
Then form the system of equations X
+
y
fact
x
2x
The system (ii)
Set
(o, 6, 0)
is consistent; in
=
xu
+
=
x{2,
-1,
the following system and reduce
+ —x + 2x
y Sy
= = =
0) it
+
4.26.
Show a
we
+
2x x
is
y(l, 3, 0)
— =
b
a
a solution.
Hence u and v generate W.
=
(2x
+ y,-x + 3y,
0)
to echelon form:
2x
a b
or
+
y
7y
= -
a a
+
2b
W
is generated by u and v. (Observe is consistent and so has a solution. Hence do not need to solve for x and y; it is only necessary to know that a solution exists.)
The system that
y
yv:
{a, 6, 0)
Form
= a = b or = = a, y = b-2a
that the vector space number of vectors.
V of polynomials
over any
field
K cannot be
generated by
finite
Any finite set S of polynomials contains one S cannot contain polynomials of degree
L{S) of
finite set S.
of maximum degree, say m. Then the linear span greater than m. Accordingly, V f^ L{S), for any
VECTOR SPACES AND SUBSPACES
78
4.27.
Prove Theorem
Let S be a nonempty subset of V.
4.5:
linear combinations of vectors in S, is a subspace of
W
is
any other subspace of
V
—
where
w^
Vi,
G S and
•
+
•
+ w =
is
Then
L{S), the set of all
Furthermore,
containing S.
if
w —
+
b^Wi
•
+
•
is
nonempty
since
S
is
6„w„
Then
+
a^vi
Also, L(S)
a subset of L{S).
and
a^nVm
are scalars.
5j
fflj,
V
+
ai^Ui
V
containing S, then L{S) C W.
If V G S, then Iv = v G L{S); hence S nonempty. Now suppose v,wGL(S); say,
V
[CHAP. 4
•
+ a^v^ +
•
+
b^Wi
•
•
+
•
6„w„
and, for any scalar k,
-
kv belong to L(S) since each
+
A;((iii;i
is
•
•
+ a^v^ -
•
+
ka^v^
•
a linear combination of vectors in S.
•
+ ka^v^
•
Accordingly, L(S)
is
a subspace
of V.
Now
suppose
multiples a^Vi is, contains
W
W
a subspace of
is
Om'^'m all
^ ^>
"where
V
containing
Oj
G
K,
S and suppose v^, and hence the sum a^v^
.
linear combinations of elements of S.
.
S cW. Then all + a^Vm ^ ^- That
.,v^E.
+
•
Consequently, L(S)
• •
c T^ as
claimed.
ROW SPACE OF A MATRIX 4.28.
Determine whether the following matrices have the same row space:
Row
reduce each matrix to row canonical form:
^
B
-
=
5
1 z
r. (^
z) 13
3
-1 -2 ; r (s -2 -3 f
I
to
(J
~
('
to
)
to
\)
;
~ 1
3>
1
3
2
6/
I
I
I
1
3
to
)
Vo
(;
/I -1 -l'
c
=
4
V
-3 -1 -1 3/
to
I
to
to
I
Since the nonzero rows of the reduced form of A and of the reduced form of C are the same, C have the same row space. On the other hand, the nonzero rows of the reduced form of B are not the same as the others, and so B has a different row space.
A
4.29.
and
Consider an arbitrary matrix A = (a«). Suppose u = (&i, ...,&«) is a linear combination of the rows Ri, .. .,Rm of A; say u = kiRi + + kmRm. Show that, for each i, bi — kiaii + feozi + + kmOmi where an, ., Omi are the entries of the ith column of A. •
•
We
are g:iven
u = (6i
•
•
ftjiJi
.
+
K)
•
•
= =
•
+
k^^R^;
fci(an. (feifflii
•
•
.
hence
•
•
• .
+
•
•
•
Om)
+
•
+ fc^Oml)
Setting corresponding components equal to each other,
•
+
•
•
we
•
• >
fem(ami. ^^l^ml
+
•
•
• .
•
•
O-mn)
+ kmO'mn)
obtain the desired result.
CHAP.
4.30.
VECTOR SPACES AND SUBSPACES
4]
A=
Prove: Let
and
let
B—
79
be an echelon matrix with distinguished entries aij^, a^^, be an echelon matrix with distinguished entries bik^, &2fc2»
(an)
(&«)
•
****** \
Olj
^
b-i
/
a2i.
4i
^
:]c
A =
•
4i
9|i
V
v
ifc
^
Osfc
•
.
.
., ttrj,,
bsk;.
.
B a,v,
A and B have the same row B are in the same position: ji —
Suppose of
32 =
fci,
Clearly
and
—
s
CiO
+
C2O
i4
We
1.
Since the
+
•
let
space.
+
•
cji
# 0.
A =
.
Ji
—
ij
row space of B, we Cj. But this contradicts the fact that and similarly fei — ij. Thus j'l = fcj.
k^,
and
only prove the theorem when r — 1 Then the j^th. column of B is zero. have by the preceding problem, Oi^^ =
if
the distinguished
for scalars
Hence
A
B = 0, and so we need = k^. Suppose ii < k^.
the
is in
of
.
A obtained by deleting the first row of A, and let B' be the obtained by deleting the first row of B. We prove that A' and B' have the same The theorem will then follow by induction since A' and B' are also echelon matrices. A' be the submatrix of
submatrix of
row
•
and only show that
if
first
row of
first
element a^
Now
=
Then the distinguished entries kz, ., jr = kr, and r = s.
space.
B
.,a„) be any row of A' and let R^, ...,B,n ^^ the rows of B. Since R is in Let R = («!, 02, .,d^ such that R — diRi + ^2^2 + + dmRm- Since the row space of B, there exist scalars d^, A is in echelon form and R is not the first row of A, the ^ith entry of R is zero: aj = for Furthermore, since B is in echelon form, all the entries in the fcjth column of B are i = ;j = fej. .
.
.
except the
= Now
6ifc
#
and so
=
but 62^1
61^.^ ¥= 0,
first:
di
=
0.
>
^>
=
Ofcj
•
.
+
difeifcj
B
Thus
is
^rnkj
=
dgO
•
•
Thus
0.
+
•
•
+
•
d„0
d,b.
a linear combination of R^,
.
.
.,Bm and so
is in
row
the
space of B'. Since R was any row of A', the row space of A' is contained in the row space of B'. Similarly, the row space of B' is contained in the row space of A'. Thus A' and B' have the same row space, and so the theorem is proved.
4.31.
Prove Theorem 4.7: Let A = {ckj) and B = (&«) be row reduced echelon matrices. Then A and B have the same row space if and only if they have the same nonzero rows. we
Suppose exist scalars
A
and
Cj,
.
.
B
. ,
have the same row space, and suppose
c^
1
but
Cfc
=
for
A;
#
+
CiRi
where the Ri are the nonzero rows of B.
=
R
¥=
+
c^Rs
««i
But by the preceding problem
C2R2
+
•
•
The theorem
•
is
proved
=
6y. is
i.e.
Cl&ljj
the
+
first
¥=
row
Then there
of A.
i,
W if
we show
«262Ji
nonzero entry of R.
+
•
•
•
+
a distinguished entry of
the only nonzero entry in the ijth column of B. Thus from oy = 1 and 6y = 1 since A and B are row reduced; hence Cj
suppose k
the ith
that
R —
R^,
or
t.
Let ay be the distinguished entry in R,
Now
is
such that
R Cj
Thus
Obviously, if A and B have the same nonzero rows then they have the same row space. only have to prove the converse.
and
b^j.
a„.
"«fc
is
=
+
Hb2j^
+
(1)
and Problem
B
and, since
(2)
=
+
4.29, (2)
C,b,j.
we
B
obtain
is
Oy^
row reduced,
=
Cjfty..
it is
However,
1.
the distinguished entry in R^. Cibij^
By
By
C,b,j^
(i)
and Problem
4.29, (S)
VECTOR SPACES AND SUBSPACES
80
B
Since
—
"ijfc
A
row reduced,
is
and the theorem
4.32.
=
row reduced, a^^
is
bj^j^ is
the only nonzero entry in the i^th column of B; hence by
Furthermore, by the preceding problem
OkHj,^-
is
Thus
0.
[CHAP. 4
=
c^b^j^
a distinguished entry of
is
a,.j
and, since
b^j^
=
1,
c,,
=
A
R = R^
Accordingly
0.
(3),
and, since
proved.
Determine whether the following matrices have the same column space: /l
3
5\
1
4
3
\1
1
9/
A = Observe that the same
A
row
space.
A'
=
112
-2 -3 -4
B =
,
3^
12
7
\
17y
and B have the same column space if and only if the transposes A* and B* have Thus reduce A' and J5' to row reduced echelon form: 1
4
3
to
1
-2
1
to
1
1
1
-2
to
0/ 'l
=
B*
-3 \s -4
12
2
to
1
17/
Since A* and B* have the same row space,
4.33.
Let
jR
defined,
.
. ,
B»
RB = = = = Thus
and
7
1
-2
B
to
0/
\0
have the same column space.
B a matrix for which RB is defined. Show that RB is a rows of B. Furthermore, if A is a matrix for which AB is
show that the row space of AB is contained in the row space of B. i? = (aj, ttg, a^) and B = (6y). Let 5i, ...,B^ denote the rows
Suppose .
A
-2
to
be a row vector and
linear combination of the
Bi,
2
\0
-2 -4/
.
.
•
. ,
B
of
and
Then
columns.
its
{R'B^.R'B^, ...,R'B^)
+ 02621 +
(aibii
ai(6ii, 612,
+
a^Bi
6i„)
. ,
.
+
a252
•
•
•
ttm^ml. ai*12
+ 02*22 +
+ a2(b2i, 622 + amBm
&2n)
•
+
•
+ am&m2. + am(&ml.
• .
•
•
•
+ 02*2n ^
«l&ln 6m2.
•
•
•
.
1"
am&mn)
bmn)
a linear combination of the rows of B, as claimed.
fijB is
By Problem result each
.
^-
row
3.27, the
of
AB
is
AB are RiB where i?j is the tth row row space of B. Thus the row space of
rows of in the
of A.
AB
is
Hence by the above contained in the row
space of B.
SUMS AND DIRECT SUMS 4.34. Let U and W be subspaces (i)
(ii)
of a vector space V.
W are contained in
U and U+W
\&
JJ
Show
that:
+ W;
the smallest subspace of V containing C/ and PF: C/ + W^ = L(C7, W).
U
and W, that
is,
U+W
is
the
linear span of (i)
a subspace of V and so d &W. Hence m = m + OS J7+W. U +W. Similarly, W is contained in U + W. Since 17 + W is a subspace of V (Theorem 4.8) containing both U and it must also contain the linear span of V and W: L{JJ,W) (ZU + W. On the other hand, if v GU +W then v — u + w = lu+lw where uGU and w &W\ Let M
G
By
[/.
Accordingly, (ii)
U
hypothesis TF contained in
is
is
Tl',
hence v
U+
W
is (Z
a linear combination of elements in
UuW
L(U, W).
The two
inclusion relations give us the required result.
and
so
belongs to L{'U,W).
Thus
CHAP.
4.35.
VECTOR SPACES AND SUBSPACES
4]
U
Suppose
W are subspaces of a vector space
and
Show
W.
{Wj) generates
^U +W.
Let V
that {Ui,W}),
= u-\-w where u G U and w G W. Since {mJ generates and since {Wj} generates W, w is a linear combination of Wj's:
Mj's;
=
u
w =
and so
4.36.
= u+w =
V {mj,
w^} generates
Prove Theorem 4.9: if and only ii (i) V
Since such a
On
-
V
(1)
uG U
u-u'GU
U
Let
Note
+
a2Ui
+
•
•
+
•
•
•
+
a„Mj
+
K'Wj^,
•
+
a^Mj
?7,
tt
is
K bj e K G
ttj
,
+
biW,
b2Wj
+
•
•
+
•
fc^Wj
V
is
sum
the direct
UnW
=
of its subspaces
U
and
vGU,
1;
GV
G W;
V - U
—
and
unique and
+ v where OGU, UCiW = {0}.
v
w
GW
UnW
= {0}. Let vGV. Since V = U + W, and u + w. We need to show that such a sum is unique. and w' G W. Then
V =
u
so
UnW
- w 6 W; hence by u — u' — 0, w' — w = is
=
v
(2)
Accordingly,
0.
+W
W
{0}.
can be uniquely written in the form v = u + Thus, in particular, V = U + W. Now suppose v G UnW. Then:
Then any
V
a
TF.
w S W such that v = = u' + w' where u' G U u + w = u' + w' and
first
is
=
{{a,
b,c): a
the yz plane.) JJnW'
that
=
b
=
=
u'
w'
—w
{0},
and
so
®
TF.
C7
—
u
=
u',
w =
w'
=
= c
=
b
=
c}
and
W
=
{(0, 6, c)}
Show that R^ = U ® W. v = (a, b, c) G UnW implies that = a = 0, 6 = 0, which implies
for
{0},
and a
=
c
(0, 0, 0).
We also claim that (a, a, a)
imply R3
4.38.
+
and w'
a
where
62WJ2
•
and
and PT be the subspaces of R^ defined by
(Note that PF
V
+
must be unique, v
U =
i.e.
+
where
Thus such a sum for f G
4.37.
b^Wj^
and
Suppose also that v
But
i7
the other hand, suppose
there exist
+
W
for «
sum
ciiUi
The vector space = U+ and (ii)
+
V
+
ttiMj
U ® W. w G W.
Suppose V = where uG U and
ffi^j
U
V, and that {im} generates generates U + W.
{Wj),
Then v
linear combination of
Thus
U
{Vn}
i.e.
81
=
S
R^
and
C7
= U + W. For if = (a, 6, c) S RS, then v = (a, a, - a, c- a) GW. Both conditions, UnW = {0} 1;
a)
(,0,b
+ (0,b-a,c- a) = U + W,
and R3
TF.
C7
W
be the Let V be the vector space of ti-square matrices over a field R. Let U and Show that subspaces of symmetric and antisymmetric matrices, respectively. iff anti-symmetric M*, and is symmetric ifi V = U ® W. (The matrix
M
M
M* = -M.)
We We
first
claim that
that
is, -J^CA
V = U + W. Let A be any arbitrary w-square matrix. A = ^(A + A*) + i(A - At) |(A + A') G 17 and that ^(A - A«) G W. For {^{A+At)y = i(A+A«)' = i(A« + A«) = ^{A+A')
show that
+ A')
symmetric.
is
(^(A-At))' that
is,
We implies
J(A
— A')
is
M
Furthermore,
= i(A-A')'
:-
i(At-A) = -^(A-A'^)
antisymmetric.
UnW = {0}. M = Hence
next show that
= -M
Note that
or
0.
Suppose
I7nW =
MGUnW. {0}.
Then
Accordingly,
M = M'
and M^
V= U®W.
= -M
which
VECTOR SPACES AND SUBSPACES
82
[CHAP. 4
Supplementary Problems VECTOR SPACES 4.39.
Let
y be
the set of infinite sequences V defined by
(a^, a^,
.
.
K
in a field
.)
with addition in
V
and scalar multi-
plication on
+
(«!, a2. •••)
where 4.40.
k
aj, bj,
G K. Show
V
that
Let V be the set of ordered pairs on V defined by
+
(a, 6)
Show
V
that
(c,
satisfies all of the
.
.
.
+
02+
62,
numbers with addition
in
=
)
(ai
6i,
ka2,
(fcaj,
.
.
.)
.
.
.
a vector space over K.
is
of real
(a, 6)
= {a+
d)
=
(6i, &2. •••)
k(ai, 02,
c,b
+ d)
and
V
=
k{a, b)
axioms of a vector space except [AfJ
:
and scalar multiplication
{ka, 0)
=
lu
u.
Hence
[^4] is not a
consequence of the other axioms.
4.41.
4.42.
Let V be the set of ordered pairs (a, b) of real numbers. Show that with addition in V and scalar multiplication on V defined by: (i)
{a,b)
(ii)
(c, 6)
(iii)
(a, b)
(iv)
(a, 6)
Let
V be
real field
+ + + +
= = = =
(c,d) (c,
d)
(c,
d)
(c,
d)
4.43.
4.44.
4.45.
(a
+ d,b + c) + c,b + d) and
(0, 0)
and
k(a, b)
and
k(a, b)
k{a, b)
and
(ac, bd)
-
k{a, b)
= =
is
z^, Z2,
Wi,
+
=
(wi, W2)
(«i
+ Wi,
not a vector space over
R
(ka, kb); (a, 6);
{ka, kb);
=
(ka, kb).
the set of ordered pairs (zi, z^) of complex numbers. Show that R with addition in V and scalar multiplication on V defined by (zj, Z2)
where
{a
V
«2
+ '"'2)
and
V
a vector space over the
is
=
k{zi, 22)
{kzi, kz2)
W2 ^ C and k GB,.
Let y be a vector space over K, and let F be a subfield of K. Show that V is also a vector space over F where vector addition with respect to F is the same as that with respect to K, and where scalar multiplication by an element k G F is the same as multiplication by k as an element of K.
Show
that [A4], page 63, can be derived from the other axioms of a vector space.
W
be vector spaces over a field K. Let V be the set of ordered pairs (u, w) where u Let U and uGU, w G W}. Show that y is a vector space over belongs to U and w to W: V = {{u, w) with addition in V and scalar multiplication on V defined by
K
:
(u,
where u, u' and W.)
G
w)
+
U, w,w'
(u',
GW
=
w')
and
(u
k
G
+ u',w + w') K.
and
(This space
V
w)
k(u,
is called
=
(ku,
kw)
the external direct
sum
of
U
SUBSPACES 4.46.
Consider the vector space that TF is a subspace of V (i)
(ii)
4.47.
Let (iii)
Problem
4.39, of infinite sequences (a^, 02,
.
.
.)
in a field K.
Show
all
Determine whether or not (ii) a ^ (i) a = 26; where k^ S R.
W
in
W consists of all sequences with as the first component; W consists of sequences with only a finite number of nonzero components.
which:
4.48.
V if:
W b
is
^
W
a subspace of RS if (iv) a (iii) ab = 0;
c;
=
consists of those vectors b
-
c;
(y)
a
^
6^;
(vi)
(a, b, c)
kia + kib
G RS for + kgfi = 0,
be the vector space of w-square matrices over a field K. Show that T^ is a subspace of V if antisymmetric (A« = -A), (ii) (upper) triangular, (i) of all matrices which are diagonal, (iv) scalar.
y
consists
CHAP.
4.49.
VECTOR SPACES AND SUBSPACES
4]
AX = B
Let
be a nonhomogeneous system of linear equations in that the solution set of the system is not a subspace of K".
Show 4.50.
V be the V in each
Let of
83
W consists of
(i)
that
all
^ M,
|/(a;)|
W consists of W consists of W consists of W consists of
(ii)
(iii)
(iv)
(v)
R
vector space of all functions from the real field of the following cases.
bounded functions. (Here /
bounded
is
Show
into R.
if
W
that
K.
a subspace
is
Af
there exists
field
GR
such
G R.)
Va;
/(— «)
=
—
1.
G
R.)
any number of subspaces of a vector space
V
is
even functions.
all
continuous functions.
all difTerentiable
R -* R
f{x), Va;
all
all
R -^ R
:
n unknowns over a
(Here /
:
is
even
if
functions.
integrable functions
say, the interval
in,
—
a;
(The last three cases require some knowledge of analysis.)
4.51.
4.52.
Discuss whether or not R^
Prove Theorem
The
4.4:
is
a subspace of R^.
intersection of
a subspace
of y.
4.53.
U
Suppose
UcW
and
W
V
are subspaces of
UuW
for which
is
also a subspace.
Show
that either
WcU.
or
LINEAR COMBINATIONS 4.54.
4.55.
Consider the vectors u
(1,
-3,
and v
2)
=
(2,
-1,
in R3.
1)
—4) as a linear combination of u and
v.
as a linear combination of u and
v.
(i)
Write
(1, 7,
(ii)
Write
(2,
(iii)
For which value of k
(iv)
Find a condition on
—5,
4)
a linear combination of u and vt
is (1, k, 5)
a, b
and
a linear combination of u and
c so that (a, 6, e) is
Write m as a linear combination of the polynomials (i)
4.56.
=
M
=
B
Write
+ 8« - 5,
3*2
where:
(ii)
M
=
4<;2
E =
Q "^
;
=
+ 3t — i
and
^ = (_j
n)
2t^
w = t^-2t—Z
where
- 6t - 1. ^ -((._,)
as a linear combination of
(i)
v
v.
(ii)
E =
(J
_l^
>
*"d
*^
~
(
)
.
LINEAR SPANS, GENERATORS 4.57.
4.58.
Show that (1, 1, 1), (0, 1, 1) and bination of the given vectors. Show
that the yz plane
(0, 2, 3)
4.59.
4.60.
Show
{(0, b, c)}
—
R*
i.e.
that any vector
generated by:
is
(i) (0, 1,
1)
(a, b, e) is
and
(0, 2,
a linear com-
-1);
(ii) (0, 1,
2),
(1
- 1)^,
and
1
—t
z
-l — 2i
and
1
generate the complex
field
C
as a
generate the space of polynomials of
3.
{{a, b, 0)},
Prove:
in
w = 2 + 3t
that the polynomials {l-t)»,
Find one vector
U = 4.62.
W=
—1) generate R*,
(0, 3, 1).
Show that the complex numbers vector space over the real field R.
degree 4.61.
and
(0, 1,
L(S)
is
in
and
R3 which generates the intersection of is the space generated by the vectors
W
the intersection of
all
the subspaces of
V
U
and
(1, 2, 3)
W
where U is the and (1, —1, 1).
containing S.
xj/ plane:
VECTOR SPACES AND SUBSPACES
84
4.63.
Show that L{S) = L(S u{0}). That change the space generated by the
4.64.
Show
that
4.65.
Show
that
Sc
if
LmS)) =
c
L(S)
then
T,
[CHAP. 4
by joining or deleting the zero vector from a
is,
set,
we
do not
set.
L(T).
L(S).
ROW SPACE OF A MATRIX 4.66.
Determine which of the following matrices have the same row space: /l -1
.-<: (3 4.67.
Let
= =
Ml vi
4.68.
-.
-4
(1, 1,
:,, 5)'
-(.ra-O'
-1),
U2
(1,-1,-3),
=
V2
=
M3
-1),
(2, 3,
'^
(3,-2,-8),
=
(3, 1,
Vs
=
Show
that if any row of an echelon (row reduced echelon) matrix is still in echelon (row reduced echelon) form.
the
-5)
(2,1,-3)
same as the subspace generated by the
that the subspace of R* generated by the
is
1:-'°
=
Show
it;
3\
is
Vj.
deleted, then the resulting
matrix 4.69.
Prove the converse of Theorem
row 4.70.
4.71.
Show Let
4.6:
Matrices with the same row space (and the same
size)
are
equivalent.
A
in the
A
that
and
B
have the same column space
and B be matrices for which column space of A.
AB
A« and B* have the same row space.
iff
Show
is defined.
that the column space of
AB
is
contained
SUMS AND DIRECT SUMS 4.72.
We
extend the notion of
a vector space
V
(iii)
(iv)
nonempty subsets
{s
+t
that for any subspace
4.74.
Give an example of a subset S,
(ii)
S+SCS
We extend the notion
sG
tG
S,
(not necessarily subspaces) T}.
of
sum
(S2
+
Show
S and T
S
of a vector space (properly contained).
of subspaces to
V
which
= W. is
not a subspace of
more than two summands as
follows.
V
but for which
If T^i,
W^,
.
.
. ,
are subspaces of V, then
Wi + Show
4.76.
W2+---+W„
=
{wi
+ Wi+'-'+w^: WiGWi)
that:
(i)
L{Wi,
(ii)
if Si
W2
W„)
generates Wi,
Suppose U,
V
aijd
of
that this operation satisfies:
S3)
W of a vector space V, W + W
Show
S+S =
:
S+
4.73.
4.75.
+ T =
associative law:
(ii)
(i)
to arbitrary
T = T + S; S2) + S3 = Si + + (Si S + {0} = {0} + S = S; S + V = V + S = V. commutative law:
(i)
sum
by defining S
^ W, + W2+ + W„; = then .,n, i 1, Si U S2 U .
.
•
W are subspaces of a vector space. (UnV) + (UnW)
c
Find subspaces of B? for which equality does not hold.
•
•
U S„ generates W^ + W2 +
Prove that
Un(V+W)
•
•
+ Wn-
T^„
CHAP.
4.77.
VECTOR SPACES AND SUBSPACES
4]
V
Let U,
W be the following subspaces of R^:
and
U = Show that 4.78.
85
a+b + c =
{(a, b,c):
= V+
R3
(i)
V,
V =
0},
(ii) B,»
(iii)
=
a
{(a, b,c):
V + W,
=^
W
c},
=
= V + W. When
R^
Let V be the vector space of all functions from the real field B into B. even functions and the subspace of odd functions. Show that V = even iff f{-x) - f{x), and / is odd iff f(-x) = -f(x).)
W
4.79.
Let Wi, W^, ... be subspaces of a vector space that is a subspace of Y.
W
Show 4.80.
4.81.
V
{(0, 0, c)
for which Wy
In the preceding problem, suppose Sj generates W^, generates W.
i
=
•
K}
direct?
Let U be the subspace of U ®W. (Recall that / is
Let PF
.
Show
1, 2,
sum
the
is
e
c
:
that
= Wj U S =
TFj
U
Sj U Sa
U
•
•
•
•
Let V be the vector space of w-square matrices over a field K. Let U be the subspace of upper triangular matrices and the subspace of lower triangular matrices. Find (i) U + W, (ii) UnW.
W
4.82.
Let Let
V
be the external direct
sum of
that
U
(i)
4.50.
(i)
Yes.
(ii)
No;e.g.
(iii)
No;e.g. (1, 0, 0), (0, but not their sum.
(1,2,3)
GW
Let f,g
(1)
\(af+bg)(x)\
That
GW
but -2(1,
G
V,
(ii)
=
2, 3)
€
W'.
T7,
+
(iv)
Yes.
(v)
No;
(vi)
Yes.
(9,3,0)eW' but 2(9,3,0)
e.g.
=
bg(-x)
+
af(x)
bg(x)
=
(af
+
No. Although one may "identify" the vector (a, b) G R2 with, say, they are distinct elements belonging to distinct, disjoint sets.
4.54.
(i)
-3m + 2v.
4.55.
(i)
u
4.56.
(i)
E = 2A- B +
4.61.
(2,
4.66.
A
4.67.
4.74.
—
-5,
2v
=
k
-8.
(iv)
o
|ff(a;)|
- 36 — 5c =
(a, b, 0)
in the
xy plane
in R3,
0.
Impossible.
(ii)
2C.
Impossible.
(ii)
and C.
Form
A whose rows are the Mj and the matrix have the same row canonical forms.
the matrix
A
(i)
InR2,let
(ii)
InR2,
and
The sum
4.78.
Hint. f(x)
(i)
(iii)
G R, ^ \a\Mf+\b\Mg
for any scalars a, 6
0).
4.77.
4.81.
— w.
that
is
Impossible,
^W.
bg)(x}
4.51.
(ii)
(See Problem 4.45.)
W.
M^ and Mg bounds for / and g respectively. Then = \af(x) + bg(x)\ ^ \af(x)\ + \bg(x)\ = |a| |/(*)| + |6| + \b\Mg is a bound for the function af + bg. af(-x)
K.
w & W}
{(0,w):
V = U®
field
Supplementary Problems
to
1, 0)
W
uGU},
W over a
and
with
+ bg)(-x) =
(af
(ii)
\a\Mf
is,
{(m,0):
W are subspaces of
and
Answers 4.47.
ys.
U =
Show
U
the vector spaces
A
is
B
let
S = S =
{(0,0), (0,1), (0,2), (0,3), {(0,5), (0,6), (0,7),
direct in
=
^(f(x)
(ii)
+
and
f(-x))
.
B
whose rows are the
(ii)
and then show
.}.
.
...}.
(iii).
+
^f(x)
-
f(-x)),
where
^(f(x)
+
odd.
V = U + W.
Uj,
J7nW
is
the space of diagonal matrices.
/(-»))
is
even and ^(f(x)
- f(-x))
chapter 5
and Dimension
Basis INTRODUCTION Some
of the fundamental results proven in this chapter are:
(i)
The "dimension"
(ii)
If
(iii)
A
V
of a vector space
has dimension n over K, then
is
V
well defined (Theorem 5.3).
"isomorphic" to K" (Theorem
is
system of linear equations has a solution if and only if the augmented matrices have the same "rank" (Theorem 5.10).
5.12).
coefficient
and
These concepts and results are nontrivial and answer certain questions raised and investigated by mathematicians of yesterday.
We will begin the chapter with the definition of linear dependence and independence. This concept plays an essential role in the theory of linear algebra and in mathematics in general.
LINEAR DEPENDENCE Definition: Let F be a vector
.,Vm&V are said space over a field Z. The vectors vi, exist scalars if there to be linearly dependent over K, or simply dependent, .,am&K, not all of them 0, such that ai, .
.
.
.
+
aiVi
aiVi
+
•
+ dmVm =
•
(*)
Otherwise, the vectors are said to be linearly independent over K, or simply independent.
Observe that the relation only in this case, that aiVi
+
(*) will
always hold
if
the a's are
If this relation holds
all 0.
is,
a^Vi
+
•
•
+
•
OmVm
-
only
ai
if
=
0,
.
.
.
Om
,
=
then the vectors are linearly independent. On the other hand, if the relation (*) also holds when one of the a's is not 0, then the vectors are linearly dependent. Observe that if is one of the vectors vi, ...,Vm, say vi = 0, then the vectors must be dependent; for Ivi
and the
+
Ov2
+
•
•
+
•
On
coefficient of Vi is not 0.
for independent; *^
,
kv
=
Ot;m
^
0,
=
1
•
+
+
•
•
•
=
+
the other hand, any nonzero vector v
V ¥=,a
t implies •
i.
A;
=
is,
by
itself,
a
Other examples of dependent and independent vectors follow. Example
5.1:
= (1,-1,0), — w = 0,
The vectors m for 3m
+
2v
3(1,
-1,
0)
1;
+
=
(1,3,-1)
2(1, 3,
86
-1)
and
-
w=
(5, 3,
(5, 3,
-2)
=
-2)
are dependent since,
(0, 0, 0)
CHAP.
AND DIMENSION
BASIS
5]
Example
87
We
show that the vectors u = (6, 2, 3, 4), v are independent. For suppose xu + yv + zw
5.2:
w=
+ y{0, 5, -3, 1) + z{0, 0, 7, + 5y, 3x-Sy + Iz, Ax + y- 2z)
-2)
-3,
(0, 5,
where
1)
x,
(0, 0, 7,
-2)
unknown
z are
Then
scalars.
= =
(0, 0, 0, 0)
and
and y and
= =
x{6, 2, 3, 4)
2x
(6a;,
by the equality of the corresponding components,
so,
=0 =0
6a;
+ hy Zx-Zy + 4a; + y — 2x
The first equation yields the third equation with
u,
w
v and
a;
= 0; = 0,
2z
the second equation with x
y
—
+ zw =
xu-'t yv
Accordingly
a;
=
lz
yields 2
Q
implies
=
0.
x
=
=
=
y
yields
and
0;
Thus y
0,
=
0,
z
=
are independent.
Observe that the vectors in the preceding example form a matrix in echelon form:
Thus we have shown that the (nonzero) rows of the above echelon matrix are independent. This result holds true in general; we state it formally as a theorem since it will be frequently used.
Theorem
The nonzero rows
5.1:
For more than one
of a matrix in echelon
form are
linearly independent.
dependence can be defined equivalently as
vector, the concept of
follows:
The vectors Vi, .,Vm are linearly dependent combination of the others. .
For suppose,
and only
if
one of them
+
UrnVm
Then by adding —Vi aiVl
=
+
aiVi
•
•
•
•
+
•
+
we
to both sides,
+
•
Ui-iVi-i
+
tti+iVi +
+
i
•
•
•
—
Oi-iVi-i
Vi
+
Ui + lVi +
i
+
•
•
•
+ amVm —
coefficient of Vi is not 0;
biVi
and so
Vj
Vj is
~
+
•
—bi^biVi
•
•
+
—
bjVj
•••
+
—
•
+ bmVm =
•
•
bf^bj-iVj-i
—
where bi^bj+iVj+i
—
Conversely,
bj ¥-
•••
— bi^bmVm
a linear combination of the other vectors.
We now make a slightly stronger statement than that above; this result has portant consequences. Lemma
5.2:
a linear
obtain
hence the vectors are linearly dependent. suppose the vectors are linearly dependent, say,
Then
is
say, Vi is a linear combination of the others: Vi
where the
if
.
Vi, .,Vm are linearly dependent if and only a linear combination of the preceding vectors:
The nonzero vectors them, say
vt, is
.
Vi
—
.
kiVi
+
kiVi
+
•
•
•
+
fci-iVi-i
many
if
im-
one of
AND DIMENSION
BASIS
88
Remark
The
1.
set 2.
vi,
.
.
.
.
two of the vectors dependent, For
If
and the
Remark
Two
3.
5
.,Vm} is called a dependent or independent set according as the .,Vm are dependent or independent. We also define the emptyto be independent.
set [vi,
vectors
Remark
[CHAP.
coefficient of
Vi,
.
,
i;i
is
- vz, - n
.,Vm are equal, say vi
.
not
a.,
.
,
n.,
then the vectors are
0.
vectors Vi and v^ are dependent
if
and only
if
one of them
is
a multiple of
dependent.
Hence any
the other.
Remark
4.
A
Remark
5.
If the set {vu
which contains a dependent subset subset of an independent set is independent. set
{Vij, Vi^,
Remark
6.
.
.
.
.
.
.
independent, then any rearrangement of the vectors also independent.
Vm}
,
Vi„} is
,
In the real space
itself
is
is
R^ dependence
of vectors can be described geometrically as
follows: any two vectors u and v are dependent if and only if they lie on the same line through the origin; and any three vectors u, v and w are dependent if and only if they lie on the same plane through the origin:
u and V are dependent.
BASIS
u,
V and
w
are dependent.
AND DIMENSION
We begin with a definition. A vector space V Definition: written dim V = which span V.
The above
Theorem
5.3:
said to be of finite dimension n or to be n-dimensional, e„ , n, if there exists linearly independent vectors ei, e2, of V. basis called a is then e„) ... The sequence {ei, 62, is
.
F be
a
.
,
definition of dimension is
Let
.
finite
well defined in view of the following theorem.
Then every basis of
dimensional vector space.
V
has the
same number of elements.
The vector space
the above definition since, vector space is not of finite Example
5.3:
have dimension 0. (In a certain sense this agrees with When a is independent and generates {0}.) by definition, dimension, it is said to be of infinite dimension.
{0} is defined to
Let
K
be any
ments of K.
field.
The vectors 62
m n0, n0, = (1, = (0,1,0,
e„
=
ei
form a
K" which
Consider the vector space
(0,0,0
basis, called the usual basis, of K".
.
.
.,
consists of n-tuples of ele-
n n\ 0, 0)
...,0,0)
0,1)
Thus K" has dimension
n.
CHAP.
AND DIMENSION
BASIS
5]
Example
Let
5.4:
U
be the vector space of
X
0\ 0/'
/I
Vo
(
2
all
89
3 matrices over a field
/O
1
i^o
10
ON
/O
0/'
Vo
Then the matrices
K.
1\ 0/'
10
1
Thus dim C/ = 6. More generally, let V be the vector space of all m X % matrices over K and let E^ S y be the matrix with ly-entry 1 and elsewhere. Then the set {ffy} is a basis, called the usual basis, of V (Problem 5.32); consequently dim V — mn. form a basis of U.
Example
W dim W = n+1. Let
5.5:
is
be the vector space of polynomials (in t) of degree linearly independent and generates W. Thus it
We
—
The
n.
is
a
set {1, t,t^,
basis
of
W
.
.
.,
t"}
and so
V
comment that the vector space of all polynomials is not finite dimensional since (Problem 4.26) no finite set of polynomials generates V.
The above fundamental theorem on dimension portant "replacement lemma":
Lemma
Suppose the
5,4:
is
a consequence of the following im-
set {vi, V2, If {wi, ., Vn} generates a vector space V. ., Wm} linearly independent, then n and V is generated by a set of the form .
.
.
.
m—
is
{Wi, ...,Wm,
Thus, in particular, any %
+1
Vij,
.
.
.
,
Vi^_J
or more vectors in
V
are linearly dependent.
m
set
Observe in the above lemma that we have replaced of the vectors in the generating independent vectors and still retained a generating set.
m
by the
Now suppose S is a subset of a vector space V. dependent subset of S if: (i)
it is
(ii)
{vi,
5.5:
call {vi,
.
.
.,
Vm} a maximal in-
an independent subset of S; and .
.
.,Vm,w}
The following theorem
Theorem
We
is
applies.
Suppose S generates of S.
w e S.
dependent for any
Then
(vi,
.
.
.
,
V
and
Vm}
is
{vi, ., Vm} a basis of V. .
.
is
a maximal independent subset
relationship between the dimension of a vector space and contained in the next theorem.
The main subsets
is
Theorem
5.6:
Let (i)
(ii)
V be
of finite dimension n.
Any Any
linearly independent set is part of a basis,
set of
n + 1 or more
its
independent
Then:
vectors
is
linearly dependent. i.e.
can be extended to
a basis, (iii)
Example
5.6:
A
n elements
linearly independent set with
The four vectors
in
is
a basis.
K* (1,1,1,1), (0,1,1,1),
(0,0,1,1), (0,0,0,1)
are linearly independent since they form a matrix in echelon form. since dim K* = 4, they form a basis of K*.
Example
5.7:
The four vectors
in R3,
(257,-132,58), (43,0,-17),
must be
Furthermore,
linearly dependent since they
(521,-317,94), (328,-512,-731)
come from a vector space of dimension
3.
AND DIMENSION
BASIS
90
[CHAP.
5
DIMENSION AND SUBSPACES The following theorems give
basic relationships between the dimension of a vector space
and the dimension of a subspace.
Theorem
W he a subspace of an n-dimension vector space dim W = n, then W — V. In particular
Let
5.7:
V.
Then dim
W -n.
if
Example
W
be a- subspace of the real space B?. Let can only be ing theorem the dimension of
5.8:
(i)
(ii)
(iii)
(iv)
Theorem
5.8:
Let
U
U+
W
W
W W W dim W
= = —
dim dim dim
and has
U + dim W
dim
Example
V
if
is
5.9:
or
3.
=
3; hence by the precedThe following cases apply:
W = {0}, a point; W a line through the origin;
then
1,
then
2,
then T^
is
a plane through the origin;
S,
then
W
is
the entire space R^.
is
W be
finite-dimensional subspaces of a vector space V.
finite
dimension and
=
+ F)
sum
the direct
(Problem
dim R^
0, 1, 2
0,
dim(C7
Note that
Now
U
of
dim
f/
+ dim
and W,
- dim{UnW)
TF
V = U ® W,
i.e.
Then
dim V =
then
5.48).
U
Suppose
W= dim W —
and
W are the xy plane and yz plane, respectively, in R^:
{(0, 6, c)}. 2.
By
Since RS = t/ + I^, the above theorem,
3
=
2
dim
+ 2-dim(f/nTF)
{U+W) =
Also,
dim{UnW) =
or
Observe that this agrees with the fact that {(0, 6, 0)}, and so has dimension 1.
B.
UnW
is
U=
dim
17
{(a, 6,0)},
=
2
and
1
the y axis,
i.e.
UnW
=
z
w ^Vr\W
^^
^^0 ^,/^ V ..^
r
RANK OF A MATRIX x w matrix over a field K. Recall that the row space of A is Let A be an arbitrary of R^ the subspace of K" generated by its rows, and the column space of A is the subspace of A space column the of and space the row of dimensions The columns. generated by its A. rank of column the rank and row the are called, respectively,
m
Theorem
The row rank and the column rank
5.9:
Definition:
of the matrix
The rank of the matrix A, written rank (A), rank and column rank.
is
the
A
are equal.
common
value of
its
row
Thus the rank of a matrix gives the maximum number of independent rows, and also of a matrix as the maximum number of independent columns. We can obtain the rank follows.
Suppose
row
/I
2
2
6
A =
operations:
\
3
-1\
-3 -3
10—6
—5/
.
We
reduce
A
to echelon
form using the elementary
CHAP.
AND DIMENSION
BASIS
5]
-1 -3 -1 4 -6 -2 2
1
A
2
1
2
to
91
to
2
-1 -3 -1
\o
row equivalent matrices have the same row space. Thus the nonzero rows of the echelon matrix, which are independent by Theorem 5.1, form a basis of the row space of A. Hence that rank of A is 2.
Recall that
APPLICATIONS TO LINEAR EQUATIONS Consider a system of
m linear equations in CziaJi
+ ai2X2 + + a22X2 +
dmlXi
+
ttuXl
am2X2
+
•
n unknowns •
+ ainXn = + ChnXn =
•
•
•
•
•
•
•
a;i,
+
OmnXn
.
.
.
,
«„ over a field K:
bi
^2
—
&m
or the equivalent matrix equation
AX = B where
A—
is
(an)
consisting of the
the coefficient matrix, and
unknowns and of the
matrix of the system
is
1.
The above
(xi)
B=
and
are the column vectors Recall that the augmented
(6i)
defined to be the matrix
{A,B)
Remark
X=
constants, respectively.
an
ai2
a2i
022
ttml
ftm2
.
ttln
bi
.
a2n
62
.
dmn
hm
linear equations are said to be dependent or independent according
as the corresponding vectors,
i.e.
the rows of the augmented matrix, are
dependent or independent.
Remark
2.
Two systems of linear equations
Remark
3.
if and only if the corresponding have the same row space.
are equivalent
augmented matrices are row equivalent,
i.e.
We
can always replace a system of equations by a system of independent The number of independent equations will always be equal to the rank of the augmented matrix.
equations, such as a system in echelon form.
Observe that the above system
is
also equivalent to the vector equation
AX
= B has a solution if and only if the column vector B is a linear combination of the columns of the matrix A, i.e. belongs to the column space of A. This gives us the following basic existence theorem. Thus the system
Theorem
5.10:
The system coefficient
matrix
A
AX — B
has a solution if and only if the and the augmented matrix (A, B) have the same rank.
of linear equations
AND DIMENSION
BASIS
92
[CHAP.
5
AX = B does have a solution, say v, then its + w: wGW} where W is the general solution AX = 0. Now W is a subspace of K" and so has a
Recall (Theorem 2.1) that if the system = {v general solution is of the form v +
W
of the associated homogeneous system
The next theorem, whose proof
dimension.
is
postponed until the next chapter (page 127),
applies.
Theorem
The dimension of the
5.11:
n-r
is equations the rank of the coefficient
In case the system (see
and
page all
AX =
say, Xi^.xi^,
21),
.
other free variables
(Problem 5.43) and so form Example
5.10:
W of the homogeneous system of linear
solution space
AX =
where n matrix A.
number
the
is
unknowns and r
of
is
echelon form, then it has precisely n-r free variables Let Vj be the solution obtained by setting aji^. = 1, .,Xi^_^. Then the solutions Vi, 0. ., v^-r are linearly independent is in
.
=
.
.
a basis for the solution space.
W
Find the dimension and a basis of the solution space equations x + 2y- Az + 3r- s =
+ +
X
2x
2y iy
-2z + 2r+ -2z + 3r +
of the system of linear
=
s
4s
Reduce the system to echelon form: X
+
-
2y
+ 2z — 6z Az
— r + 3r +
3r
= = =
s
2s 68
+
X
4z
2y
and then 2z
+ -
3r
-
s
r
+
2s
There are 5 unknowns and 2 (nonzero) equations in echelon form; hence — 2 = 3. Note that the free variables are y, r and s. Set:
= = dim
W
-
5
(i)
J/
=
1,
r
=
0, s
=
0,
(ii)
=
3/
0,
r
=
s
1,
=
0,
y
(iii)
=
0,
r
=
0, s
=
l
to obtain the following respective solutions: vi
The
=
(-2,1,0,0,0),
=
V2
(-1,
0,
f
vs
1, 0),
set {^i, V2, V3} is a basis of the solution space
= (-3,0,-1,0,1)
W.
COORDINATES be a basis of an n-dimensional vector space V over a field K, and be any vector in V. Since {ei} generates V, 1; is a linear combination of the d: Let
{ei,
.
.
.
,
e„}
+
ttiCi
aiCi
+ dnen,
+
OH
let
v
:K
i.e. the n are independent, such a representation is unique (Problem 5.7), We call {ei}. basis the and v vector the by determined completely scalars ai, ., a„ are coordinate the a„) «-tuple (ai, the call we ., and of in {ei}, v these scalars the coordinates vector of v relative to {ei} and denote it by [v]e or simply [v]:
Since the
d .
.
.
[V]e
Example
5.11
:
Let
V
—
(tti,
ai,
.
.
.,
On)
be the vector space of polynomials with degree
V -
{af2
+
.
5t
+c
a, 6, c
:
e
^
2:
R}
The polynomials ei
=
1,
form a basis for V.
= t-1 and = 2t2-5t + 6. Let v
02
relative to the basis {cj,
e^, 63}.
= (t- 1)2 = i^ - 2t + 1 Find [v\„ the coordinate vector of v
63
CHAP.
AND DIMENSION
BASIS
5]
Set V as a linear combination of the yCi
using the unknowns
e;
-
5«
+
Then
set the coefficients of the
same powers of
—
solution of the above system
=
3ei
Consider the real space the basis
/i
=
-
62
+
t
/a
=
+
y
-2z = -5
is
z
=
a:
6
=
3,
j/
2
=
—1, z
=
[v]^
=
and so
2e3,
(0, 1, 1),
= ^ =
=
/s
Set r as a linear combination of the
(3,1,-4)
+
equal to each other:
—
y
Thus
2.
/;
-1,
(3,
=
Find the coordinate vector of r
tt?.
(1, 1. 1),
xe^
3/
z
V
=
x(\)
X
The
v
z:
+ y(t - 1) + z(f2 - 2« + 1) = + 3/f - + «t2 _ 2zt + z = zfi + (y- 2z)t + (x-y + z) =
6
a;
5.12:
y and
x,
+ zeg. 2«2
Example
93
(3, 1,
-4) relative to
(0, 0, 1).
using the unknowns
+ j/(0, 1, 1) + z(0, 0, + (0, y, y) + (0, 0, 2)
a;(l, 1, 1)
(x, X, x)
2)
x,
y and
v
z:
=
xfi
+
1)
+ y,x-\-y + z)
(x,x
Then
set the corresponding components equal to each other to obtain the equivalent system of equations X
having solution
We
x
=
remark that
y
3,
=
X
+
y
X
-\-
y
-2, z
=
-5.
+
z
= —4 Thus
relative to the usual basis
the coordinate vector of v
(0, 0, 1),
=3 =1
We
is
[v]f
e^
=
=
-2, -5).
(3,
(1, 0, 0),
identical to v itself:
[v]^
ej
=
=
(0, 1, 0),
(3, 1,
-4)
=
eg
=
v.
vGV
have shown above that to each vector there corresponds, relative to a given ., e„}, an n-tuple [v]e in K\ On the other hand, if (ai, .,a„) G j?«, then there exists a vector in V of the form aiCi + + a„e„. Thus the basis {d} determines a one-toone correspondence between the vectors in V and the w-tuples in K". Observe also that if basis
{ei,
.
.
.
•
= w = V
an
ttiei
biBi
+ +
•
•
•
.
•
+ a„e„ + 6„e„
•
•
•
corresponds to
(ai,
.
.
.
corresponds to
(&i,
.
.
.
,
a„)
,
6„)
then t;
+
and, for
=
+ &,)ei + any scalar k G K, i«
(ai
kv
That
=
is,
•
(A;ai)ei
[v
•
•
+
+
•
(a„
•
•
+ 6„)e„
+
{kan)e„
+ w]e = M« +
[w]e
corresponds to
(ai,
corresponds to
and
[kv]e
=
.
.
k{ai,
.
a„)
,
.
.
.
,
+
(bi,
.
.
.
,
b„)
a„)
k[v]e
Thus the above one-to-one correspondence between V and K" preserves the vector space operations of vector addition and scalar multiplication; we then say that V and K" are isomorphic, written
Theorem
5.12:
Let
V ^ K"". We
V
state this result formally.
be an «-dimensional vector space over a
isomorphic.
field
K.
Then
V
and K^ are
AND DIMENSION
BASIS
94
The next example gives a Example
[CHAP.
5
practical application of the above result.
Determine whether the following matrices are dependent or independent:
5.13:
^=a
"^G
I'D-
-4\
3
AJ'
5
6
The coordinate vectors of the above matrices page
5.4,
=
[A]
Form
(1,2,-3,4,0,1),
the matrix
reduce
=
[B]
~"
U6
10
9
relative to the basis in
(1,3,-4,6,5,4),
Example
[C]
=
(3,8,-11,16,10,9)
M whose rows are the above coordinate vectors: =
M to echelon form:
40l\ M
^
89, are
M Row
^
c = f
to
1-1
I
2
5
4
10
/12-3
3
to
6/
1-1
40 2
5
\00000
Since the echelon matrix has only two nonzero rows, the coordinate vectors [A], [B] Accordingly, the [C] generate a space of dimension 2 and so are dependent. original matrices A, B and C are dependent.
and
Solved Problems
LINEAR DEPENDENCE 5.1.
Determine whether or not u and v are linearly dependent
=
(i)
u =
(3,4), V
(ii)
u =
(2,-3), V
(V)'
« =
(1,-3)
=
/l-2
4N
[so
-1/
(6,-9)
(iv)
u = (-4,6,-2), v = (2,-3,1)
M = 2 -
(viii)
u = l-St + 2t^- St\ V = -S + 9t-6t^ +
Two
+
6*2
i3,
V
=
+ 2t-4t^ +
3
vectors u and v are dependent if and only
=
3m. No. (ii) Yes; for v No. (viii) Yes; for v = —3m.
(i)
(vii)
5.2.
u - (4,3,-2), v = (2,-6,7)
2-3N /l (,i), = -5 i) ^6
/2-4 8\ 0-2/ [g -
(iii)
(vii)
5*
(iii)
No.
if:
(iv)
if
^/6-5 [l
2
4
-3
5t^
one
Yes; for
9«'3
is
a multiple of the other.
u =
—2v.
(v)
Yes; for
v
=
2m.
(vi)
No.
Determine whether or not the following vectors in R^ are linearly dependent: (i)
(1,-2,1),
(ii)
(1,
(i)
Method
-3,
(2,
1,-1),
(7,
-4,1)
-6),
(3,
-1, -1),
7), (2, 0, 1.
(2, 4,
-5)
(iii)
(1,2,-3), (1,-3,2), (2,-1,5)
(iv)
(2,
-3,
7), (0, 0, 0), (3,
-1, -4)
Set a linear combination of the vectors equal to the zero vector using unknown
scalars x, y and
z:
x(l,
-2,
1)
+
2/(2, 1,
-1)
+
2(7,
-4,
1)
=
(0, 0, 0)
CHAP.
BASIS
5]
Then
(x,
0^
-2x,
(x
x)
AND DIMENSION
+
(2v, y,
-y)
+
(Iz,
+ 2y + 7z, -2x + y - 4:Z,
95
-4z,
- y + z)
x
=
z)
-
(0, 0, 0)
(0, 0, 0)
Set corresponding components equal to each other to obtain the equivalent homogeneous system, to echelon form:
and reduce
o,+ -2x
2/
+ 7z - 4z =
X-
y
+
X
+
2y
X
.„ z
+
or
2y 5j/
=
-3j/
+
7z
=
+ -
lOz
=
6z
=
X
+
7z
=
y\-2z
=
2y
+
or
The system, in echelon form, has only two nonzero equations in the three unknowns; hence the system has a nonzero solution. Thus the original vectors are linearly dependent. Method
2.
Form
the elementary
the matrix whose rows are the given vectors, and reduce to echelon form using operations:
row
5-3
to
to
Since the echelon matrix has a zero row, the vectors are dependent. generate a space of dimension 2.)
(ii)
Yes, since any four (or more) vectors in R3 are dependent.
(iii)
Form
the matrix whose rows are the given vectors, and
(The three given vectors
row reduce the matrix
to echelon form:
to
Since the echelon matrix has no zero rows, the vectors are independent. (The three given vectors generate a space of dimension 3.)
(iv)
5.3.
Let
V be
A,B,C
(i)
=
Since
(0, 0, 0)
is
one of the vectors, the vectors are dependent.
the vector space of 2 x 2 matrices over R. are dependent where:
GV
Set a linear combination of the matrices A, scalars x, y and z; that is, set xA + yB + zC
C
:)
and 0.
HID- c
/x + y + \
B =
X
z
x x
+ z\ + yj
Determine whether the matrices
C equal to the zero matrix using unknown Thus:
:)
^
_ ~
f^ \0
c ^\ 0)
:)
AND DIMENSION
BASIS
96
[CHAP.
5
Set corresponding entries equal to each other to obtain the equivalent homogeneous system of equations:
+ +
X X
y
+
= =
z
z
=0 =0
X
+
X
y
Solving the above system we obtain only the zero solution, x = 0, y = 0, z = 0. We have shown that xA + yB + zC implies a; = 0, y = 0, z = 0; hence the matrices A,B and C are linearly independent. (ii)
Set a linear combination of the matrices A,B and C equal to the zero vector using scalars x, y and z; that is, set xA + yB + zC = 0. Thus:
2x\
X
/
°^
/3 -1\
2\
'1
x)
\3x
X 3x
z
-5z\
\-Az
/
/ '^
2y)
\2y
+ 3y + + 2y-
-5\
-y\
/Zy '^
1
/
2x-y -
z
+
x
4:Z
/O 1,0
/O
5z\
2y
ON
/O
_ ~
unknown
~
J
\0
Set corresponding entries equal to each other to obtain the equivalent homogeneous system of and reduce to echelon form:
linear equations
X
2x 3x
x
+ + +
3y
y 2y
= 5z -4z =
+ —
X
z
X
+
or
+ +
Sy
y
The system j^
5.4.
= =
2,
y
0, z
z
-
z
=
7z 7z z
= = = =
form has a free variable and hence a nonzero solution, for example, does not imply that x = 0, zC =
in echelon
= -1, z = 1. We have shown that xA + yB + = 0; hence the matrices are linearly dependent.
V
be the vector space of polynomials of degree are independent or dependent where: u,v,w
Let
3y+z
-ny -ly -y —
=0
2y
or finally
X
+
^3
over R.
Determine whether
gV
(i)
u ^ i^-Sf^ +
(ii)
u =
(i)
1^
+ + 4t^-2t +
l,
V
S,
V
Bt
= t^-t^ + St + 2, w = = t^ + 6t^-t + 4, w =
Set a linear combination of the polynomials u,v and scalars x, y and z; that is, set xu + yv + zw
unknown
x(t»
or
or
The
xfi
-
(x
- 3*2 + 5t + l) +
3xt^
+
5xt
+
+ y + 2z)t? +
coefficients of the
x
+
{-3x
powers of
y(t^
yt^
-
- t2 + 8t + 2) + yt^
+
Syt
-y- iz)t^ + t
+
(5*
must each be
2/
Solving the above homogeneous system, hence u, v and w are independent.
we
B1^
9t St
+5 +7
equal to the zero polynomial using Thus:
0.
z{2fi
+
- 4*2 + + St^ -
-U^ + 9t + 5) =
2zt»
-
+ 8y + 9z)t +
izt^ {x
+
9zt
+
=
5z
+ 2y + 5z) =
0:
+ —
+ y 2z 4z —3a; — 5x + 8y + 9z x + 2y + 5z X
2y
w =
2*8
= = =
obtain only the zero solution:
x
=
0,
y
=
0,
z
=
0;
CHAP.
(ii)
AND DIMENSION
BASIS
5]
97
Set a linear combination of the polynomials u,v and scalars x, y and z\ that is, set xu + yv + zw =
unknown
x{ti
or
xt^
or
+
(x
+ 4«2 - 2t + 3) +
-
4a;t2
2xt
+
+ y + 3z)i3 +
+
Zx
(4a;
+
yt»
6yfi
2(3*3
-
+
X
+ + -2x -
y
4a;
62/
+
yt
+
^y
y
+ + -
8z
Ay
+
7z
= = = =
3z
82
or finally
+
a;
3zt^
+
y
2y or
y y
y
+
-
izt^
=
7)
+
Szt
=
7z
+ 4y + 7z) =
{3x
and reduce the system X
3/
+ §(2 _ gt +
-y- 8z)t +
(-2x
each equal to
t
equal to the zero polynomial using Thus:
0.
- f + 4) +
6f2
+ 62/ + Sz)t^ +
Set the coefficients of the powers of
3x
+
3/(t3
w
to echelon form:
= 4:Z -2z = - 2z = + -
3z
+ 82 = -2z =
The system in echelon form has a free variable and hence a nonzero solution. We have shown xu + yv + zw = does not imply that x = 0, y = 0, z - 0; hence the polynomials are
that
linearly dependent.
5.5.
Let V be the vector space of functions from R into R. Show that f,g,hGV are independent where: (i) f{t) = e^, g{t) = t\ h{t) = t; (ii) f{t) = sint, g{t) = cost, h{t)
—
t.
In each case set a linear combination of the functions equal to the zero function scalars x, y and z: xf -\- yg + zh = 0; and then show that a; = 0, j/ = 0, z = 0. xf + yg -\- zh - {) means that, for every value of t, xf{t) + yg(t) + zh(t) = 0.
using unknown emphasize that
We
(i)
In the equation
+ yt^ + zt = 0, substitute t = to obtain xe" + j/0 + zO = * = 1 to obtain xe^ + y =0 z t = 2 to obtain xe* + 4y + 2z =
xe^*
or
=
a;
-\-
=0
X Solve the system
xe^
-
xe*
Hence (Ii)
/,
Method
y Ay
+ +
z
2z
= =
to obtain only the zero solution:
x
=
0,
y
=
0,
z
—
0.
g and h are independent. In the equation
1.
t
= =
t
=
*
a;
sin
t
+
cos
3/
+
«
=
zt
substitute
0,
to obtain
x-Q + y •! + z'Q =
tt/2
to obtain
a; •
TT
to obtain
a; •
V Solve the system
X -y
/,
+ +
= + jrz/2 = + vz =
1
+ +
+
j/ •
Zff/2
+
y(—l)
z
• jr
Q
= =
=
or
y
or
a;
or
—y +
Trz
=
y
to obtain only the zero solution:
x
+
Q
— =
7rzl2
0,
=
0, z
=
0.
Hence
g and h are independent.
Method
Take the
2.
respect to
t
first,
second and third derivatives of
x sin
t
+
2/
cos
t
+
z«
=
with
to get
X cos
—X
—X
t
—
2/
sin
t
+
sin
t
—
y cos
cos
i
+
2/
sin
=
z
(1)
t
=
(2)
t
=
(5)
AND DIMENSION
BASIS
9g
Add
and
(1)
(5) to
obtain
=
z
Multiply
0.
t
X
(2):
—
a;
sin^
t
—
y sin
cos
t
X
(3):
—X
cos^
t
+
y sint cos
-x(sin2
1
by
— cos t
and
x sint
/,
5.6.
Let
y cost
+
+
cos2
sin
—
zt
=
t
=
f)
=
sin2
and then add:
t,
=
cost
or
and then add
t;
by cos
(3)
5
or
=
to obtain
=
y
implies
a;
=
a;
0,
=
j/
0,
a
=
and h are independent.
fir
V and
u,
+
by
(5)
2/(cos2 e
+
t
and
t
sin
Lastly, multiply (2)
Since
by sin
(2)
[CHAP.
w
Show
be independent vectors.
u-v
u + v,
that
and
u-2v + w
are
also independent.
+ w) -
Suppose x(u + v) + y(u -v) + z(u-2v or xu + XV + yu — yv + zu — 2zv + «w =
+ y + z)u+
{x
But
u,
w
V and
— y — 2z)v + zw =
(x
+
=
2
=
x
0,
0:
+ z = -2z =
y
x-y solution to the above system is
Then
y and z are scalars.
x,
are linearly independent; hence the coefficients in the above relation are each
X
The only
where
=
y
-
0, z
+
Hence u
0.
u-v
v,
and
u-2v + w
are
independent.
5.7.
..,Vm be independent vectors, and suppose m is a linear combination of + OmVm where the ai are scalars. Show that the the Vi, say u = aiVi + onVz + above representation of u is unique.
Let
vi, V2,
.
•
Suppose u
=
+
biVi
b2V2
+
•
•
= u—u = But the
Vi
of the
5.8.
a,
+ b„v„
•
(»!
where the
— bi)vi +
(a2
=
61,
02
=
62,
—
bi
...,
=
a^
=
a2—
0,
=
b2
6j
are scalars.
— 62)^2 +
+
(«m
Subtracting,
~
bm)Vm
0,
.. .,
««
~"
*m
0:
=
b^ and so the above representation of m as a linear combination
unique.
Vi is
w
= {l,l+i) in C^ are linearly dependent that the vectors v = (l+i, 2i) and R. over the complex field C but are linearly independent over the real field
Show
of
w
Recall that 2 vectors are dependent iff v is 1, I* can be a multiple of
w
overR. Since and
5.9.
•
each are linearly independent; hence the coefficients in the above relation are ai
Hence
•
1
+tG
C,
Suppose S
S
is
=
(i+i)w
= (l
one (l
is
a multiple of the other. Since the first coordinate But 1 + i € R; hence v and w are independent
+ i)w.
+ i)(l,
1
+ i) =
(1
+ 1,
2i)
=
v
they are dependent over C.
{vi,
contains a dependent subset, say {vi, ...,Vr}. Show that Hence every subset of an independent set is independent.
...,Vm}
also dependent. Since
=
iff
{v^, ...,v^} is
dependent, there exist scalars a^vi
+
a^v^
+
a^, ...,a^,
not
all 0,
such that
CHAP.
Hence there
exist scalars aj
flr.
+
aiVi
S
Accordingly,
5.10.
AND DIMENSION
BASIS
5]
•
.
•
.
.
not
0,
,
+
+
a^Vr
all 0,
such that
O^r+j
+
{vi, .,Vm} is independent, but a linear combination of the i;;.
is
Method
.
•
+ Ov^ =
•
dependent.
is
Suppose
0,
•
99
{vi,
.
Since {vj,
w}
v^,
.
.
.,Vm,w}
Show
dependent.
is
that
w
dependent, there exist scalars «!,..., «„. ^. "o* *11 0, such that = 0, then one of the aj is not zero and Oi^yj + + a^v^ = 0. But this contradicts the hypothesis that {v^, .... i>„} is independent. Accordingly, 6 # and so
+
Oj^x
1.
•
•
+
•
.
.
.
w = That
is,
w
,
+ 6w =
(im'"m
is
If
0.
6
•
a„0
6-i(-aiVi
a linear combination of the
is
•
5.2,
.
of the
.
-
-b-^a^Vi
•
- b-^a^Vm
Vi.
2. If w = 0, then w = Ovi + one of the vectors in {dj, v^, w} vector cannot be one of the v's since {vj,
Method
=
•
•
+
•
is
. ,
.
.
,
.
Ov^. On the other hand, if w ^^ then, by Lemma a linear combination of the preceding vectors. This v^} is independent. Hence w is a linear combination
v^.
PROOFS OF THEOREMS 5.11.
Prove Lemma 5.2: The nonzero vectors Vi, .,Vm are linearly dependent if and only if one of them, say vi, is a linear combination of the preceding vectors: Vi = .
.
+
ait;i
•
•
+
•
Suppose
=
Vj
aj'Ui
+
aiVi
and the 0,
+
•••
+
•
aj-x-yj-i.
ai_iDi_i
-
Vi
Hence the
Vj
are linearly dependent.
+
•
coefficient of vi is not 0.
Then
+
Oi>j+i
+
•
•
•
+
Ov,n
=
Conversely, suppose the Vj are linearly dependent. Then there exist scalars Oi, such that ai^i + + a^Vm = 0. Let k be the largest integer such that o^ ¥= 0. •
+
a^Vi
Suppose fe > 1 and
=
•
•
•
+
•
then
1;
Ofc'Ufc
a^Vi
+ =
Ot)fc
+i
+
•
•
+ Ov^ —
and so
0, a^ ¥=
or
=
v^
0.
a^Vi
But the
+
Vi
•
•
+
•
a^v^
.
.
. ,
a^, not
all
Then
—
are nonzero vectors; hence
k
That
5.12.
a linear combination of the preceding vectors.
is, -y^ is
Prove Theorem
5.1:
The nonzero rows
Ri,
.
.
.,Rn of a matrix in echelon form are
linearly independent.
Suppose {RntRn-i,
,Ri}
dependent. Then one of the rows, say R^,
is
of the preceding rows:
is
a linear combination
"
(*) + <''n^n ^m — "m+l^m + l + «m + 2-'^m + 2 + Now suppose the fcth component of R^ is its first nonzero entry. Then, since the matrix is in echelon form, the fcth components of Rm+u -y^n ^^^ ^^ 0, and so the feth component of (*) is a^+i* + + Ore* = 0. But this contradicts the assumption that the feth component of B„ is "m+2* + •
not
5.13.
0.
•
Thus
Suppose
•
i?i,
{vi,
.
.
.
w GV,
.
.,Rn are independent.
.,
Vm} generates a vector space V.
then {w,
(i)
If
(ii)
If V{ is a linear
vi,
.
.
.,
Vm)
is
Prove:
linearly dependent
and generates V.
combination of the preceding vectors, then
[vi,
.
.
.,Vi-uVi+i,
.
.
.,
Vm]
generates V. (i)
If
we
{w, Vi,
.
y, .
.,
then
w
a linear combination of the Vj since {iij} generates V. Accordingly, dependent. Clearly, w with the Vj generate V since the Vf by themThat is, {w,Vi, .,v^} generates V. is
Vj^} is linearly
selves generate V.
.
.
(ii)
Suppose
=
Vi
bination of the
u =
Thus
{vi,
.
.
k^Vi + say,
Lemma
•
ttiVj
{ai
+
•
+
•
•
•
+
•
1,
•
•
•
•
•
+
•
+
ai_ii;j_i
+ aiki)vi +
..t'i-i, 1^1+
+
a^Vi
•
aj(fciVi
{ai^i
+
•
•
+ ki^^Vi-i) +
•
+ a^k^_l)v^_l +
generates V.
•,^'ot}
[CHAP.
5
Let uGV. Since {Vii generates V, m is a linear com+ a^v^. Substituting for Vj, we obtain
fei_i'yj_i.
u -
Vi,
erating set and
5.14.
AND DIMENSION
BASIS
100
aj+iVj+i
+
a^+iVi + i
•
•
•
+ a^Vm
+ a^Vm
•
we can
In other words,
+
delete Vj
from the gen-
retain a generating set.
still
If [wi, .,Wm} form set of the generated by a is linearly independent, then are or more vectors in V l particular, + any » .,Vi^_^}. Thus, in {wi, .,Wm,vi^, linearly dependent.
Prove
5.4:
Suppose
{vi,
.
.
.,Vn} generates a vector space V.
m — n
.
.
.
V
and
.
.
is
.
prove the theorem in the case that the generates V, we have by the preceding problem that
V;
It suffices to
are
all
not
0.
(Prove!)
Since the {Vj}
{wi, Vi, ..., -yj
{1)
linearly dependent and also generates V. By Lemma 5.2, one of the vectors in (1) is a linear combination of the preceding vectors. This vector cannot be Wj, so it must be one of the v's, say Vj. Thus by the preceding problem we can delete Vj from the generating set {!) and obtain the generating is
Now we
That
repeat the argument with the vector Wj{Wi, W2, Vi,
.
is,
since
(2)
generates V, the set
vJ
..,Vj-i,Vj+i, ...,
(5)
dependent and also generates V. Again by Lemma 5.2, one of the vectors in (3) is a linear combination of the preceding vectors. We emphasize that this vector cannot be Wj or Wj since w^} is independent; hence it must be one of the v's, say v^. Thus by the preceding problem {wi, we can delete v^ from the generating set (3) and obtain the generating set
is linearly
.
.
.
,
{Wi, W2, Vi,
.
.
.,
Uj_i, Hj+i,
We repeat the argument with Wg and so w's and delete one of the y's in the generating
.
set.
Wm.
V
Vfc_l, Vfc +
.,
At each
forth.
set of the required form:
{Wl, ...,
.
If
l,
step
m ^ n,
then
.
.
.,
-vJ
we are able to add one of the we finally obtain a generating
••'"*n-m^
m>n
is not possible. Otherwise, after n of the above steps, we obtain ., w„ which implies that w„+i is a linear combination of Wj, This w„}. ., the generating set {wi, contradicts the hypothesis that {wj} is linearly independent.
Lastly,
we show
that
.
.
5.15.
.
.
Prove Theorem 5.3: Let 7 be a finite dimensional vector space. V has the same number of vectors.
Then every
basis of
Since •) is another basis of V. Suppose {ei,e2, ...,e„} is a basis of V, and suppose {fufi, dependent by the •} must contain n or less vectors, or else it is generates V, the basis {fi.fz, } contains less than n vectors, then preceding problem. On the other hand, if the basis {fiJ^, ej is dependent by the preceding problem. Thus the basis {fiJz, ••} contains exactly n {ej •
{e,}
•
vectors,
5.16.
and so the theorem
is
•
•
true.
Prove Theorem 5.5: Suppose {vi, ...,Vm} is a maximal independent subset of a S which generates a vector space V. Then [vi, ..,Vm} is a basis of V.
set
.
weS. Then, since {vj} is a maximal independent subset of S, {Ui, ...,«„, w} is dependent. By Problem 5.10, w is a linear combination of the i^j, that is, w G L(Vi). Hence linearly is mS C L(Vi). This leads to V = L{S) C L(vi) C V. Accordingly, {vJ generates V and, since it dependent, it is a basis of V. Suppose
CHAP.
^
5.17.
AND DIMENSION
BASIS
5]
V
Suppose
generated by a finite set S. S is a basis of V.
is
Show
101
V
that
of finite dimension and, in
is
particular, a subset of
Method 1. Of all the independent subsets of S, and there is a finite number of them one of them is maximal. By the preceding problem this subset of S is a basis of V.
S
since
is finite,
Method 2. If S is independent, it is a basis of V. If S is dependent, one of the vectors is a linear combination of the preceding vectors. We may delete this vector and still retain a generating set. We continue this process until we obtain a subset which is independent and generates V, i.e. is a basis of V.
^
5.18.
Prove Theorem
V
Let
5,6:
be of
finite
(iii)
A linearly independent set with n elements is a basis. {ei,
(i)
Since
(ii)
Suppose
.
.
e„} is a basis of V.
,
.
{v^,
.
.
.
,
By of (iii)
5.4,
V
is
dependent by
is
(ii),
an independent set T with n elements Thus,
T
Whe a subspace
5.7:
dim W^n. In
particular, if
V
Since
is
W
is
Let
basis
.,v^} as a subset.
But every basis of
part of a basis.
W = n,
dim
V
of an 7i-dimensional vector space V.
contains
Then
W=V.
then
In particular, if {w^, . , w„} is a basis of W, then since = V when dim = n. a basis of V. Thus .
Prove Theorem
.
W
U nW
{^1
and
W respectively.
Note that B has exactly basis of U-\-W. Since
U +W.
7111,
Thus
Uj}
and
...,«„_,}
n elements
V—
5.6(ii),
W
= n and m, dim we can extend {«{}
v„M)i, ..., w„_,}
{vi, ...,
Let
{Vi,
+n—r
{v^,
By Theorem
. ,
.
Vr, Ml,
B -
set with
dim{Ur\W).
Suppose dim
is
.
an independent
W-
+ W) = dim U + dim
dim(C7
5.8:
it is
W
a subspace of both U and W. dim (Ur\W) = r. Suppose {v^, vj is a basis of UnW. to a basis of U and to a basis of W; say,
Observe that
generates
.
W — n.
it is also
U
.
of dimension n, any w + 1 or more vectors are linearly dependent. Furthermore, since consists of linearly independent vectors, it cannot contain more than n elements.
Accordingly, dim
are bases of
n elements and every
a basis.
is
Prove Theorem
a basis of
5.4.
generated by a set of the form
the preceding problem, a subset of S^ is a basis. But S contains contains n elements. Thus S is a basis of V and contains {vi,
elements.
Lemma
{vi, ...,Vr,e,^, ...,ei^_^}
V
By n
or more vectors
By Lemma
independent.
v,.} is
+1
any m
...,«„} generates V,
{ei,
S =
5.20.
Then:
n.
(ii)
Suppose
5.19.
dimension
Any set of n + 1 or more vectors is linearly dependent. Any linearly independent set is part of a basis,
(i)
...,V„
Ml,
elements.
U
generates
it suffices to
.
..,«„_„ Wi,
.
.
.,
W„-r}
Thus the theorem is proved if we can show that B is a and {v^, w^) generates W, the union B = {vj, Uj, w^}
show that
B
is
independent.
Suppose aiVi
where
Oj, bj,
+
•
•
•
+
a^v^
c^ stte scalars.
V
+
61M1
+
•
•
+
6„_,.M„_,.
+
Ci^i
+
•
•
•
+
c„_,.m;„_,.
=
(1)
Let
=
aiVi
+
•
•
•
+
a^Vr
+
61M1
+
•
•
•
+ bn-^u^_^
(2)
AND DIMENSION
BASIS
102
By
(1),
we
[CHAP.
have that
also
^n — r *^n — r
Since {vi,Uj} cU, v e Now {Vi) is a basis of
Thus by
(^)
U
hy
UnW
c„_, = 0.
+
= 0,
j,
Uj} is a basis of
61
v
TF,
•
+
•
d^V^
+
+
CiWi
•
W
+
aiVi
a,
and since
so there exist scalars di,
W
€: .
. ,
.
by
d,
vGUnW.
Accordingly,
(5).
which v
fo""
=
dj^i
+
= 0,
.
.
6to_^
,
.
U
•
•
•
and so
+
a^Vr
+
+
bjUi
•
•
•
•
+
d^v^.
=
C„_rW„_r
+
•
Cx
= 0,
.
.
«!
= 0,
.
.
.
—
b^n-rUm-r
Hence the above equation forces
independent.
is
+
•
Hence the above equation forces
and so is independent. a basis of Substituting this into (1), we obtain
{vi, Wfc} is
But {f
{w J c
(2);
and
(S)
we have diVi
But
5
.
= 0.
Since the equation {1) implies that the is proved.
and
aj, 6^
c^ are all 0,
B=
{vi, Uj,
w^}
is
independent
and the theorem
5.21.
Prove Theorem Let
A
The row rank and the column rank
5.9:
_
A
jail
ai2
<*21
"'22
•
•
•
<'m2
•
•
•
denote
its
i?j
Suppose the row rank Si
=
is
=
fcy
each other,
Thus for
i
'^r
rows: (fflu,
O12,
•
•
• .
•
•
^m —
• >
(<*ml> '*m2>
(611, 612,
•
• ,
is
f>i„),
S2
-
(621. 622.
•
•
•
>
^2n).
•
a linear combination of the
R^
^^
^11*^1
"" '^i2'^2
~r
/?2
—
'l'21'S'l
^" fe22'^2
+
fin»
~
"'ml"'l
"^"
+
"'m2"2
'
'
'
'
"
"
•
•
.
^r
•
•
>
'*mn)
row
=
i^rV ^r2.
•
•
• >
space:
*m)
S;:
"T"
/CjjOj.
+
^2r"r
+
l^mr^r
are scalars. Setting the ith components of each of the above vector equations equal to w: , obtain the following system of equations, each valid for i = 1,
we
=
•
r and that the following r vectors form a basis for the
Then each of the row vectors
where the
«in ^2n
I
[O'ml
Bm
equal.
m X w matrix:
be an arbitrary
Let Ri, B2
any matrix are
of
1,
.
.
.
. ,
^12621
+
•
•
+
^Ir^ri
+
*'22&2i
+
•
•
+
k2rbri
+
fcm2*2i
+
'
"
+
k^rbr
«li
=
^^ll^li
+
ttgi
=
fe21*li
«m«
—
^mlbii
'
.
w:
In other words, each of the columns of
A
is
a linear combination of the r vectors
I'l
.
CHAP.
AND DIMENSION
BASIS
5]
Thus the column space of the matrix — row rank.
A
103
has dimension at most
r, i.e.
column rank
—
Hence, column
r.
rank
Similarly (or considering the transpose matrix A*)
row rank and column rank are
the
obtain
row rank — column rank.
Thus
AND DIMENSION
BASIS 5.22.
we
equal.
Determine whether or not the following form a basis for the vector space (i)
(1, 1, 1)
and
(1,
(1, 2, 3), (1, 0,
(ii)
and
(2, 1,
and
(ii).
(i)
-1,
-1),
5) (3,
-1,
0)
(iii)
(1, 1, 1),
(1, 2, 3)
and
(2,
(iv)
(1, 1, 2),
(1, 2, 5)
and
(5, 3, 4)
1)
-2)
No; for a basis of R3 must contain exactly 3 elements, since R^
The vectors form a basis if and only if they are independent. rows are the given vectors, and row reduce to echelon form:
(iii)
-1,
R^:
is
of dimension
3.
Thus form the matrix whose
to
The echelon matrix has no zero rows; hence the three vectors are independent and
so
form a
basis for R^.
Form
(iv)
the matrix whose rows are the given vectors, and
row reduce
to echelon form:
to
The echelon matrix has a zero row, i.e. only two nonzero rows; hence the three vectors are dependent and so do not form a basis for R3.
5.23.
Let
W be the subspace of R* generated by the vectors
(1,
—3, —5). (i) Find a basis and the dimension of W. to a basis of the whole space R*. (3, 8,
(i)
Form
/l 1
-2
2
-18
4
14
The nonzero rows space, that (ii)
is,
of
(1,
—2,
5,
W. Thus,
—3) and
3\ 3
5
-9 7-9 7
(0, 7,
in particular,
—9,
2)
dim
-3), (2, 3, 1, —4) and Extend the basis of
5,
W
(ii)
row reduce
the matrix whose rows are the given vectors, and
to
—2,
to echelon form:
1
-2
5
-3
7
-9
2
to
of the echelon matrix
W=
form a basis of the row
2.
We seek four
independent vectors which include the above two vectors. The vectors (1, —2, 5, —3), —9, 2), (0, 0, 1, 0) and (0, 0, 0, 1) are independent (since they form an echelon matrix), and so they form a basis of R* which is an extension of the basis of W.
(0, 7,
5.24.
Let
W be the space generated by the polynomials V2
=
2«3 -st^'
+ Qt-l
Vi
=
2t^
~5P + 7t +
5
Find a basis and the dimension of W. The coordinate vectors of the given polynomials [vi]
[v^]
= =
(1,-2,4,1)
(2,-3,9,-1)
relative to the basis {f3,
K] = K] =
(1,0,6,-5) (2,-5,7,5)
t^, t,
1} are respectively
AND DIMENSION
BASIS
104
Form
[CHAP.
row reduce
the matrix whose rows are the above coordinate vectors, and 1
to
„
I
-2
4
1
1
1
-3
to
_
5
to echelon form:
The nonzero rows (1, —2, 4, 1) and (0, 1, 1, —3) of the echelon matrix form a basis of the space generated by the coordinate vectors, and so the corresponding polynomials
_
«3
W=
form a basis of W. Thus dim
5.25.
+
2*2
+
4t
and
1
+
t^
-
t
3
2.
W of the system
Find the dimension and a basis of the solution space x
+ 2y + 2z-
s
+
St
=
x
+ 2y +
3z
+
s
+
i
=
+
8z
+
s
+
5t
=
+
Sx
6y
Reduce the system to echelon form:
+
X
+
2y ^
2z z
2z
The system
in echelon
solution space PT is
y
(i)
+ +
s
2s 4s
+ 3t = -2t = - 4« =
+
x
+ 2zz +
2y
or
s
2s
+ -
3t 2t
= =
form has 2 (nonzero) equations in 5 unknowns; hence the dimension of the 3. The free variables are y, s and t. Set
—2 =
5
= l,s = 0,t-0,
(ii)
=
J/
0, s
=
=
t
1,
y
(iii)
0,
=
0,
s^O, t-l
to obtain the respective solutions vi
The
5.26.
=
(-2,
=
V2
1, 0, 0, 0),
-2,
(5, 0,
Find a homogeneous system whose solution -2,
0, 3), (1,
set
-1, -1,
W
M
„ _
1
-2 -1 -1
4
-2
/I
3\
r
-1 -2
2
2x
\Q
y
\x
3
1
IO-25II0
I
+
first
0, 2, 0, 1)
-2, 5)}
rows are the given vectors and whose
/I
\
,.0
1
-2
3
-1
1
1
\^\(iQ2x + y + z-hx-y + w^
2
-3x + w/
z
y
(-7,
generated by
is
4), (1, 0,
whose Method 1. Let v = (x, y, z, w). Form the matrix last row is v; and then row reduce to echelon form: /I
=
W.
set {v^, V2, f 3} is a basis of the solution space
{(1,
^3
1, 0),
\0
v&WH
W
and only if the addihas dimension 2. Thus original first three rows show that row does not increase the dimension of the row space. Hence we set the last two entries to obtain the required homogeneous system in the third row on the right equal to
The
tional
Method
2.
erators of
We know
=
that v
W: (x, y, z,
The above vector equation
w)
in
2x
+
y
5x
+
y
(x,y,z,w)
=
r(l,
unknowns
G
-2, r,
+
P7 0, 3)
s
and
=0
z
—w = if
+
and only 8(1,
t is
if
-1, -1,
d
is ,
4)
+
a linear combination of the gen-
„
„
t(l, 0,
„
-2,
,.v
5)
equivalent to the following system:
CHAP.
AND DIMENSION
BASIS
5]
= X = y -2r -s -s-2t = z 3r + 4s + 5t = w r
+
+
s
Thus V
W
E.
ii
+
°^
and only
if
= = 2x + y = z = w — 3x
+ t 8 + 2t -s-2t 8 + 2t
r
t
8
5x
y
y
U
W be the following subspaces
and
+c+d =
{{a,b,c,d): b
Find the dimension and a basis of
We
2x 5x
+ + +
y y y
+z —w
^^^
i.e. if
the required homogeneous system.
is
U =
(i)
x 2x
+ z =0 —w =
Remark: Observe that the augrmented matrix of the system
Let
= = = =
t
2t
°^
b
(1)
a
=
=
c
1,
+
0,
+
c
and
a, c
d
=
W
U,
seek a basis of the set of solutions
The free variables are
of R*:
0},
(i)
=
W,
(ii)
or
{{a,b,c,d): a
+b =
=
0, c
2d}
UnW.
(iii)
of the equation
(a, 6, c, d)
=
d
the transpose of the matrix
(1) is
M used in the first method. 5.27.
+ +
s 8
the above system has a solution,
+ +
+
r
oe
2x
The above
105
0-a +
b
=
=
+
c
+
=
d
Set
d. 0,
=
a
(2)
0, c
1,
d
0,
=
a
(3)
0, c
=
0,
d
=
1
to obtain the respective solutions
=
^•l
The (ii)
We
set {v^, v^,
v^
seek a basis of the set of solutions
+
a
c
The
free variables are 6 and d.
(a, 6, c, d)
= =
6
(0,-1,1,0),
dim U =
a basis of U, and
is
=
V2
(1,0,0,0),
=
1^3
(0,-1,0,1)
3.
of the system
+ 6 = - 2d =
a or
2d
c
Set
=
6
(i)
d
=
0,
(2)
6
1, 0, 0),
V2
=
1,
=
0,
=
d
1
to obtain the respective solutions
=
Vy
The (iii)
set {Ui,
U r^W
v^
is
W, and dim
a basis of
consists of those vectors
ditions defining
W,
i.e.
(-1,
W=
(a, b, c, d)
2.
which satisfy the conditions defining
U
and the con-
the three equations
6+c+d=0 =
c
The free variable is d. Set d = 1 of VnW, and Aim (V nW) = \.
=0
a+b
=0
0+6
5.28.
(0, 0, 2, 1)
or
6
+
v
=
c+d
2d
c
to obtain the solution
-
2d
= =
(3, -3, 2, 1). Thus {v}
is
Find the dimension of the vector space spanned by: (i)
(ii)
(1, (3,
-2,3, -1) and -6,
+
3,
-9) and (-2,
+
(iii)
t^
2f2
+
3i
(iv)
i»-2f2
+
5 and
1
-2, 3)
(1, 1,
and i^
+
4,
2t^
34
-2,
+
_
4
(v)
6)
4*^
+
/ 6f
+
2
and
{\
^J 1
l\
^^^^
(^-i -i^
(vii)
3 and
-3
^
/-3 -3
a basis
AND DIMENSION
BASIS
106
[CHAP.
5
W
of dimension 2 if they are independent, and of dimension nonzero vectors span a space they are dependent. Recall that two vectors are dependent if and only if one is a multiple of the other. Hence: (i) 2, (ii) 1, (iii) 1, (iv) 2, (v) 2, (vi) 1, (vii) 1.
Two
1 if
5.29.
V
Let
Show that be the vector space of 2 by 2 symmetric matrices over K. 3. (Recall that A = (ay) is symmetric iff A = A* or, equivalently, an = an.)
dim 7 =
arbitrary 2 by 2 symmetric matrix is of the form (Note that there are three "variables".) Setting (i)
we
=
a
1,
=
6
=
a
(ii)
0,
6
0,
-. =
= (;:)•
show that {E^, E^, E^) (1)
is
a basis of V, that
A
For the above arbitrary matrix
A =
=
c
1,
where
1
a, 6, c
G A.
^
=
a
(iii)
0,
6
0,
=
0, c
=
l
is,
^\
(2)
Suppose xEi
Ki
=
o)
+
+
yE^
+ ^(i +
xEi Accordingly, {EijEz.Ea}
=
zE^
where
0,
K2
+
bE2
V
and
(2)
is
independent.
is,
suppose
cEs
is
+
yE2
{1,
t^
«,
.
.
.
,
scalars.
That
Oj
zEs
=
we
=
x
=
x
obtain
implies
zj
\y 0,
y
0,
=
y
=
0,
\0
0, z
z
=
0-
In other words,
=
independent.
«"->, «»},
Thus dim V = n +
unknown
x, y, z are
1)
V
and so the dimension of
Let V be the space of polynomials in is a basis of V:
of degree
- 1,
{1, 1
(ii)
t
(1
- tf,
V
is 3.
^ n. Show that each .
.
.
,
(1
- «)"-s
(1
of the following
- «)"}.
1.
...,t"-i and t». Furthermore, Clearly each polynomial in y is a linear combination of l,t, none is a linear combination of the preceding poly1, t, ., t"-i and t» are independent since t«} is a basis of V. nomials. Thus {1, t .
(ii)
(::)
generates
(1)
-
+
J)
{El, E2, E3} is a basis of
Thus
(i)
it
+
aEi
Setting corresponding entries equal to each other,
(i)
that
we have
in V,
("'
-• =
(?;)
{^1, E^, Eg} generates V.
Thus
5.30.
=
(
obtain the respective matrices
-.
We
=
0, c
h\
a
I
A =
An
.
form a basis of (Note that by (i), dim V = w+ 1; and so any m-l- 1 independent polynomials (1 - t)» is of degree higher than the Now each polynomial in the sequence 1, 1 - t w + 1 polypreceding ones and so is not a linear combination of the preceding ones. Thus the (1 — t)» are independent and so form a basis of V. nomials 1, 1 — t, V.)
.
5.31.
.
. ,
the vector space of ordered pairs of complex numbers over the real (see Problem 4.42), Show that V is of dimension 4.
Let
V be
We
claim that the following
is
Suppose u e V. Then v = («, w) where o, 6, c, d are real numbers. Then V
Thus
B
=
R
a basis of V:
B =
generates V.
field
{(1, 0), z,
ail, 0)
w
+
(i,
0), (0, 1), (0, t)}
are complex numbers, and so v
6(t, 0)
+
c(0, 1)
+
d(0,
t)
= (a+bi,e + di)
where
+
CHAP.
BASIS
5]
The proof
complete
is
if
we show
+
xi(l, 0)
where
Xi, x^, x^,
X4
5.32.
x-^
=
B
is
X2(i, 0)
+
that
107
Suppose
independent.
+
a;3(0, 1)
=
0:4(0, i)
G R. Then {xi
Accordingly
AND DIMENSION
0, a;2
=
+ x^,
Xs
=
as
0,
Let Y be the vector space of with 1 as the i^-entry and
+ x^x) = 0, a;4
=
Xx
-'r
x^i
«3
-V
x^i
and so
(0, 0)
and so
B
is
Q
independent.
m x w matrices over a field K. Show
elsewhere.
= —
Let E^ G V be the matrix Thus is a basis of 7.
that {E-,^
dimV = mn. We
need to show that {By} generates
A=
Let
Now
(tty)
Thus {By}
is 0.
is
and
2
=
ajjii^ij
Thus
asy
=
where the 0,
i
=
1,
.
independent.
is
A = 2
be any matrix in V. Then
suppose that
the y-entry of independent.
V
.
.,
«{.-
w,
Hence {By} generates V.
o,^E^^.
The y-entry of
are scalars. j
=
m.
1
2 «ij^ij
is ««.
and
By
are
Accordingly the matrices
a basis of V.
Remark: Viewing a vector in K» as a 1 X w matrix, we have shown by the above result that the usual basis defined in Example 5.3, page 88, is a basis of X" and that dim K" = w.
SUMS AND INTERSECTIONS 5.33.
Suppose sion 6.
W
are distinct 4-dimensional subspaces of a vector space and Find the possible dimensions of TJV^W. TJ
Y
of dimen-
W
are distinct, V -VW properly contains 17 and W; hence dim(f7+W)>4. Since V and But dim(?7+W) cannot be greater than 6, since dimV = 6. Hence we have two possibilities: = 5, or (ii) dim (U + PF) = 6. Using Theorem 5.8 that dim(f7+ T^) = dim U (i) dim(U+T7) — dim (Un TF), we obtain dim
W
That
5.34.
Let
is,
J]
5
=
4
+
4
-dim(f/nW)
or
dim(t7nW)
=
3
(ii)
6
=
4
+
4
-dim(?7nW)
or
dim(t/nTF)
=
2
the dimension of
and
{(1, 1, 0,
TJ
r\'W must be either 2 or
3.
W be the subspaces of R* generated by
-1),
respectively. (i)
(i)
(1, 2, 8, 0), (2, 3, 3,
Find
(i)
-1)}
dim (C/ +
TF),
and
{(1, 2, 2,
(ii)
dim(C7nW).
-2),
(2, 3, 2,
-3),
(1, 3, 4,
-3)}
TJ-^W is the space spanned by all six vectors. Hence form the matrix whose rows are the given six vectors, and then row reduce to echelon form:
to
to
to
Since the echelon matrix has three nonzero rows, dim
iJJ
-VW)
—
Z.
AND DIMENSION
BASIS
108
(ii)
[CHAP.
First find dim U and dim W. Form the two matrices whose rows are the generators of respectively and then row reduce each to echelon form:
U
5
and
W
1 1
-1
1
2
2
3
1
3
to
3
-1
-2 -3 -3
-1
1
1
to
3
1
1 0.
and 2
2
2
3
2
1
3
4
to
Since each of the echelon matrices has two nonzero rows, dim V - dim (UnW), 5.8 that dim (V +W) = dim U + dim
5.35.
Let
U
2
+2-dim(!7nW)
let
-2,
-3,
2, 3), (1, 4,
2
and dim
Aim{Ur\W) =
or
W=
2.
Using
we have 1
-1, -2, 9)}
4, 2), (2, 3,
W be the subspace generated by {(1, 3, 0, 2, 1), (1, 5,
Find a basis and the dimension of (i)
1
be the subspace of R^ generated by {(1, 3,
and
—
W
=
-2
2
to
Theorem
3
2
-1 -2
'l
U +W
is
XJ
(i)
-6,
6, 3), (2, 5, 3, 2, 1)}
+ W,
f/n W.
(ii)
Hence form the matrix whose rows are the
the space generated by all six vectors. and then row reduce to echelon form:
six vectors
1
4
2
3
-2 2 -3 4 -1 -2
1
3
2
1
6
3
1
3
1
5
-6
9 *° '
2-440
1
\0
-1
2
3
1
3
2
-1
2
5
3
2
1
3
-2 -1
1
2
3-2 2 3\ 1-1 2-1 0-3 3-6 0-2 2
/I
3
1
7
-2 -1
-2 -5/ 2
3
2
-1 -2
2 to
2
The
set of
(ii)
a basis
-2
2
6
-6
nonzero rows of the echelon matrix, {(1, 3,
is
to
-2
oiV+W;
-2,
thus dim
(t/
2, 3), (0, 1,
+
TF)
=
-1,
2,
-1),
(0, 0, 2, 0,
-2)}
3.
W respectively.
First find homogeneous systems whose solution sets are U and first rows are the generators of U and whose last row is
whose
(», y, z, s, t)
Form the matrix and then row reduce
to echelon form:
-3x
4a;
+
y
-2 -1
2
3
2
-1
3
-6
2x
+
z
-2x +
3 s
Sx +
t j
CHAP.
AND DIMENSION
BASIS
5]
Set the entries of the third row equal to set is U:
-X + y +
Now
=
z
4a;
Q,
109
homogeneous system whose solution
to obtain the
-
22/
+
=
8
+
-6a;
0,
y
-\-
t
W
form the matrix whose first rows are the generators of and then row reduce to echelon form:
=
and whose
last
row
is
(x, y, z, 8, t)
to
-9aj
+
row equal
Set the entries of the third
—9x + 3y +
=
z
+
3y
z
4x
—
+
2y
—
2y
+
=
s
2x
s
to obtain the
to
0,
'Ix
—
+
y
t
homogeneous system whose solution 2x
0,
-
y
+
=
t
Combining both systems, we obtain the homogeneous system whose solution
=0 =0
—x+y + z -2y -6a; + y -9x + 3y + 4a; — 2j/ 2a; -
+8
4x
+ +
+
+
4z 8z 4z
+ + +
There
is
+ -5y -6y 2y + ^ + 2y
= =
=0 =0
8
+
58
2t
-x
+
-
-
2t
Az
+
s
+
6z
t
iz
+
s
+
Az 8z
— =
t
=
2f
= = =
+
2z
U nW:
=0 =0
8z
=0
y+z
2y
=0
3s s
solution
t
y+z
2y
—x+y+z
=0
s
J/
—x +
t
z
set is
+ +
8
5s 8
=
+ -
2t
one free variable, which is t; hence dim(l7nT^ = 1. Setting t = 2, we obtain the Thus {(1, 4, -3, 4, 2)} is a basis of UnW. a; = 1, 2/ = 4, z = -3, 8 = 4, t = 2.
COORDINATE VECTORS 5^6.
Find the coordinate vector of v relative to the basis where (i) v = (4, -3, 2), (ii) v = (a, 6, c).
{(1, 1, 1), (1, 1, 0), (1, 0, 0)}
In each case set v aa a linear combination of the basis vectors using
=
V
a;(l, 1, 1)
and then solve for the solution vector
+
{x,y,z).
j/(l, 1, 0)
+
unknown
of R^
scalars x, y and
z:
z(l, 0, 0)
(The solution
is
unique since the basis vectors are
linearly independent.)
(i)
(4,-3,2)
= = =
+ j/(l, 1, 0) + z(l, 0, X, x) + {y, y, 0) + (z, 0, 0) (x + y + z,x + y,x)
a;(l, 1, 1)
0)
(a;,
Set corresponding components equal to each other to obtain the system
x
+
y
+
z
=
A,
X + y
=
—3,
a;
=
2
Substitute « = 2 into the second equation to obtain y = —5; then put x = 2, y = —5 into the first equation to obtain z = 7. Thus x = 2, y = -5, z = 7 is the unique solution to the system and so the coordinate vector of v relative to the given basis is [v] = (2, —5, 7).
-
{a, b, c)
(ii)
+
«(1, 1, 1)
Then
=
from which x {e, b — c, a — b).
5JS7.
AND DIMENSION
BASIS
110
y
c,
[CHAP.
+ z{l, 0, 0) = (x + y + z,x + y,x) x + y + z = a, x + y = b, x — c — b — c, z = a—b. Thus [v] — (c,b — c,a—
5
1/(1, 1, 0)
Let V be the vector space of 2 x 2 matrices over R. matrix A relative to the basis
that
b),
—
[(a, b, c)]
is,
Find the coordinate vector of the
GV
iHri)^{i-iHi
{{I
A
Set
w- -(!-?
I)}
as a linear combination of the matrices in the basis using
scalars x, y,
"I i)*'(:-i)-'(i-i)*
J)
C
-
unknown
z,
w:
:
w '
'^
\y
+w
X
x)
\X X
+
z
X
+
—
y
0/
\0
0/
—
z^
X
y
Set corresponding entries equal to each other to obtain the system
x — —1 x + y =^ A, + w = 2, X — y — z = 3, = = -21, (Note ll, -21, w 30. Thus [A] 30). -l, y z(-7, 11, of A must be a vector in R* since dim V = 4.)
X
from which
x
ordinate vector
5^8.
Let
+
z
=
W be the vector space of 2 x 2
symmetric matrices over R.
^ —
Find the coordinate vector of the matrix 1
-2\
-2
/2
1/' vi
A
Set
„
, ^
[
4 -1\1
/
sy
\-i -5/j
Problem
(See
relative to the basis )
•
as a linear combination of the matrices in the basis using unknown scalars 4
-11\
^ = (_n
-7J
/
/I %-2
=
-2\ ij +
5.29.)
^
^
1\
that the co-
/2
1\
^l
3J
,
,
+
/
4
%-l
x,
y and
z:
/x + 2y + 4:Z -2x + y - z -1\ -5) = (,-2x + y-. . + 3,-5.
Set corresponding entries equal to each other to obtain the equivalent system of linear equations and reduce to echelon form:
+
Az
—2x + y -
z
-2x + y -
z
X
X
We
5.39.
—2,
Let
+
2y
Sy
1).
-
5z
= 4 — -11 = -11 = -7
X
2y 5y J/
+ + -
iz 7z
9z
= 4 - -3 = -11
X
+
2y 5y
or
+ +
iz Iz
52z
- 4 - -3 - 52
=
1
first equation.
{ei, 62, es}
W
and
{/i,
h,
fa}
be bases of a vector space ei
62 63
Let
+
from the third equation, then y = —2 from the second equation, and then a = 4 Thus the solution of the system is x = 4, y — —2, z — 1; hence [A] — = 3 by Problem 5.29, the coordinate vector of A must be a vector in K*.) (Since dim
obtain «
from the (4,
+
= — =
Ci/i
+
ttaA
hifi+b^fi Ci/i
+
C2/2
V
(of
+ 0-3/3 + hifz + C3/3
P be the matrix whose rows are the coordinate vectors of
relative to the basis
{fi}:
dimension
3).
Suppose
(i)
ei, ea
and
es respectively,
'
,
CHAP.
AND DIMENSION
BASIS
5]
{tti
02
as
61
&2
63
Cl
C2
C3
111
\
^
Show that, for any vector v GV, [v]eP = [vy. That is, multiplying the coordinate vector of v relative to the basis {ei} by the matrix P, we obtain the coordinate vector of V relative to the basis {/<}. (The matrix P is frequently called the change of basis matrix,) Suppose V V
=
+
rei
=
seg
then
teg;
[v]^
=
Using
(r,8,t).
we have
(i),
+ O2/2 + agfa) + si^ifi + ^2/2 + ^3/3) + *(«i/i + "2/2 + "3/3) = (roi + s6i + tci)/i + {ra2 + S62 + fc2)/2 + {ras + sb^ + tcs)^ r(aJi
Hence
[v]f
On
+
—
{rai
+ sbi + tc^,
the other hand,
=
a2
61
62
63
Cl
C2
C3
+ s6i + tCj,
ra2
(r, s, t) \
= —
Accordingly, [v\eP
(rtti
+ sb2+
„ ai
/
[v],P
ra2
tC2,
ra3
+ sb3+
tcg)
(I3
+ 862 + tc2,
ra^
+ S63 + tcg)
['"]/•
Remark: In Chapter 8 we Then, by above,
column vectors rather than row
shall write coordinate vectors as
vectors.
'ax
=
QMe
a2 I
Ci\/r\
61
62
«2l|S
&3
Cg/
/
rai
ra2
\ t
irfflg
+ + +
+ 362 + S&3 + s6i
tCx
f
*C2
tCgy
H
Q the matrix whose columns are the coordinate vectors of ©i, e^ ^^^d respectively, relative to the basis {/J. Note that Q is the transpose of P and that Q appears on the left of the column vector \v\g whereas P appears on the right of the row vector [1;]^. where
is
RANK OF A MATRIX 5.40.
Find the rank of the matrix
-2 -1 4 3 3 -4 -7 8 1 -7
/I (i)
A =
3
u l^ \3
(i)
Row
1
A
where:
-3 -4 -3 -8
(ii)
A =
(iii)
reduce to echelon form: /I
3
1-2
1
4
3-1-4
2
3
3
8
-3\
-4 -7 -3 1-7 -8/
to
to
Since the echelon matrix has two nonzero rows, rank (A) (ii)
A
Since row rank equals column rank, reduce to echelon form:
to
Thus rank (A)
=
3.
it is
easier to
=
2
form the transpose of
to
A
and then row
The two columns are linearly independent since one rank (A) = 2.
(iii)
5.41.
AND DIMENSION
BASIS
112
[CHAP.
is
not a multiple of the other.
AB
Let A and B be arbitrary matrices for which the product rank (AB)^ rank (J?) and rank (AB) ^ rank (A).
is
defined.
5
Hence
Show
that
By Problem 4.33, page 80, the row space of AB is contained in the row space of B; hence rank (AB) — rank (B). Furthermore, by Problem 4.71, page 84, the column space of AB is contained in the column space of A; hence rank (AB) — rank (A).
5.42.
A
Let
be an n-square matrix.
Show
that
A
and only
invertible if
is
rank (A)
if
= n.
Note that the rows of the w-square identity matrix /„ are linearly independent since /„ is in echelon form; hence rank (/„) = n. Now if A is invertible then, by Problem 3.36, page 57, A is row equivalent to /„; hence rank (A) = n. But if A is not invertible then A is row equivalent to a matrix with a zero row; hence rank (A) < n. That is, A is invertible if and only if rank (A) = n.
5.43.
Let JCij, Xi^, a;i^ be the free variables of a homogeneous system of linear equations with n unknowns. Let Vj be the solution for which: x^ — 1, and all other free variables = 0. Show that the solutions Vi, V2, .,Vk are linearly independent. .
.
.
,
.
ii,
.
Let A be the matrix whose rows are the Vi respectively. We interchange column 1 and column then column 2 and column ^2, ., and then column k and column i^l and obtain the kXn matrix .
.
1
=
(/,
C)
.
\o
"l.k +
.
.
1
''2,
.
1
...
k
Cfc,fc
Ci„
l
+
+i
•
...
C2„
c„„/
The above matrix B is in echelon form and so its rows are independent; hence rank (B) = k. Since and B are column equivalent, they have the same rank, i.e. rank (A) = k. But A has k rows;
A
hence these rows,
i.e.
the
iij,
are linearly independent as claimed.
MISCELLANEOUS PROBLEMS 5.44.
The concept
of linear dependence is extended to every set of vectors, finite or infinite, the set of vectors A = {vi} is linearly dependent iflf there exist vectors a„ S Z, not all of them 0, such that .,Vi^G A and scalars ai,
as follows: Vi^,
.
.
.
.
aiVi^
.
+
,
aiVi^
+
•
•
+
a„i7i„
=
are linearly Otherwise A is said to be linearly independent. Suppose that Ai, A2, independent sets of vectors, and that AiCAzC---. Show that the union A = A1UA2U is also linearly independent. .
•
o„
G
•
a^Vi
A = uAj
and the
Wj
S A,
+
a2V2
+
•
•
•
maximum
that each Aj.
is
index of the sets
v^& A and
Thus
.
.
.,
A;^ such that
v„eAi^
...,
A
linearly independent.
is
.
•
«!,...,
(1)
= max (ij, -.-yn 6 -^k
k
scalars
=
a^Vn
•Ui,'y2,
Aj.:
contained in A^. Hence
which contradicts our hypothesis.
+
there exist sets A,,
v^eAi^, v^eAi^, Let k be the
.
•
Suppose A is linearly dependent. Then there exist vectors v^ K, not all of them 0, such that
Since
.
.
. ,
i„).
and
It follows then, since so,
by
{1),
A^
is
Ai C Ag c
•
•
linearly dependent,
CHAP.
5.45.
AND DIMENSION
BASIS
5]
113
Consider a finite sequence of vectors S — {vuV2, .,v„}. Let T be the sequence of vectors obtained from S by one of the following "elementary operations": (i) interchange two vectors, (ii) multiply a vector by a nonzero scalar, (iii) add a multiple of one vector to another. Show that S and T generate the same space W. Also show that T is independent if and only if S is independent. .
.
Observe that, for each operation, the vectors in T are linear combinations of vectors in S. On the other hand, each operation has an inverse of the same type (Prove!); hence the vectors in S are linear combinations of vectors in T. Thus S and T generate the same space W. Also, T is inde— n, and this is true iff S is also independent. pendent if and only if dim
W
5.46.
A—
Let vi,
.
.
(ay)
.,Vn be
= —
Ml
Ui
anVi
a^iVl
+ +
a22V2
+ +
OmlVl
+
am2V2
+
ttiiVi
Um =
Show
and B = (by) be row equivalent mXn matrices over a any vectors in a vector space V over K. Let
•
•
•
+
a2nVn
= W2 =
•
•
•
"I"
OmnVn
Wm =
•••-!- ainVn
Wi
same
that [Vd] and (w.} generate the
field
K, and
6iit;i
-I-
bi2V2
+•••-!- binVn
b2lVl
+
b22V2
-I-
bmlVl
+
bm2V2
+
•
•
•
•
•
-I-
•
+
let
&2nVn
bmnVn
space.
Applying an "elementary operation" of the preceding problem to {mJ is equivalent to applying an elementary row operation to the matrix A. Since A and B are row equivalent, B can be obtained from A by a sequence of elementary row operations; hence {tyj can be obtained from {mJ by the corresponding sequence of operations. Accordingly, {mj} and {wj generate the same space.
5.47.
Let
Vi,
.
= W2 = Wi
Wn — where
Oij
S K. Let P
over a
P
(ii)
Suppose
(iii)
Suppose {wi)
(i)
Since
P
is
-I-
ai2V2
+
a2iVi
+
a22V2
+•••-!- a2nVn
dnlVl
+
an2V2
other
Show
•
•
•
"I"
"I"
ainVn
annVn
coefficients, i.e. let
Show that P
independent.
is
that {wi)
P=
{an).
is
dependent.
is
invertible.
row equivalent to the identity matrix /. Hence by the preceding generate the same space. Thus one is independent if and only if the
is invertible, it is
{vi^
is.
P is not invertible, it is row equivalent to a matrix with a zero row. This means that generates a space which has a generating set of less than n elements. Thus {wj is {wit dependent.
(ii)
Since
(iii)
This
is
the contrapositive of the statement of
Suppose
V
is
if
(Ml,
.
.
independent; (i)
•
+
Let
aiiVi
be the n-square matrix of
not invertible.
problem {wj and
(i)
K.
field
Suppose P is invertible. Show that {wi) and {vi} generate the same space; hence {wi) is independent if and only if {vi) is independent.
(i)
5.48.
V
.,Vn belong to a vector space
.
Suppose
the direct
and
.,tfot}cC/ (ii)
a^Ui
sum
of its subspaces {wi,
dim V = dim
+
•
•
•
+
=
(i„,m„,
+
(ajMi
-I-
.
.
.
,
Wn)
(ii),
U
CW
and so
it
and W,
follows
i.e.
from
(ii).
V = II®W. Show
are independent, then {Vn,Wj)
is
U + dim W. 6iWi +•••-!- b„w„ •
•
-h ttrnWrn)
+
=
(^I'^l
where
0,
+
•
•
•
ttj,
bj
+ b„Wn) =
are scalars.
+
Then
that:
also
BASIS
114
where
a^u^
0,
+
•
•
+ a^Mm ^ U
•
AND DIMENSION and
+
h^Wi
0,
•
•
+
•
[CHAP.
6n'"'n
^
sum
Since such a
'^^
for
5
is
unique, this leads to axUi
+
•
•
+ a„Mm =
•
+
bjWi
0,
•
•
•
+
—
h^w^^
The independence of the tij implies that the Oj are all 0, and the independence of the Wj implies that the 6j- are all 0. Consequently, {mj, Wj} is independent. (ii)
=
dimy
-U ®W,
V
Method 1. Since 5.8, page 90,
+
dimC7
diml?F
=U+W
we have V
-
=
dim(f/nW')
Ur\W -
and
dim
1/
+ dim W -
= dim U +
W
w^ are bases of U and 2. Suppose {u^, ...,u^) and {t«i, respectively, {mj, Wj) generates V = V +W. they generate TJ and Thus {u^, w^ is a basis of V; hence dim V (i), (mj, Wj) is independent.
Method
.
.
.
W
5.49.
W
Let
{mj,
.
.
.
,
m,} be a basis of U. .
y =
?7
©
On
respectively. Since the other hand, by
= dim ?7 +
Show
of finite dimension.
dim T^
dim W.
that there exists
U®W.
basis of V, say, {mi, {Mj,Wi} generates V, V
5.50.
V
Let [/ be a subspace of a vector space oiV such that F = a subspace
Thus, by Theorem
{0}.
.
Since {mJ
is linearly
independent,
it
can be extended to a
W
Since w,}. be the space generated by {w^, wJ. Let On the other hand, C/ n H' = {0} (Problem 5.62). Accordingly,
m„ ^i = V + W. .
,
.
.
. ,
T17.
X
E
(or: £? is an extension of K), then is a subfield of a field Recall (page 65) that if may be viewed as a vector space over K. (i) Show that the complex field C is a vector space of dimension 2 over the real field R. (ii) Show that the real field R is a vector space of infinite dimension over the rational field Q.
E
(i)
claim that {l,i} is a basis of C over R. For if v&C, then v = a + hi = a'l-\-b'i that is, {1, i} generates C over R. Furthermore, if x'l + y'i = Q or a, 6 e K; - 0, where a;,j/ S R, then a; = and y = 0; that is, {l,i} is linearly independent over R. Thus {1, t} is a basis of C over R, and so C is of dimension 2 over R.
We
where x + yi
(ii)
For suppose is linearly independent over Q. claim that, for any n, {1, Tr.tr^, ., tj-"} + a„7r" = 0, where the aj G Q, and not all the aj are 0. Then ;r is a aol + ffliff + a^ifi + + o„x". But it can be root of the following nonzero polynomial over Q: a^ + a^x + ajOi^ + is a transcendental number, i.e. that ir is not a root of any nonzero polynomial shown that .,jr" are linearly independent over Q. real numbers 1, jr, jt-^, over Q. Accordingly, the Thus for any finite n, R cannot be of dimension n over Q, i.e. R is of infinite dimension over Q.
We
.
•
.
•
•
•
•
tt-
w+1
5.51.
.
.
KgLcE.
L and L
a subfield of a field E: Suppose that E is of dimension n over L and subfield of E.) over K. Show that E is of dimension mn over K.
Let
2«:
be a subfield of a
Suppose that {aiVj-. elements.
{vi,
=
i
.
.
.,i;„} is
l,...,m,j
field
.
Since
{i;i,
.
.
.
,
a^)
K is
a
m
.
...,v„} generates
W = 6i1)i + b2V2 + + generates L over K, each 64 S L •
Since {oi,
(Hence
of dimension
.,a™} is a basis of L over K. We claim a basis of E over L and {a^, l,...,n} is a basis of E over K. Note that {aji^j} contains mn
w
efficients in
is
=
be any arbitrary element in E. bination of the Uj with coefficients in L:
Let
L
K:
•
fen'Wn.
•
61
=
fciitti
+
fejafta
+
bn
-
'^nl«l
+
'^n2«2
+
'
•
"
over L,
w
is
a linear com-
'
+
fcim^m
+
Km"'n
W
S^
a linear combination of the
is
•
&{
E
a,-
with co-
CHAP.
AND DIMENSION
BASIS
5]
e
•where k^
w = -
K. Substituting in
we
obtain
+ ki^ajvi + (fcaitti + + A;2„o„,)v2 + + (Kia^ + + k„^ajv„ + ki^amVl + 'C21«l'"2 + + fe2m«m''2 + + k^iaiV^ + + fc„„a„V„
(kiitti H
•
+
^^U"!"!
(1),
••
115
•
•
•
•
•
*
"
•
•
'
•
"
'
•
•
"
•
•
•
i.}
where erates
e
fc^j
E
K. Thus
The proof Xj,
w
a linear combination of the ajVj with coefficients in K; hence {ojVj} gen-
is
over X.
2
e X,
is
complete
«n(aiVj)
{xiidiVi
or
=
that
+ Xi2a2Vi +
(ajiitti
we show
if
0;
•
+ Kiattj +
•
•
+ Xi,„a„,Vi) +
•
•
that {a^Vj} is linearly independent over K.
•
•
+ Xi,^a„)vi +
•
.
Kiiffli
But
{ffli,
.
.
.
,
«„}
+
+
•
=
{ttjVj} is
+
=
a;im«m
=
Xi2
0,
•
+ x^^ch.i'n +
{x^iaiV^
+ x„2a2 +
•
+
L
and since the above
0,
.
.
(a;„ieii
+
x^^Ui
.,
»im
=
«„!
...,
0,
K
linearly independent over
=
•
+ a;„ma„v„) =
•
+ x„^a„)v„ =
coefficients of the Vj belong to
•
•
•
+ »„„«„ =
£ K,
ar^i
=
a;„2
0,
•
•
•
+
x„2a2
independent over K; hence since the
is linearly
«ii
Accordingly,
Xi2a2
+
•
•
Since {vi, ., v„} is linearly independent over L, each coefficient must be 0: .
Suppose, for scalars
is,
...,
0,
and the theorem
a;„m
=
proved.
is
Supplementary Problems LINEAR DEPENDENCE 5.52.
5.53.
Determine whether u and v are linearly dependent where:
=
(i)
u
(ii)
u=
(vii)
u = -fi +
(1, 2, 3, 4),
(-1,
6,
V
=
-12), V
=
(|,
- 16,
V
=
|t2
-3,
^f3
6)
- :^t2 + 8
m=
(iv)
u
(viii)
m=
(ii)
(1,-2,4,1),
(2,
1,0,-3),
Let V be the vector space of 2 X 3 matrices over R. are linearly dependent or independent where: 1
-2
„ _ /I -1
3\
4
v
=
(0,
-3)
(1, 0, 0),
v
=
(0, 0,
(0, 1),
=
t^
+ 3t + 4^
-y
=
t^
-3)
+ 4t + 3
u
=
fS
-
4*2
(ii)
u
-
t3
_
5^2
+ _
2t
+
3,
V
2t
+
3,
V
= -
—
ts
+
4t
-
1,
t^
-4t^ -3t
+
A,
+
—1,
4),
2
10
A,B,C B V
7
-1
^=(r4-s
I'D-
2*2
Determine whether the matrices
I
V
(i) (1, 3,
-6,1,4).
'
Let be the vector space of polynomials of degree linearly dependent or independent where: (i)
(3,
/3 -8
4\
-2 7
5
-(4
<"'^=a4-:). 5.55.
(iii)
Determine whether the following vectors in R* are linearly dependent or independent: (3,8,-5,7), (2,9,4,23);
5.54.
(4, 3, 2, 1)
3 over R.
w w =
Determine whether u,v,w
2fi
-
fi
2fi
-
7fi
-
3t
-
+
It
5
+
9
GV
are
.
5.56.
Let V be the vector space of functions from R into R. ent where: (i) f{t) = e«, g(t) = sin t, h(t) = th (ii)
=
g{t)
5.57.
AND DIMENSION
BASIS
116
Show
sin
that:
C
field
cos
the vectors
(i)
)
Show
that f,g,h
=
/(*)
g(t)
e«,
=
eV
are linearly independ-
hit)
e^*,
5
=
t;
/(f)
(iii)
=
e«,
t.
(1
—
and
i, i)
(2,
—1 + i)
in C^ are linearly dependent over the complex
but are linearly independent over the real
+ 2'\/2
(7, 1
=
h(t)
t,
[CHAP.
field
R;
+ \/2)
the vectors (3 -I-V2, 1
(ii)
R
in R2 are linearly dependent over the real field
and
but are linearly independent over the
rational field Q. 5.58.
v and w are linearly independent vectors. Show that: — u + V 2w, u — v — w and u + w are linearly independent; u + V — 3w, u + 3v — w and v + w are linearly dependent.
Suppose (i)
(ii)
5.59.
u,
Prove or show a counterexample: If the nonzero vectors a linear combination of u and v.
u,
v and
w
are linearly dependent, then
w
is
5.60.
Suppose
Vi, v^,
{oi'Wi, a2>'2<
(ii)
{vi,
.
and 5.61.
•
•
5i #
Suppose
Ui,
.
.,
(c,
d)
v„}
Prove the following:
where each
linearly independent
is
eij
where
# 0. w = b^v^ +
•
•
•
+
b^Vi
+
•
w—
and
{ui,
.
.
n L{Wj) =
.
,
m^,
{0}.
Show
belong to K^.
that {v,
w}
is linearly
dependent
if
5.65.
.
.
.
,
.
5.66.
5.67.
.
=
.
.
.
+
au^'i
•
.
•
.
+
•
ffliB^n,
.
.
Wm =
.,
a^i^i
Determine whether or not each of the following forms a basis of (1, 1)
and
(ii)
(2, 1),
(1,
(3, 1)
-1) and
(0, 2)
+
•
•
•
+
«mn1'K
(iii)
(0, 1)
and
(iv)
(2, 1)
and (-3,
(0,
R^:
-3) 87)
Determine whether or not each of the following forms a basis of R^: (i)
(1, 2,
-1) and
(ii)
(2, 4,
-3),
(0, 1, 1)
and
(iii)
(1, 5,
-6),
(2, 1, 8),
(3,
(1, 3,
-4),
(0, 3, 1)
(1, 4,
(0, 1,
-1,
-3) and
4)
-1)
and
(2, 1, 1)
-11)
(2, 3,
Find a basis and the dimension of the subspace (i)
(1, 4,
(ii)
(1,
V
-1,
3),
-4, -2,
-3, -1) and
(2, 1,
1),
(1,
-3, -1,
2)
(0, 2, 1,
and
(3,
be the space of 2 X 2 matrices over
ui).
(-:
W of R* generated by:
-5)
-8, -2,
R
and
7)
let
W be the
\y (-r:)
subspace generated by
-
(4-;;
Find a basis and the dimension of W. 5.68.
Let
if
, a^^) a^'e linearly independent vectors in Z", and suppose (a^, ..., ai„), , (ttmi, v„ are linearly independent vectors in a vector space V over K. Show that the vectors
(i)
Let
b„v„
.
AND DIMENSION
(iv)
+
and only
are also linearly independent.
BASIS
•
Show that is a linearly independent subset of a vector space V. , Wg} Wi, (Recall that L(Mj) is the linear span, i.e. the space generated by the Mj.)
Wi
5.64.
•
0.
Suppose .
linearly independent
"n'^'n} is
•>
be
L(Mi) 5.63.
f „ are linearly independent vectors.
. ,
.
.,v^-l,w,v^^l,
.
= (a, 6) = 0.
Let V
ad— 5.62.
.
(i)
W be the space generated by the polynomials tt
=
(3
+
2*2
-
2t
+
1,
t)
Find a basis and the dimension of W.
=
t*
+
3*2
-
«
+
4
and
w =
2*3
+
t2
-
7t
-
7
CHAP.
5.69.
AND DIMENSION
BASIS
5]
X X 3x
+ + +
Sy 5y 5y
+ + +
= =
2z z
8z
X
2y
2x
+ -
2x
3y
+ -
2z
y
+
z
(i)
5.70.
W of each homogeneous system:
Find a basis and the dimension of the solution space
-
X 2x
= =
7z
+ 2x+ X
+ 2y ~2z + 2s - t = + 2y - z + 3s - 2t = + 4y ~ Iz + s + t =
W of each homogeneous system: + + +
x 2x 2x
-
2y 4y iy
z
2z 2z
V =
{1, 1
(ii)
{1
-3),
2, 5,
V,
(i)
U
W
0},
W,
(ii)
(iii)
-2,
(1,
=
and
W are
subspaces of
—
and that dim
U be
and
let
W be
5.78.
Let {t3
V
dim
dim
-3, -1, -4),
{U+W),
+ 4*2 - t + 3,
Let
U
and
let
(1, 4,
+t"-l +
•••
b
=
2c}
that
t«}
VnW
^
{0}.
U=
4,
dim
W=5
and dim
U-
1,
dim
W=2
and
-1, -2, -2),
-2,
3), (2, 8,
-1, -6, -5),
V-
7.
Find the
UlW.
Show
that
(2, 9, 0,
-5, -2)}
+ 5*2 + 5^
t3
Find
(i)
dim
(1, 3,
-1, -5, -6)}
dim (t/n VT).
(ii)
be the vector space of polynomials over R.
respectively.
5.79.
d,
the subspace generated by {(1, 6, 2,
(i)
=
the subspace of Rs generated by {(1, 3,
Find
a
U nW.
Let U and P7 be subspaces of R3 for which Rs = r; w'.
Let
-2)}
Determine whether or not each of the
n.
Show
subspaces of R3.
V
{{a,b,c,d):
©
5.77.
1, 2,
VnW.
of degree
t
l
possible dimensions of
5.76.
generated by
is
+ t+t2, l+t+t2 + t3, ..., l + t+t2+ t+ fi, t^ + t», ..., t"-2 + t"-i, t»-i + t"}.
+ t,
+ t,
Suppose
-3,
(2,
Let V be the vector space of polynomials in following is a basis of V: (i)
W
set
6-2c + d =
{(a,b,c,d):
SUMS AND INTERSECTIONS 5.74. Suppose V and W are 2-dimensional 5.75.
-1),
0, 3,
Find a basis and the dimension of 5.73.
+ 3s-4t = - s + 5t = + 4:S -2t -
W be the following subspaces of R*:
V and
Let
-2,
5z
(ii)
Find a homogeneous system whose solution {(1,
5.72.
y
= =
2z
(iii)
(i)
5.71.
+ +
4:y
(li)
Find a basis and the dimension of the solution space X
117
3*3
(f/
+
+
10(2
W'),
- 5t + 5} (ii)
Let
and
U and W be the subspaces generated by {t^ + U^ + &,t^ + 2t^-t + 5, 2*3 + 2*2 - 3* + 9}
i\m.(VnW).
be the subspace of RS generated by {(1,
W be the
-1, -1, -2,
0), (1,
-2, -2,
0,
-3),
(1,
-1, -3,
2,
-4),
(1,
-1, -2, -2,
1)}
subspace generated by {(1,
-2, -3,
0,
-2),
(1,
(i)
Find two homogeneous systems whose solution spaces are
(ii)
Find a basis and the dimension of
U r\W.
-1, -2,
U
2,
-5)}
and W, respectively,
BASIS
118
AND DIMENSION
[CHAP.
5
COORDINATE VECTORS 5.80.
Consider the following basis of B^: {(2, 1), (1, -1)}. Find the coordinate vector of vSU^ v = (a,b). to the above basis where: (i) i; = (2,3); (ii) v = (4,-1), (iii) (3,-3); (iv)
5.81.
- t, (1 - t)^, In the vector space V of polynomials in t of degree - 3, consider the following basis: {1, 1 basis if: (i) v = 2 - 3t + t* + 2t^; (1 _ t)3}. Find the coordinate vector of v S y relative to the above (ii)
5 82
i;
= 3 - 2t - ^2;
(iii)
v
= a + bt + ct^ + dt^. X
In the vector space PF of 2
2 symmetric matrices over R, consider the following basis:
AGW
Find the coordinate vector of the matrix
-
5.83.
=
(4
<"'
-I)
^
:)•(-' 1)}
a
{(-11).
...
relative
relative to the above basis
if:
I)
Consider the following two bases of R*: {ei
=
(1, 1, 1),
02
=
(0, 2, 3),
63
=
(0, 2,
(i)
Find the coordinate vector ot v
(ii)
Find the matrix
P
=
and
-1)}
(3,5,-2)
{/i
=
(1, 1, 0), /z
=
(1,
relative to each basis: [v]^
-1.
0), fs
and
[v]y.
whose rows are respectively the coordinate vectors of the
e,
=
(0, 0, 1)}
relative to the
basis {/i, /a, /a)(iii)
5.84.
5.85.
Verify that [v]eP=[v]f.
(of dimension n). Let P be the .,e„} and {fu ••../„} are bases of a vector space V Suppose {e^, e's relative to the basis {fih Prove the of vectors coordinate the respectively are rows matrix whose Problem 5.39 in the case n - 3.) that for any vector veV, [v]^P = [v]f. (This result is proved in .
.
SF
that the coordinate vector of
Show
relative to
V
any basis of
is
always the zero w-tuple
(0,0, ...,0).
RANK OF A MATRIX 5.86.
Find the rank of each matrix:
5.87.
Let
5.88.
Give examples of 2
A
and
B
(i)
rank (A
(iii)
rank (A
be arbitrary
X
+ B)< + B) >
mXn matrices.
2 matrices
rank
(A),
rank
(A),
A
and
rank
B
Show
rank (A
+
^
B)
rank (A)
+
rank
(B).
such that:
{B)
rank
that
(ii)
rank (A
+
B)
=
rank (A)
=
rank (B)
(B)
MISCELLANEOUS PROBLEMS 5.89.
W
be the vector space of 3 X 3 symmetric matrices over K. Oy hibiting a basis of W. (Recall that A = (ay) is symmetric iff
Let
Show
=
that
dimW =
5.90.
Show that be the vector space of 3 X 3 antisymmetric matrices over K. iff «« - -a^j.) antisymmetric = is A (a„) that (Recall W. exhibiting a basis of
5.91.
Suppose dim y = 5.6(iii), page 89).
Let
W
n.
Show that a generating
set with
w elements
is
a
6
by ex-
a^;.)
basis.
dim
II'
=
3
by
(Compare with Theorem
CHAP.
5]
5.92.
Let •
AND DIMENSION
BASIS
•
tj, «2,
•
+ a„t„
•
• .
K
where
be symbols, and let be any field. aj G K. Define addition in V by (ajfj
+ 02*2 +
V
Let
119
be the set of expressions
+ a„tj + (biti + 62*2 + + = (Oi + 6i)ti + (02 + 62)«2 + •
•
•
•
•
•
•
V
Define scalar multiplication on
•
that y is a vector space over basis of y where, for t = 1, , n,
Show
.
5.94.
+
•
K+
«>„)««
•
+ o„<„) =
•
K
+
ka^ti
+
ka^t^
•
with the above operations.
•
•
+
fea„t„
Also show that
{t^,
.
.
.
t„} is
,
a
Oti
+
•
+
•
•
Oti_i
+
l«i
+
Ofj +
i
+
•
•
+
ot„
K
Let y be a vector space of dimension n over a field K, and let he a, vector space of dimension over a subfield F. (Hence V may also be viewed as a vector space over the subfield F.) Prove that the dimension of V over F is ww.
U
Let
W
=
Let
t7
field K, and let V be the external direct sum of U and and Jy be the subspaces of V defined by C7 = {(m, 0) m G 17} and :
w G W). Show that U is isomorphic
(i)
(ii)
y
w):
{(0,
morphic to
W
Show
dim
that
V =
dim
V+
W. Let V be
Answers (i)
no,
5.53.
(i)
dependent,
(ii)
independent.
5.54.
(i)
dependent,
(ii)
independent.
5.55.
(i)
independent,
5.57.
(i)
5.59.
The statement
(2,-l
(iii)
(ii)
+ t) =
(l
w
yes,
(iv)
u
(m, 0),
and that
W
is iso-
*-> (0, w).
dim W.
= u+w
to
5.52.
yes,
under the correspondence
the external direct product of
under the correspondence v
(ii)
U
to
under the correspondence
V = U®
Suppose to
m
W be vector spaces over the same
and
TF (see Problem 4.45).
5.95.
6„«„) •
.
=
ti
5.93.
.
+ 03*2 +
by
+ «2*2 +
fc(«i*i
ajti
U and
W. Show that
V is isomorphic
{u,w).
Supplementary Problems
no, (v) yes,
(vi)
no,
(vii) yes,
(viii)
no.
dependent.
+ i)(l-t,i);
(ii)
(7, l
+ 2\/2) = (3-v^)(3 + -\/2, 1 + -/2).
=
—
w=
Lemma
5.2
requires that one of the nonzero vectors u,v,w is a linear combination of the preceding ones.
In
v
this case,
=
is false.
Counterexample:
u
(1, 0),
no,
(iii)
no,
(iv) yes.
yes,
(iii)
no,
(iv)
(ii)
dim
W=
(i)
yes,
5.65.
(i)
no,
5.66.
(i)
5.67.
dim
5.68.
dim Ty
5.69.
(i)
5.70.
(i)
basis, {(2,-1,0,0,0), (4,0,1,-1,0), (3,0,1,0,1)};
(ii)
basis, {(2, -1, 0, 0, 0),
dim
(ii)
W=
W=
2.
=
2.
3,
(2, 0)
and
(1, 1)
in R*.
2m.
5.64.
(ii)
v
basis, {(7, -1, -2)};
no.
2.
dim ly
=
1.
(ii)
(1, 0, 1, 0, 0)};
dim ly
dim ly
=
=
2.
0.
(iii)
dim Ty
basis, {(18, -1, -7)};
=
3.
dim
W=
1.
+ +
5x 5.71.
{ 5.72.
AND DIMENSION
BASIS
120
X
—z— —z—
y y
t
basis, {(1, 0, 0, 0), (0, 2, 1, 0), (0,
(ii)
basis, {(1,0,0,1), (0,2,1,0)};
(iii)
basis, {(0,2,1,0)};
yes,
=
=
3.
2.
VnW
Hint.
1.
V=
dim
0, 1)};
dim Tf
dim (VnH^)
V = n + 1,
For dim
no.
(i)
5.75.
dimd/nW') =
2,
3 or 4.
5.77.
dim {U+W)
=
3,
dim
5.78.
dim
(U+W) =
3,
dim (t/n W)
5.79.
(i)
(ii)
—
+ \4x + [30:
{(1,
-1,
=
1.
\Ax
=0
s
M
=
(5/3, -4/3),
5.81.
(i)
M
=
(2,
5.82.
(i)
[A]
=
(2,
-1,
1),
5.83.
(i)
[v],
=
(3,
-1,
2),
5.86.
(i)
5.88.
(i)
A
(ii)
A
7,
-2),
=
[v]
(ii)
(ii)
(ii)
but the set contains only n elements.
-8 + 2y 2 + + + 2j/ X^x -1)}. dim {VnW) = 2
=
t
0, 0), (0, 0, 1, 0,
(i)
-5,
2.
+
2y
-2, -5,
(UnW)^
z
5.80.
[v]
[A]
=
(1, 2),
(iii)
=
-1,
(0, 4,
(3, 1,
M=
[v]
(iii)
0),
fl 5.89.
<;io \0
(ii) 2,
(iii) 3,
M; =
B (J (o
P'
o\
/o
o)'
1
0,1 0/
= =
[v]
=
(a
=
((a
+
6)/3,
(a-26)/3).
+ h + c + d, -b-2c-3d, c + 3d,
-d).
-2).
-1, -2);
(ii)
P =
1
1\
-1
3
.
(iv) 2.
\o
CI
-I)
c
:) /o
o\
<""^=a i>
l\
0,0
1
\l
0/
-1
,0
0/
l\
/O
0/
\o -1
0,0
5.90.
5.93.
(4,
t
(iv)
(0, 3),
/I
3,
satisfy all three conditions on a,b,c
must
d.
5.73.
(ii)
5
= =
s
(i)
and
[CHAP.
0\
/O
0/
\0
"h:
°)
0'
0,0
1 |, 1
0;
0^ 1
o;
Hint. The proof is identical to that given in Problem 5.48, page 113, for a special case (when an extension field of K).
V
is
chapter 6
Mappings
Linear MAPPINGS Let A and 5 ment from
be arbitrary
Suppose to each
sets.
of B; the collection, /, of such assignments A into B, and is written
f:A-^B
aGA
there
is
assigned a unique ele-
a function or mapping
is called
(or-
map)
A^B
or
We write f{a), read "/ of a", for the element of B that / assigns to a e A; it is called the value of fata or the image of a under /. If A' is any subset of A, then /(A') denotes the set of images of elements of A'; and if B' is any subset of B, then f-'{B') denotes the set of elements of A each of whose image lies in B': f{A')
=
{/(a)
aGA'}
:
f'W) = {aGA:
and
f{a)
G
B'}
We call f{A') the imxtge of A' and /-^(jB') the inverse image or preimage of B'.
In particular, the set of all images, i.e. f{A), is called the image (or: ranflre) of /. Furthermore, A is called the domain of the mapping f:A^B, and B is called its co-domain.
f:A^B
To each mapping {{a, f{a))
:
aGA}. We
call this set
there corresponds the subset of A x B given by the graph of /. Two mappings f:A-*B and g:A-^B
are defined to be equal, written f = g, if /(a) = 5r(a) for every aGA, that is, if they have the same graph. Thus we do not distinguish between a function and its graph. The negation of f = g is written f ¥- g and is the statement: there exists an for which f(a)
aGA
^ g{a).
Example
6.1
:
A = {a, b, e, d} A into B:
Let
B
and
{x, y, z, w}.
The following diagram
defines a
mapping
f from
Here
f(a)
=
The image Example
6.2:
Let
/:R
-»
y,
(or:
R
/(6)
=
x,
f{c)
=
z,
and
f(d)
=
faa,b,d})
=
{f (a), fib), fid)}
range) of /
is
the set [x,y,z}:
Also,
y.
=
{y,x,y}
f(A)
=
=
{x,y}
{x,y,z}.
be the mapping which assigns to each real number x
X
Here the image of —3
is
9 so
V^
x^
we may
121
or
write
= a;2 /(— 3) = 9. f{x)
its
square
a;^:
[CHAP.
LINEAR MAPPINGS
122
We use the arrow
eA
image of an arbitrary element x
to denote the
i^
/:A->5 bywriting
^
x
6
under a mapping
fix)
as illustrated in the preceding example. Example
6.3:
Consider the 2 X 3 matrix
A =
R2 as column vectors, then
A
V
Av,
\-*
'1
-3
5'
c,2
4
-1,
determines the that
T{v)
is,
:
Thus
V
if
then
,
= Av =
T{v)
(
-2/
Every
Remark:
mxn
defined
matrix
A
-3
/I
1
write the vectors in R^ and
mappmg T RS -> R2 v& R3 - Av,
3\
=
we
If
4 -_i
2
|
K
field
^
\-2/
^
over a
^^
5\/
defined
<
determines the mapping
by
-10 12
T X" -» K™ :
by
v v^ Av written as column vectors. For convenience are where the mapping by A, the same symbol used for the above the we shall usually denote vectors in if"
and
li:"'
matrix. Example
6.4:
Example
6.5:
real field R. be the vector space of polynomials in the variable t over the for any polynomial / G V, where, D:V-^V mapping a defines derivative the Then we let D(f) = df/dt. For example, D(3t^ - 5t + 2) = 6t - 5.
Let
V
Let
V
Then
preceding example). be the vector space of polynomials in t over R (as in the V -* R where, for any to 1 defines a mapping the integral from, say,
^
polynomial
f&V, we
^(/)
let
^(3(2-5* + Note that this
map Example
6.6:
map
= =
2)
r
(3t^-5t
from the vector space example is from V into
is
in the preceding
f:A^B
Consider two mappings
For example,
f(t) dt.
f
V
+ 2)dt
=
i
into the scalar field R,
illustrated below:
-©—^-©
0.
Let a G A; then /(a) G B, the domain of g. under the mapping g, that is, g(f{a))- This
a
K
Hence we can obtain the image of
g{f(a))
{9°f){a)
theorem
first
Theorem
6.1:
Let
tells
=
denoted by
g(f(o))
f.A^B, g.B^C and h.C^D. Then If a
G A,
ho{gof)
=
(hog)of.
then
{ho{gof)){a)
=
h{igof){a))
=
h{g{f{a)))
({h°g)of){a)
=
{hog){f{a))
=
Hg{f{a)))
Thus iho{gof)){a) = {ihog)of){a) for every a G A, and
Remark:
is
us that composition of mappings satisfies the associative law.
We prove this theorem now. and
/(a)
map
and from A into C is called the composition or product of / and g, is the mapping defined by (g°f):A-^C words, other In g°f.
Our
whereas the
itself.
g.B^C
and
:
so ho{gof)
=
{hog)of.
of a G A Let F:A-^B. Some texts write aF instead of F{(i) for the image and F:A-*B functions under F. With this notation, the composition of text. in this G-.B^C is denoted by F o G and not by G o F as used
CHAP.
LINEAR MAPPINGS
6]
We
123
next introduce some special types of mappings.
Definition:
A
mapping f:A-*B
if
different elements of
is
or, equivalently.
A
Definition:
GB
is
A
have distinct images; that
if
a
is
implies
v^ a'
if /(a)
mapping f:A-^B
every b
said to be one-to-one (or one-one or 1-1) or injective
=
6.7:
Let h(x)
/:R^R, g:B,-*R = x^. The graphs
f(x)
=
and
fc
:
of these
a
=
maps
said to be onto (or: /
the image of at least one a
A mapping which is both one-one and onto is Example
/(a) ¥- f{a')
implies
f{a')
is,
a'
A
onto B) or surjective
if
G A.
said to be bijective.
R -» B
be defined by f(x)
mappings
2=^
g(x)
=
-
2'',
g{x)
—
x^
-x
and
follow:
h(x)
X'
=
a;2
The mapping / is one-one; geometrically, this means that each horizontal line does not contain more than one point of /. The mapping g is onto; geometrically, this means that each horizontal line contains at least one point of g. The mapping h is neither one-one nor onto; for example, 2 and —2 have the same image 4, and —16 is not the image of any element of R. Example
Example
6.8:
6.9:
Let A be any set. The mapping /:A-»A defined by f{a) = to each element in A itself, is called the identity mapping on 1^ or 1 or /. Let
f:A-*B.
We
call
g-.B^A f°g =
the inverse of
Ifl
/,
g°f =
and
a,
i.e.
A
and
written /~i,
which assigns is denoted by
if
1a
We
emphasize that / has an inverse if and only if / is both one-to-one and onto (Problem 6.9). Also, if then /"'(ft) = a where a is the unique element of A for which f(a) = 6.
6GB
LINEAR MAPPINGS Let V and U be vector
A
mapping F:V -* U is called a spaces over the same field K. linear mapping (or linear transformation or vector space komomorphism) if it satisfies the following two conditions: (1)
For any
v,wGV,
(2)
For any
kGK
+ w) =
+ F{w). and any vGV, F{kv) = kF{v). F{v
F{v)
In other words, F:V^ U is linear if it "preserves" the two basic operations of a vector space, that of vector addition and that of scalar multiplication. Substituting k = into (2) we obtain F{0) the zero vector into the zero vector.
=
0.
That
is,
every linear mapping takes
[CHAP.
LINEAR MAPPINGS
124
Now
for any scalars a,b
gK
eV
and any vectors v,w
we
6
by applying both
obtain,
conditions of linearity,
=
F{av + bw)
More
aiGK
any scalars
generally, for
+
F{av)
=
F{bw)
+ bF{w)
aF{v)
viGV we
and any vectors
obtain the basic
property of linear mappings: F{aiVi
+ anV„) = aiF{vi) + a^FiVi) + condition F{av + hw) = aF{v) + bF{w)
+ aiVi +
We linear
•
•
•
•
remark that the mappings and is sometimes used as
Example
6.10:
Let
A
mX%
be any
completely characterizes
We
claim that
T
y->
previously, A determines a (Here the vectors in K" and K""
Av.
For, by properties of matrices,
is linear.
+ w) - A(v + w) = Av + Aw =
T(v
=
T(kv)
and
anF(Vn)
As noted
matrix over a field K. by the assignment v
are written as columns.)
+
•
its definition.
T:&^K^
mapping
•
A(kv)
= kAv =
+
T{v)
T{w)
kT{v)
where v,w G K^ and kE. K.
In comment that the above type of linear mapping shall occur again and again. finite-dimensional one from mapping linear every that show fact, in the next chapter we of the above type. vector space into another can be represented as a linear mapping
We
Example 611-
Let
We
z) B» be the "projection" mapping into the xy plane: F(x, y, Then b', c'). = {a', w = and b, c) v (a, Let show that F is linear.
i^
R^
:
-^
F{v and, for any
fc
€
Example 612-
Example
6.13:
is,
=
F{,ka, kb, ke)
F.V^U
=
k(a, b, 0)
any
v,weV + w) =
F
is
=
linear.
=
+
We
F(v)
F
call
v,wGV
and any
+
+
the zero
Consider the identity mapping
Thus
G
be the mapping which assigns and any kG K, we have
/(av
6.15:
=
(ka, kb, 0)
kF(v)
linear.
Let
for any
Example
=
=
(x
:
Thus 6.14:
is
6', 0)
by F(x,y) Let F R2 ^ R2 be the "translation" mapping defined zero vector Observe that F(0) = F(0,0) = (1,2) ^ 0. That is, the linear. not onto the zero vector. Hence F is
F{v
Example
F
{x, y, 0).
R,
F(kv)
That
+ w) = F{a + a',b + 6', c + c') = {a + a', b + = (o, b, 0) + (a', b', 0) = F(v) + F(w)
=
F(kv)
and
F(w)
mapping and
to every
[7
which maps each we have
a,beK, = av + bw =
+
veV.
=
=
al{v)
+
vGV
l,y
+ 2)
not mapped
fcO
shall usually denote it
I.V^V
bw)
is
Then, for
=
kF(v)
by
into itself
0.
Then, .
bl(w)
/ is linear.
R JtV^B
t over the real field be the vector space of polynomials in the variable mtegral mappmg the and D:V-^V mapping differential Then the any and 6.5 are linear. For it is proven in calculus that for
V
Let
defined in
u,v
SV
Examples 6.4 and fe e R,
dv du + v) dt- = dt+dl
d{u
that
is,
D(u + v)=D(u) + D(v)
f
that
{u(t)
+
is,
J{u +
•»)
=
SM + SM
^""^
=
and D(ku)
v(t)) dt
f
and
j^du
djku)
,
=
ku{t)dt
=
and ^(few)
k D(u); and u(t) dt
\
k
=
"dt
dt
\
+
u{t) dt
k^iu).
\
also,
v(t) dt
CHAP.
LINEAR MAPPINGS
6]
Example
F:V ^ U
Let
6.16:
be a linear mapping which
mapping F"! ping
is
125
:
C/ ->
y
We
exists.
will
both one-one and onto. Then an inverse 6.17) that this inverse map-
is
show (Problem
also linear.
When we investigated the coordinates of a vector relative to a basis, we also introduced the notion of two spaces being isomorphic. We now give a formal definition. Definition:
A
linear
mapping
Example
Let
6.17:
y
F:V^U U
vector spaces V, V onto U.
is
called
an isomorphism
are said to be isomorphic
be a vector space over
K
its
K
[v]^,
coordinate vector relative to the basis {e},
is
one-to-one.
if it is
there
n and
of dimension
Then as noted previously the mapping v
if
let {e^,
.
The
an isomorphism of
is
.
.,e„} be a basis of V.
eV
which maps each v into an isomorphism of V onto K". i.e.
Our next theorem it tells
gives us an abundance of examples of linear mappings; in particular, us that a linear mapping is completely determined by its values on the elements
of a basis.
Theorem
V and U be vector spaces over a field K. Let {vi,'i;2, .,Vn} be a basis V and let Ui, Ui, .,Un be any vectors in V. Then there exists a unique linear mapping F:V-^U such that F{vi) = Ui, F{v2) = ..., F{vn) = th. Let
6.2:
.
of
.
.
.
112,
We
emphasize that the vectors Mi, zt„ in the preceding theorem are completely armay be linearly dependent or they may even be equal to each other. .
.
.
,
bitrary; they
KERNEL AND IMAGE OF A LINEAR MAPPING We
begin by defining two concepts.
Definition:
Let of
F:V-^U
be a linear mapping.
image points
{uGU
Im F =
The kernel of F, written KerF,
OGU:
Theorem
6.3:
Example
F:V->U
Let of
6.18:
is easily
U
:
is
=u
F{v)
for some v
is
proven (Problem
F{v)
=
V
which map into
0}
6.22).
be a linear mapping. Then the image of /?" is a subspace of V.
/^ is
a subspace
and the kernel of
Let F:R3-^H^ be the projection mapping into the xy plane: F(x, y, z) — (x, y, 0).
entire
Clearly the image of
F
is
the
xy plane:
l
|
Im
F =
{(a, 6, 0)
KerF =
:
o, b
F
is
{(0, 0, c):
c
Note that the kernel of
G R} the z axis:
G R}
since these points and only these points into the zero vector = (0, 0, 0).
map
the set
G V}
the set of elements in
F = {vGV:
Ker The following theorem
The image of F, written Im F,
in U:
||llill
l
||
•
»
=
(a, 6, c)
[CHAP.
LINEAR MAPPINGS
128
Theorem
The dimension
5.11:
AX
6
of the solution space
W of the homogeneous system of linear
=
is
n-r where
equations rank of the coefficient matrix A. is
n
the
number
of
unknowns and r
is
the
OPERATIONS WITH LINEAR MAPPINGS We are able to combine linear mappings in various ways to
obtain new linear mappings. These operations are very important and shall be used throughout the text. field K. Suppose F:V-*U and G:V-^U are linear mappings of vector spaces over a F(v) assigns + G{v) to We define the sum F + G to he the mapping from V into U which
^^^'
(F + G){v)
=
F{v)
+
Giv)
mapping from Furthermore, for any scalar kGK, we define the product kF to be the into U which assigns k F{v) to i; e F: ikF)iv) = kF{v) show that if F and G are linear, then i^^ + G and kF are vectors v,w GV and any scalars a,h GK,
We
{F
{kF)(av
and
F+G
Thus
and kF are
kF{av
Let
The space
in the
V
above theorem
is
usually denoted by [/)
comes from the word homomorphism. dimension, we have the following theorem.
Hom
6.7:
Suppose dim 7
G:U^W
=
m
W
In the case that
GoF
and
U ^ n. Then dim Hom(V,
mapping Recall that the composition function Goi?' is the whenever linear is that = show {GoF){v) G{Fiv)). and any scalars a,b GK, for any vectors v,w
We
V
U
are of finite
= mn. spaces over the same field K, and that F:V-*U
and dim
are vector V, U and are linear mappings:
Now suppose that and
b(kF){w)
applies.
Hom(7,
Theorem
=
of all and U be vector spaces over a field K. Then the collection and addition of operations linear mappings from V into U with the above K. scalar multiplication form a vector space over
6.6:
Here
+ bF{w)) a(kF)(v) +
k{aF{v)
linear.
The following theorem
Theorem
have, for any
F{av
+ bw) = akF{v) + bkF(w)
= =
+ bw)
We
+ bw) + Giav + bw) aF{v) + bF{w) + aG(v) + bG{w) a{Fiv) + G{v)) + b(F{w) + G{w)) a(F + G){v) + b{F + G){w)
= = = =
+ G){av + bw)
also linear.
F
from and
G
V
U)
into
W
are linear.
defined by have,
We
gV
{GoF)iav
That
is,
V
+ bw)
G o F is linear.
= =
+ bw)) = G{aF{v) + bFiw)) = aiGoF){v) + b{GoF)(w) aG{F{v)) + bGiF{w))
G{Fiav
CHAP.
LINEAR MAPPINGS
6]
The composition
of linear
129
mappings and that of addition and scalar multiplication are
related as follows:
Theorem
6.8:
U and W be vector spaces over K. Let F, F' be linear mappings from U and G, G' linear mappings from U into W, and let k&K. Then:
Let V,
V
into
= GoF + GoF' + G')oF = GoF + G'oF
(i)
Go(F +
(ii)
(G
(iii)
k{GoF)
F')
= {kG)oF =
Go(kF).
ALGEBRA OF LINEAR OPERATORS Let F be a vector space over a field K. We novir consider the special case of linear mappings T:V^V, i.e. from V into itself. They are also called linear operators or linear transformations on V. We will write AiV), instead of Horn (V, V), for the space of all such
mappings.
By Theorem 6.6, A{V) is a vector space over K; it is of dimension n^ if V is of dimension Now if T,SgA{V), then the composition SoT exists and is also a linear mapping from V into itself, i.e. SoTgA(V). Thus we have a "multiplication" defined in A{V). (We shall write ST for SoT in the space A{V).) n.
We
is
(i)
F{G + H) =
(ii)
(G
(iii)
k{GF)
K
over a field is a vector space over in which an operadefined satisfying, for every F.G,H and every kGK,
GA
FG + FH + H)F = GF + HF
If the associative (iv)
X
A
remark that an algebra
tion of multiplication
=
{kG)F
=
G(kF).
law also holds for the multiplication,
i.e. if
for every
F,G,H gA,
{FG)H = F{GH)
A
then the algebra is said to be associative. Thus by Theorems 6.8 and 6.1, A{V) is an associative algebra over with respect to composition of mappings; hence it is frequently called the algebra of linear operators on V.
K
Observe that the identity mapping /
:
7 -> F
belongs to A{V).
V(x)
we can form
=
tto
+
aix
+
+
a2X^
•
•
+
any T G A{V), use the notation
Also, for
we have TI - IT - T. We note that we can also form "powers" of T^^ToT,T^ = ToToT, .... Furthermore, for any polynomial
T;
we
aiGK
a„x",
the operator p{T) defined by p{T)
=
aol
+
aiT
+ a^T^ +
•
+
•
a„r"
(For a scalar kGK, the operator kl is frequently denoted by simply k.) In particular, = 0, the zero mapping, then T is said to be a zero of the polynomial p{x).
if
V{T)
Example
6.21:
Let
T R3 ^ R3 be :
defined
by
=
T(x,y,z)
(0,x,y).
Now
if
{a,b,c)
is
any element
of R3, then:
{T +
and
T^(a, b, c)
is
= =
(0, a, b)
T^0,
+
a, b)
(a, b, c)
=
= (a,a+b,b + c)
T{0, 0, a)
see that rs = 0, the zero mapping from a zero of the polynomial p{x) = v?.
Thus we
T
I)(a, b, c)
V
=
(0, 0, 0)
into itself.
In other words,
»
LINEAR MAPPINGS
130
[CHAP.
6
INVERTIBLE OPERATORS
A
T -.V^ V
linear operator
e A{V)
exists r-i
Now T
is
such that
invertible if
suppose over,
T
is
nonsingular,
V
has
dimF
= =
assuming
Then
ImT -V,
and so
i.e.
A
6.9:
i.e.
finite
6.22:
if
there
Thus in particular, if T is if it is one-one and onto. map into itself, i.e. T is nonsingular. On the other hand, Ker T = {0}. Recall (page 127) that T is also one-one. More-
dimension,
we
T
=
T)
6.4,
dim(Imr) + dim({0})
dim (Im T)
V; thus
is
Theorem
have, by
=
dim (Im T) +
T
Hence T
onto.
is
is
both one-one and onto
have just proven
linear operator
vertible if
Example
i.e.
I.
dim(Imr) + dim (Ker
We
has an inverse,
if it
can
the image of
is invertible.
Theorem
TT-^ = T-^T =
and only
gV
invertible then only
said to be invertible
is
and only
T:V-*V if it is
on a vector space of finite dimension nonsingular.
is
in-
Let T be the operator on R2 defined by T(x, y) = (y, 2x-y). The kernel of T is hence T is nonsingular and, by the preceding theorem, invertible. We now Suppose (s, t) is the image of {x, y) under T\ hence (a;, y) find a formula for T-i. T(x,y) = (s,t) and T-'^(s, t) = (x, y). We have is the image of (s, «) under r-i; {(0, 0)};
T(x, y)
-
2x
(y,
— y) =
Solving for x and y in terms of given by the formula T~^(s,
is
The finiteness of the dimensionality of the next example. in Example
6.23:
Let
V
V
T(ao
and
T is
and
=
we
t,
obtain
+ it,
(|s
y
so
in the preceding
+ ajn) =
+ ait-\
increases the exponent of
nonsingular.
However, T
t
a^t
=
a;
2x-y =
s,
is
theorem
+
Uit^
+
term by
1.
not onto and so
is
in each is
=
+
li,
2/
=
t
s.
Thus T^'
s).
be the vector space of polynomials over K, and
defined by
i.e.
s t)
and
(s, t)
let
•
•
is
necessary as seen
T
be the operator on
+
a„<» +
V
Now T is a linear mapping not invertible.
We now give an important application of the above theorem to systems of linear equations over K. Consider a system with the same number of equations as unknowns, say n. We can represent this system by the matrix equation Ax =
b
(*)
K which
we view as a linear operator on K". Suppose has only the zero solution. equation Ax = matrix the matrix A is nonsingular, i.e. the This means that the onto. and one-to-one Then, by Theorem 6.9, the linear mapping A is suppose the matrix hand, other the G K". On system (*) has a unique solution for any b Then the linear solution. = nonzero has a A is singular, i.e. the matrix equation Ax does not have a which for b G K" exist (*) mapping A is not onto. This means that there proven the have Thus we unique. is not Furthermore, if a solution exists it solution. where
A
is
an n-square matrix over
following fundamental result:
Theorem
6.10:
Consider the following system of linear equations:
a2lXl
+ ai2X2 4+ a22X2 +
a„lXl
+
anXi
an2X2
+
•
•
•
•
•
•
•
•
+ amajn = + a2nXn = +
annXn
=
bi b2
&«
CHAP.
LINEAR MAPPINGS
6]
131
(i)
If the corresponding homogeneous system has only the zero solution, then the above system has a unique solution for any values of the bu
(ii)
If the corresponding (i)
homogeneous system has a nonzero solution, then: 6i for which the above system does not have whenever a solution of the above system exists, it is
there are values for the
a solution; not unique.
(ii)
Solved Problems
MAPPINGS 6.1.
State whether or not each diagram defines a {x,y,z}.
mapping from
B=
(i)
6.2.
(i)
No. There
(ii)
No.
(iii)
Yes.
Two
is
(i)
(ii)
(iii)
(ii)
are assigned to c
z,
(i)
Since / assigns to any /(4)
(ii)
number z
=
43
Since g assigns 5 to any
number
4,
—2 and
=
Two
its
cube
number
x,
we can
=
5,
0,
=
by
-8,
g by g{x)
define
/(») /(O)
=
= =
a:^.
0^
Also:
=
Thus the value of g at each
5.
=
42
=
16.
On
-
5,
ff(0)
=
5
follows:
=
'a;2
6
>
0.
define /
(-2)3
fl-(-2)
h as
h{x)
h(4)
into R.
is 5:
different rules are used to define
Since 4
R
and to each nonpositive number
square,
we can
=
from
5.
—2 and
a^,
/(-2)
64,
flr(4)
(iii)
A.
to define each of the following functions
Also, find the value of each function at 4,
into
G A.
G
To each number let / assign its cube. To each number let g assign the number To each positive number let h assign its let h assign the number 6.
{a, b, c)
v(iii)
nothing assigned to the element 6
elements, x and
Use a formula
A=
the other hand,
if if
-2,
X a;
> =s
^
and so ^-2)
=
6, fe(0)
=
6.
LINEAR MAPPINGS
132
6^.
6
Let A = {1,2,3,4,5} and let f:A-^A be the mapping defined by the diagram on the right, (i) Find the image of /. (ii) Find the graph of /. (i)
(ii)
The image /(A) of the mapping / consists of all the points assigned to elements of A. Now only 2, 3 and 5 appear as the image of any elements of A; hence /(A) = {2,3,5}.
The graph of / consists of the ordered pairs where a&A. Now /(I) = 3, /(2) = 5, /(3) = 5,
=
/(5)
3;
(a, /(a)),
=
/(4)
2,
hence the graph of
=
/
6.4.
[CHAP.
{(1,3),(2,5), (3,5), (4,2),(5,3)}
Sketch the graph of
(i)
f{x)
^x^ + x-6,
(ii)
g{x)
= x^-^x^-x + Z.
Note that these are "polynomial functions". In each case set up a table of values for x and then find the corresponding values of f{x). Plot the points in a coordinate diagram and then draw a smooth continuous curve through the points. (i)
m
X
-4
(ii)
6
-3
X
9{x)
-2
-15
-1
-2
-4
-1
-6
1
-6
2
1
3
-4
3
2 3
6.5.
6
A
(i)
f.A^B
and
g.B^C
be defined by the diagram
B
f
C
g
Find the composition mapping {gof):A-*C. ping: f,g and go f.
We
15
4
Let the mappings
(i)
-3
(ii)
Find the image of each map-
use the definition of the composition mapping to compute:
= g(f(a)) = = ig°f)ib) gim) = (9° me) = g(f(c)) = (gofXa)
Observe that
we
arrive at the same answer if
a
-*
y
-*
t,
b -*
we X
g(y) g(x)
g{y)
= = =
t
s t
"follow the arrows" in the diagram:
-* s,
e
^ y -*
t
CHAP.
LINEAR MAPPINGS
6]
(ii)
By
the diagram, the image values under the
g are
and
r, s
t\
6.6.
mapping
/ are x
and
and the image values under
y,
hence
=
image of /
By (i), 0°f =
I33
=
image oi g
and
{x, y}
{r, s, t}
the image values under the composition mapping gof are {s, *}• Note that the images of g and g°f are different.
and
t
Let the mappings / and g be defined by f{x) = 2x + 1 and g{x) = x^-2. (i) Find (sro/)(4) and (/osr)(4). (ii) Find formulas defining the composition mappings gof and fog. (i)
= 2^4 + 1 = 9. = 42 - 2 =
14.
Hence
Compute the formula
for
g o f as
/(4)
g(A) (ii)
Hence
(ff°f){x)
=
°/)(4)
(ff
=
=
p(/(4))
=
(/ofl')(4)
=
^(9)
=
/(fir(4))
-
92
=
/<14)
2
2
•
=
79.
14
+
1
=
+
4a;
(f°g)(«)
^
f{g(x))
=
=
g(f{x))
fif(2a;
+ l) =
(2a;
+ 1)2 -
=
/(a;2-2)
+
2(a;2-2)
1
=
2a!2
f:A-^B,g:B-^C and h:C-*D
Let the mappings
A
B
f
29.
follows:
=
2
4a;2
Observe that the same answer can be found by writing y = f(x) = 2a! and then eliminating y: z - y^-2 = {2x-¥ 1)^ -2 - 4x^ + 4a! - 1. j/2 - 2,
6.7.
image of
hence
s;
-
+1
1
and
z
=
g(y)
=
fog^g°f.
Observe that
3.
be defined by the diagram
c
g
-
D
h
a
Determine (i)
The mapping /
g.B->C is (ii)
(iii)
each mapping
if
is
A
B
is
(i)
one-one,
(ii)
is
onto,
(iii)
has an inverse.
one-one since each element of A has a different image. The mapping not one-one since x and z both map into the same element 4. The mapping h:C -* D :
->
is
one-one.
The mapping f -.A^B is not onto since « e B is not the image of any element of A. The mapping g:B-*C is onto since each element of C is the image of some element of B. The mapping h-.C^D is also onto.
A
mapping has an inverse
if
and only
if it is
both one-one and onto.
Hence only h has an
inverse.
6.8.
Suppose f:A-*B and g.B^C; hence the composition mapping {gof):A^C exists. Prove the following, (i) If / and g are one-one, then gof is one-one. (ii) If / and g are onto, then
gof
is,
onto,
(iii)
If
gof
is,
one-one, then /
is
one-one.
(iv) If
gof
is
onto, then g is onto. (i)
Suppose (g°f)(x) is
x
one-one,
=
y.
=
(g°f){y).
We
Then
g(f{x))
have proven that
=
g{f(y)).
Since g
=
(g ° f){y)
{g ° f){x)
one-one, f{x) = f(y). Since / implies x = y\ hence flro/ is
is
one-one. (ii)
exists (iii)
is onto,
f(a)
-
there exists b.
Thus
(g
h ''
e.
B for which g(h) = c. Since = g(f(a)) = g(b) = c; hence
f){a)
/
flr
o
is
/
onto, there is onto.
Suppose /
=
f{y).
Thus
it
g°f
is (iv)
c&C. Since g ae.A for which
Suppose
is not one-one. Then there exists distinct elements x,y G A for which /(a;) {g°f)(x) = g{f{x)) = g(f(y)) = (g ° f)(y); hence g°f is not one-one. Accordingly one-one, then / must be one-one.
then (g ° f){a) = g(f{a)) G g{B); hence (g° f){A) C g(B). Suppose g is not onto. properly contained in C and so (g ° f)(A) is properly contained in C; thus g°f is not onto. Accordingly Hgofia onto, then g must be onto. If
aGA,
Then g(B)
is
LINEAR MAPPINGS
134
6.9.
f:A-*B
Prove that a mapping
Suppose / has an inverse,
/0/-1
=
Since 1a
ig.
onto by Problem
Now in
A, say
B
A
6.8(iv).
suppose /
Thus
6.
is
both one-to-one and onto.
is if
defined by
=
f{a)
b,
We
6 1-^6.
then
a
=
and only :
B
by Problem and onto.
Then each
hence /(6)
b;
SB
6
=
-^
is
one-to-one and onto.
f~^°f = 1a and
for which
and since 1b
6.8(iii);
Now
b.
if it is
A
is
onto, / is
the image of a unique element let
g denote the mapping from
have:
(i)
(«?
° /)(<»)
-
fir(/(o))
=
S(b)
=
6
(ii)
{f°ff)(b)
=
f{g{b))
-
fib)
=
6,
Accordingly, / has an inverse.
6.10.
if
there exists a function /-i
one-to-one, / is one-to-one That is, / is both one-to-one
A.
to
i.e.
has an inverse
[CHAP. 6
fix)
every a
a, for
for every 6
Its inverse is the
Let / R ^ R be defined by has an inverse mapping /"^ :
=
G
G A;
hence s
B; hence
mapping
° /
=
1a-
f°g -Ib-
g.
= 2x-S. Now
and onto; hence
/ is one-to-one
Find a formula for
/
f~^.
the Let y be the image of x under the mapping f: y - f(x) =2x-S. Consequently x will be equation: above the of in terms in for x y image of y under the inverse mapping f-K Thus solve X = {y + 3)/2. Then the formula defining the inverse function is f-Hy) = {y + 3)/2.
LINEAR MAPPINGS 6.11.
Show
that the following mappings
^ R2
(i)
F
(ii)
F:R^-*R
(i)
Let v
R2
:
=
defined
defined
(a,b)
and V
We
F
are linear:
+ y, x). = 2x-Sy + 4z.
by F{x, y)=^{x
by F{x,y,z)
w = (a',b'); hence + w - (a + a',b +
=
kv
and
b')
{ka, kb),
= (a' + b', a'). Thus + h+b' a+ F(v + w) = F(a + a',h + b') - (a.-\= (a + 6, a) + (a' + b', a') = F{v) + F(w) F(kv) - F(ka, kb) := (ka + kb, ka) = k(a + b,a) = =
have F(v)
(a
-t-
and F(w)
6, a)
a.'
and Since (ii)
v,
w
and
A;
were arbitrary,
F
,
w = (a',b',c'); hence kv - (ka, kb, and = (a + a',b + b',c + e') V + w We have F(v) = 2a - 36 4c and F(w) = 2a' - 36' 4c'. Thus F(v + w) = F(a + a',b + b',c + c') = 2(a a') - 3(6 + = (2a - 36 + 4c) + (2a' - 36' 4- 4c') = F(v) + F(kv)
and
Accordingly,
Show
F
=
F(ka, kb, kc)
-
=
2ka
F
are not linear:
3kb
that the following mappings
(i)
Let w
:
+
=
4kc
k(2a
is linear.
(iii)
F
R2
=
We
(l,2)
and
have F(v)
w = (3,4); = 1*2 = 2
kc),
fc
eE
-t-
^ R defined by F(x, y) = xy. FrB?^B? defined by F{x, y) = {x + 1, 2y, F:W-^B? defined by F{x, y, z) = (\x\, 0).
(ii)
kF{v)
Let v-(a,b,c) and
-1-
(i)
a')
is linear.
-I-
6.12.
k&R
then v
+w = =3
and F(w)
•
x
+ y).
(A,6).
4
=
12.
Hence
6') -h
4(c
-I-
c')
F(w)
- 36 + 4c) =
kF(v)
CHAP.
LINEAR MAPPINGS
6]
+ w) =
F(v Accordingly,
F
is
(ii)
Since F{0,
=
(1, 0, 0)
(iii)
Let v
=
We
0)
=
have F{v)
^
6.13.
V be
Let
T
=
6
^
24
+
F{v)
F{w)
=
cannot be linear.
-3; hence
= (-3, -6, -9). = -S(1,0) = {-S,0).
kv
and so kF (v)
{1,0)
=
=
F{-3, -6, -9)
Then
#
(3, 0)
fcF('y)
not linear.
M
AM
:
AeV.
is linear.
A,BGV
For any
and any k G K, we have
= (A + B)M + M{A + B) = AM + BM + MA + MB = {AM + MA) + {BM + MB) = T{A) + T{B)
T{A +B)
and
= {kA)M + M{kA) = k{AM) + k{MA) = k{AM + MA) ^
T{kA)
T
Accordingly,
6.14.
•
the vector space of n-square matrices over K. Let be an arbitrary matrix r F -* 7 be defined by T{A) = + MA, where Show that
Let
in F.
F
(0, 0, 0),
Fikv)
F is
and hence
4
not linear.
and k
(1, 2, 3)
=
F(4, 6)
135
kT{A)
is linear.
Prove Theorem 6.2: Let V and U be vector spaces over a field K. Let {^'i, ...,?;„} be a basis of V and let Mi, «„ be any arbitrary vectors in U. Then there exists a unique linear mapping F:V-^U such that F{Vi) = Ui, F{v2) = %2, ., F{Vn) = Un. .
.
,
.
.
There are three steps
F{v^
=
=
Mi, i
Step
Let V
{1).
for which
v
(Since the
osj
...,n.
l,
=
a^v^
to the proof of the theorem:
Show
(2)
eV. + a^v^
F
that
Since {v^,
Show
(3)
that
F
.
F .V ^ U
mapping
a^,
+ a„i;„. We define F:V ^ U by F{v) = a^Ui + a^u^ mapping F is well-defined.) Now, for i= \, ..., n, + Ivi + + Ov„ Vi - Ovi +
-]
are unique, the
such that
unique.
is
a basis of V, there exist unique scalars
.,i;„} is
.
.
linear.
is
Define a
(1)
.
.
.,a„
GX
h a„M„.
-\
•
Hence
Thus the
first step
Step
{2).
and, for any
Suppose
kG K,
F{v)
=
Hence
=
v
kv
OiMi
F{v
a^Vi
= +
ka^v^
{a^
f
1-
•
•
+
•
+
Imj
•
+
•
Om„
=
m;
•
•
+
•
•
•
F{v)
(fci;)
+
+
b2)v2
and
{a^
fc(oiMi
w=
and
+ ka^v^. By
•
+ ajMj + + F(w)
=
{a2
a„M„
+ h-i)ui +
(«!«,
+
hi)Vi
+ kazVz +
+ a„'U„
•
•
•
+ •
•
+
(a„
+
h^Vi
+
=
h^u^
•
•
•
+
Then
h^v^.
6„)v„
mapping F,
+
+
h^Vi
•
•
+
6„m„
+•••+(«„ + 6„)m„
62)^2
+ a„M„) +
•
•
+
definition of the
F{w)
+ O2M2 +
b^Vi
•
•
(6iMi
+ 62M2 +
+ o„mJ =
•
•
+
•
6„M„)
^^(1;)
linear. (3).
a^nVn,
Now
G(t))
suppose
G:V ^V
is
linear and
G{v^
=
M;, t
=
1,
.
.,
.
m.
If
v
=
Oi^i
then G(t))
Since
+
(tti
+
a^u^
+ w) =
and
Step
+ a^v^ +
+w =
= =
f is
+
Omj
of the proof is complete.
V
Thus
=
F{Vi)
=
= =
+ a^v^ + + a2M2 +
G(aiVi OiMj
F{v) for every
•
i?
G
V,
•
•
•
•
+ a„v„) + «„«*„ =
•
G=
F.
Thus
a^G{v-^
+
a2G{v^
+
•
•
•
+
a„G(v„)
ii'(t))
F is
unique and the theorem
is
proved.
+ a^v^ +
LINEAR MAPPINGS
136
6.15.
Let r R2 :
R
-»
be the linear mapping for which
=
r(l, 1)
(Since {(1,1), (0,1)}
Theorem First
write
and
3
as a linear combination of
(a, 6)
=
(a, 6)
(x, x)
+
=
X using
(1)
and
Finally, using (3)
6.16.
unique by
is
we
obtain
=
a
and
+
3/(0,
b,
+
using unknown scalars x and
(0, 1)
(2)
j/(0, 1)
+ y)
x-a,
and so
=
y
b
—
x
+ y=^b
a
(3)
=
T{x(l, 1)
we have
-
T(a, b)
3x
-
D)
=
-
2y
xT{l,
3(a)
-
+
1)
yT(0,
- a) =
2(6
=
1)
-
5a
3x
-
2y
26.
Let T -.V^U be linear, and suppose Vi, ...,Vn&V have the property that their .,Vn images T{vi), .... T{vn) are linearly independent. Show that the vectors vi, are also linearly independent. .
Suppose that, for scalars
=
r(0)
=
ai,
Tia^vi
.
. ,
.
+ ag^a + h a„i'„) =
a^v^
o„,
+ a.2V2 +
Since the T{Vi) are linearly independent, independent.
6.17.
y:
we have
(2)
T(a, b)
(x,x
and
(1, 1)
x{l, 1)
=
(0, y)
Solving for x and y in terms of a and
Now
{1)
T{a, b).
{a, b)
Then
= -2
r(0, 1)
a basis of R^, such a linear mapping exists and
is
Find
6.2.)
we
[CHAP. 6
F:V^U
Suppose the linear mapping
mapping F~^:U -*V
Suppose M.w'G U. which F{v) = u and F(v')
By
=
+ v') -
F
fc'U.
=
•
+
a„r(v„)
Thus the vectors Vi,...,v„ are
0.
Show
one-to-one and onto.
linearly
that the inverse
one-to-one and onto, there exist unique vectors Since F is linear, we also have
+
= u+
F(v')
mapping,
=
F-i(tt)
F-i(m')
v.
=
F{kv)
and
u'
=
kF(v)
-
•«
+
•"'
=
F-i(m)
+
F-HM
and
F-Mm')
=
fei)
=
=
v
+ v'
feF-Htt)
is linear.
IMAGE AND KERNEL OF LINEAR MAPPINGS Let F R* -» R5 be the linear mapping defined by 6.18. F{x,y,s,t) = {x-y + s + t,x + 2s-t,x + v + ?>s-Zt) image U of F, (ii) kernel W Find a basis and the dimension of the :
(i)
(i)
The images of the following generators of R* generate the image F(l, 0,0,0) F(0,
Form
1, 0, 0)
=
(1,1,1)
F(0, 0,1,0)
(-1,
F(0, 0,
the matrix whose rows are the generators of
to
Thus
0, 1)
{(1, 1, 1), (0, 1, 2)} is
„
.
a basis of V; hence dim
U
0, 1)
to
=
of F.
of F:
(1,2,3) (1,
-1, -3)
and row reduce to echelon form:
«
C/
= =
V
2.
for
ku
F-i-{u-\-u')
f',
v,v'BV
Then
F-Mm + m') = and thus F"'
is
O;
•
is
u'.
F{v)
definition of the inverse
F-^ku) =
the
+ a„Vn = 0. Then aiT(vi) + azTiv^) +
•
is also linear.
Since
F{v
all
•
.
and
CHAP.
LINEAR MAPPINGS
6]
(ii)
We
seek the set of
(x, y, s, t)
such that F{x,
= (x-y + s +
F(x, y,s,t)
t,x
y, s,
137
=
t)
(0, 0, 0), i.e.,
+ 2s-t,x + y + Bs-St) =
Set corresponding components equal to each other whose solution space is the kernel of F:
(0, 0, 0)
form the following homogeneous system
to
W
X
—
y+s+t
x
+
y
-
X
= + 3s-3t =
+
X
=
2s
The free variables are
Thus which
6.19.
Let
and
s
(a)
s
=
—1,
f
=
(6)
s
=
0, t
=
t;
hence
to obtain the solution
=
{x
+ 2y — z,
y
+ z,
2t
=
+
C/
dim IT
=
2
+
2
=
4,
+ y — 2z)
X
of T,
The images of generators of R^ generate the image
U
of T:
=
r(i,o,o)
(i)
=
r(o,i,o)
(1,0,1),
U
1
1
{(1, 0, 1), (0, 1,
1
1
1
1
-2
-1)}
is
-1 1-1
T(x, y,z)
=
{x
=
+ 2y - z,
y
i,
-2)
to echelon form:
-1
1
-1
a basis of U, and so dim T(x,y,z)
(-1,
1
to
1
1
seek the set of {x,y,z) such that
=
1
'\
°
to
W of T.
kernel
and row reduce
1
(" 2
(ii)
r(o, o, i)
(2,i,i),
the matrix whose rows are the generators of 1
We
dim
(Observe that
U
Thus
-
s
0),
image
Form
+
be the linear mapping defined by
Find a basis and the dimension of the
(ii)
y
to obtain the solution (1, 2, 0, 1).
l
-1, 0), (1, 2, 0, 1)} is a basis of W. the dimension of the domain R* of F.)
T:W^W
or
—1,
(2, 1,
{(2, 1, is
=
y
T{x, y, z)
(i)
y+s+t
+ s - 2t = 2y + 2s - 4t = dim W = 2. Set
or
t
—
U=
(0,0,0),
+ z,
X
2.
i.e.,
y -2z)
-\-
=
(0, 0, 0)
Set corresponding components equal to each other to form the homogeneous system whose solution space is the kernel of T:
W
X
+
2y y
a;
+
y
—
z
=
+ z = — 2z =
x
+
+ —y —
a
W
= 1. is z; hence dim a basis of W. (Observe that dim sion of the domain R3 of T.)
The only free variable {(3,
6.20.
—1,
1)} is
Find a linear map Method
F
:
R^
-
y
or
a
2y
=
z
X
= =
z
z
Let
U+
+
2y
-
z
y
+
z
or
z
—
dim
1;
W
=
then y = 2 + 1 =
—1 and 3,
which
^ R* whose image is generated by (1, 2, 0, —4) and
= = a;
is
=
3.
Thus
the dimen-
(2, 0,
—1, —3).
1.
Consider the usual basis of R^: e^ = (1, 0, 0), eg = (0, 1. 0), eg = (0, 0, 1). Set F(ei) = (1, 2, 0, -4), F(e2) = (2, 0, —1, —3) and F{eg) = (0, 0, 0, 0). By Theorem 6.2, such a linear map F exists and is unique. Furthermore, the image of F is generated by the F(ej); hence F has the required property. We find a general formula for F(x, y, z): F(x, y,
z)
= = =z
F{xei
+ ye^ + zeg) =
x(\, 2, 0, (x
+
-4)
2y, 2x,
+
xFie-^)
2/(2, 0,
—y, —4x
+
-1, -3)
— 3y)
yF{e2)
+
+
2^(63)
2(0, 0, 0, 0)
'
LINEAR MAPPINGS
138
Method
A
a 4 X 3 matrix
whose columns consist only of the given vectors; say, 2
1
Recall that
6.21.
Let
V
A
determines a linear
-1 -1 -4 -3 -3
map A R3 ^ B^ whose image
map
seek the set of
^\
(^
Fr'
such that
:)
-
C
_
/x \s
X 2x 2s
/-2s \-2s Thus
2x
The free variables are y and (a)
y
—
(6)
y
—
^^"^{(o
—1,
t
0,
—
o)'
Prove Theorem is a subspace of
t
=
2x
-2t = 2s =
hence dim
St
i«
Let
and
(ii)
/X
)
2t\
x
y
2s
+
y
-
x
+
W set
To obtain a basis of
=
=
1,
1,
y
=
y
=
—1, s
0, s
2t
= =
t
s
2.
Let
/« \0
or
W—
.
j
3(
_ ~
y
|
t
3s
\
+ 2y-
+
=
Find a basis and the
DC
{I
_
3y\
M
let
•
q)
2s
to obtain the solution
;)}
6.3:
U
t;
2y
+ +
(p
-
I)
to obtain the solution x
1
G
+
=
J
and
AM — MA.
by F{A^ =
defined
W of F.
K:
(i)
R
be the vector space of 2 by 2 matrices over
be the linear dimension of the kernel
6.22.
generated by the columns of A.
is
:
satisfies the required condition.
F:V^Y We
2\
2
=
A
A
6
2.
Form
Thus
[CHAP.
=
=
0,
0, t
=
t
=
0;
1.
a basis of T^.
F.V^U
be a linear mapping. Then F is a subspace of V.
(i)
the image of
F
the kernel of
GlmF
and a,b& K. Since u and u' belong to G Im F. Now suppose u, u' Since F(Q) = 0, such that F(v) = u and F(v') = u'. Then the image of F, there exist vectors v,v'
GV
F{av
Thus the image of (ii)
is
+
bw)
Thus the kernel of
=
F is
aF(v)
+
hF(v')
=
au
+
e Im F
bu'
a subspace of U.
G Ker F. Since F(0) = 0, to the kernel of F, F{v) = F(av
6.23.
F
+ bv') -
Now
suppose
=
and F(w)
aF(v)
+
bF{w)
==:
v,wG Ker F 0.
Thus
aO
+
60
=
and a,b e K. Since v and
and so
av
w
belong
+ bw S KerF
a subspace of V.
F:V-^U be W = dim V.
Prove Theorem 6.4: Let V be of finite dimension, and let ping with image U' and kernel W. Then dim U' + dim Suppose dim V = n. Since Thus we need prove that dim U'
W
ia
a.
subspace of V,
= n — r.
its
dimension
is finite;
say,
a linear mapdim
W = r — n.
CHAP.
LINEAR MAPPINGS
6]
Let {wi,
.
.
,
.
We
Wr) be a basis of W.
extend {wj to a basis of V: Wr,Vi, ...,i;„_J
{w'l
B =
Let
The theorem
is
Proof that
proved
B
{Wj, Vj} generates
if
we show
{F{Vi),F(v2), ...,F(v„^r)}
B is a basis of the image u S U'. Then there exists
U' of F.
that
generates U'. Let and since v S V,
&V
v
such that F(v)
+ b^-r'^n-r — since the Wj belong to the kernel = F(aiici + + a^Wf + biv^ + + b„^^v„-r) u = = aiF{wi) + + 6„_^F(i;„_,) + a^(Wr) + b^Fivi) + = OjO + + bn-rF(Vn-r) + a^O + biF(Vi) + = b,F(v,) 6„_,FK_,) —
+
OjWj
•
•
•
+
+
a,Wr
^l'"!
+
•
•
u.
Since
•
Note that F(Wi)
are scalars.
ftj
—
V
V
where the a„
139
jF'(t')
•
•
•
•
•
++
•
•
•
•
Thus
of F.
•
•
Accordingly, the F{Vf) generate the image of F.
B
Proof that
is
+
a^Fivi)
Then F(aiVi of F.
Since
Suppose
linearly independent.
+ 02^2 +
•
•
+
•
•
+
•
and so a^Vi + + a„_,T;„_^ a^^^v^-r) there exist scalars 61, 6^ such that •
{wj generates W, a^Vi
or
+
=
a„_ri^K_,.)
=
+
•
a.2F(v2)
.
+
a2^'2
ail^i
+
•
•
•
•
•
+
•
+
an-r'Un-,
—
an-r'Wn-r
=
.
.
belongs to the kernel
•
W
,
+
b^Wi
b^Wi
•
—
•
•
62^2
—
+
•
•
•
+
b^Wr
=
fe^w^
(*)
of the W; and
Since {tWj, «{} is a basis of V, it is linearly independent; hence the coefficients Accordingly, the F(v^ are linearly independent. are all 0. In particular, Oj = 0, ., a„_r = 0.
Thus
6.24.
B
is
a basis of
V, and
Vj in (*)
.
so
dim
V —n—r
and the theorem
is
proved.
f:V-*U is linear with kernel W, and that f{v) = u. Show that the "coset" + W = {v + w: w e W} is the preimage of u, that is, f~^{u) — v + W. Suppose v + T^ c/-i(m). We first prove f~Hu)cv + W and We must prove that v'Gf-Hu). Then f(v') = u and so f(v' - v) = f(v') - f{v) = u-u = 0, that is, v'-vGW. Thus Suppose
V
(i).
(ii)
(i)
= V + (v' — v) €. V + W and hence f~Hu) Cv + W. Now we prove (ii). Suppose v' G v+W. Then v' = + w where w G W. = kernel of /, f(w) = 0. Accordingly, f{v') = /(-u + w) = f(v) + f(w) = /(t)) + v' e /-i(m) and so v + Wc f-^(u). v'
Since
1;
SINGULAR AND NONSINGULAR MAPPINGS 6.25. Suppose F:V ^U is linear and that V is of finite dimension. Show image of F have the same dimension if and only if F is nonsingular. nonsingular mappings
By Theorem mension T.
6.26.
if
6.4,
and only
if
T
:
R*
f(v)
=
W m.
is
the
Thus
V and the Determine all
that
^ R^. + dim (Ker/i^). Hence V and ImF KerF = {0}, i.e. if and only if F is
dim F = dim (Im/f) dim (Ker F) = or
have the same dinonsingular.
Since the dimension of R^ is less than the dimension of R*, so is the dimension of the image of Accordingly, no linear mapping T B* -» R^ can be nonsingular. :
Prove that a linear mapping F:V-*U an independent set is independent. Suppose
F is nonsingular
and suppose
F
•
a^Vy
+
021^2
•
•
+
nonsingular
if
and only
if
the image of
We claim that ., v^} is an independent subset of V. Suppose aiF{Vi) + a<^{v^ +;•••+ a„F{v„) - 0, where + o^vj = 0; hence
{v^,
F(vJ are independent. the vectors F(vi) is linear, F(ajVi + a^v^, + a, e X. Since
is
.
.
•
•
•
+
On^n
^ Ker F
LINEAR MAPPINGS
140
[CHAP.
6
+
But F is nonsingular, i.e. Ker F = {0}; hence a^v^ + a^v^ + a„v„ = 0. Since the i;; are linearly independent, all the a^ are 0. Accordingly, the F(v>i are linearly independent. In other words, the image of the independent set {v^, i)„} is independent. .
.
.
,
On
the other hand, suppose the image of any independent set is independent. If v G nonzero, then {v} is independent. Then {F{v)} is independent and so F(v) 0. Accordingly, nonsingular.
#
V F
is is
OPERATIONS WITH LINEAR MAPPINGS 6.27.
F:W^W
Let (x -z,y). (F
W
and G:W^ be defined by F{x, y, z) = {2x, y + z) and G{x, Find formulas defining the mappings F + G,ZF and 2F - 5G.
+ G)(x,
y, z)
=
(3F)(a;, y, z)
(2F
6.28.
- 5G){x,
= =
z) + G(x, y, z) + z) + (x — z,y) =
{y, x)
(2x,
ZF(x,
=
y, z)
= =
y, z)
y
2F(x, y,
2y
(Ax,
3(2*, z)
z)
(-5a;
F°G
The mapping
Show:
ment
-F =
(i)
{Qx,
z)
=
+ 5z,
+
y
2(2a;,
=
-5y)
-
z)
+ 5z,
(-x
=
=
G(F{x,y,z))
G{2x, y
G
not defined since the image of
is
+
- z,
5{x
-2,y
=
y)
+ 2z)
{2x,y
+ z) and
FoG. =
z)
(y
+ z,
2x)
not contained in the domain of F.
is
U);
0, defined by 0{v) = for every v GV, is the zero elethe negative of F G Hom(7, U) is the mapping {-1)F, i.e.
(ii)
(-l)F.
F G Hom
Let
(i)
-
+ 3z)
By
mapping
the zero
Hom(F,
of
G{x,y)
-z,2v + z)
=
5G{x, y,
+ 22) +
(GoF){x,y,z)
6.29.
-
+
y
(3x
and G/R'^W be defined by F(x,y,z) Derive formulas defining the mappings G°F and
.
=
F(x, y,
F:W-^W
Let
y, z)
{V, U).
+ Q){v) =
{F Since
(F
+ 0)(v) =
GV, + 0{v) = eV, F + =
Then, for every
F(v)
for every
v
v
=
+
F{v)
F(,v)
F(v)
F.
For every v G V,
(ii)
+ {-l)F){v) = F{v) + {-l)F{v) = + {-l)F){v) = 0(v) for every vGV, F + [F
Since
{F
F{v)
-
(-l)F
F{v}
=
=
Thus (-l)F
0.
=
0{v) is
the negative
of F.
6.30.
Show
{aiFi
By Thus by
mapping
aiFj,
...,a„GK, and for any
ai,
aiFi{v)
(a^F^iv)
=
+ aJFiiv) +
•
+
vGV,
ajf'niv)
hence the theorem holds for
a^F^{v);
n =
1.
induction,
Let /^:R3^R2,
+ (I2F2 +
G.W^B?
•
•
+ a„F„)(i;) = =
and
G{x, y, z) = {2x + z,x + y) and are linearly independent. Suppose, for scalars
(Here
U) and
+ a„F„)(i;) =
+ a2F2 H
definition of the
(aiFi
6.31.
Hom {V,
that for Fi, ...,F„G
is
a,b,c
the zero mapping.)
(aF
aiFiCv)
HrR^^R^
•
be defined by
=
i?(a;, y, z)
+ {a^F^ + + a^F^iv) +
(a^F^)(v)
{2y, x).
+ a„F„)(i;) + a„F„(D)
•
•
•
•
= {x + y + z,x + y), F,G,H G Hom (RS R2)
i^'Cx, i/, 2)
Show
that
G K,
aF + bG + cH = For e^ = (1, 0, 0) G R3, we have
+ bG + cH)(e{) = =
aF(l,
0, 0)
a(l, 1)
+
+
bG(l,
6(2, 1)
+
0, 0)
+
c(0, 1)
{1)
cH(l,
=
(a
0, 0)
+ 2b,a + b + c)
i
CHAP.
LINEAR MAPPINGS
6]
and
0(ei)
=
Thus by
(0, 0).
a
=
Similarly for eg
(aF
(0, 1, 0)
(a
+ 2b, a+b + e) =
+
26
{!),
e
aF(0,
Since
6.32.
+
a (2)
(1)
+
we
obtain
implies
(4),
the mappings F,
Prove Theorem Suppose
mapping
in
elements
Vj
{vi,
.
6G(0,
+
=
+
a
+
6
G
+
1, 0)
=
c
=
6
0,
cH(0,
=
c(2, 0)
(2)
=
=
0(62)
(0,0)
=
6
(5)
=
c
0,
1, 0)
(a+2c, a+6)
+
a
and
=
a
(*)
and
H
are linearly independent.
=
m
and dim
Suppose dim y
6.7:
.
+
6(0, 1)
2e
(5)
and
and
1, 0)
a(l, 1)
Thus Using
and so
(0, 0)
we have
R3,
+ bG + cH){e2) = =
=
141
U = n. Then dim Hom {V, U) -
mn.
.,m„} is a basis of V. By Theorem 6.2, a linear .,v„} is a basis of V and {mj, is uniquely determined by arbitrarily assigning elements of t/ to the basis .
.
Hom {V, V)
We
of V.
define
F^ e Hom {V,U),
i
=
1,
.
.
m,
,
.
j
=
...,n
1,
Uj, and Fij(Vk) -0 for fe # i. That is, Fy maps Vi to be the linear mapping for which Fij{v^ theorem into Mj and the other v's into 0. Observe that {Fy} contains exactly mn elements; hence the is proved if we show that it is a basis of Hom {V, U).
=
Proof that {Fy} generates W2, ..., F(Vm) = Wm- Since w^
Wk =
afclMl
Hom (F, G
U,
+
+
«fc2«*2
We now
Hom (V,
=
compute G(Vk), k
t7) is
m
Thus by
F=
G{v^,)
(1),
=
Proof that {Fy}
=
+
ak2'"-2
fc
=
1,
.
.
is linearly
But
=
0(v^)
=
6.33.
22 = l j
all is
Prove Theorem mappings from k&K. Then:
the ay
=
a basis of
6.8:
V
fcCGoF)
(i)
For every v
=
Oy
G
X
(i)
a linear combination of the Fy, the
F=
that
=
G.
k^i
for
=
OfciJ^)cj(vic)
1
+
•
and
^^((Vfc)
=
Mj,
t
2 =
3
2 =
Ofci«j
1
»fcnMn
= w^
for each
Accordingly, by Theorem
fe.
Suppose, for scalars ay
6.2,
G
K,
-
»«^«
1
3
GV,
+
Let V,
U
and
17);
ak2M2
Go(F + F') Go(fcF).
is
+ =
•
•
1,
+
•
.
2 —
.
a^jF^j(v^)
=
1
3
2 —
aicjMi
fflfen^n
.,m,
we have
a^i
—
0,
0^2
=
0,
.
.
.
,
ajj„
=
0.
linearly independent.
hence dim
Hom {V,
U)
=
mn.
W
be vector spaces over K. Let F, F' be linear G, G' be linear mappings from U into W; and let
and let
= 3
hence for k
Hom (V,
= {kG)oF =
(iii)
•
«ii^ij(^ic)
and so {Fy}
into f7
(i)
=l
afcl^l
Mj are linearly independent;
Thus {Fy}
•
^(1;^)
independent.
i
In other words,
m,
. ,
w^, F(v2)
.,w,
=
But the
is
.
.
=
Hom (V, U).
2 2 =
i;^,
Fy('Ufc)
3
i=l
For
G
1,
n
for each k.
w^.
=
we show
if
Since
OiiF«('yic)
+
a^iMj
G; hence {Fy} generates
complete
Since
1
3
fc
fflfc„Mn>
n
i=l
=
+
•
•
...,m.
l,
22=
=
G(i;k)
•
F{vi)
u's; say,
n
TTi
proof that {Fy} generates
Suppose
a linear combination of the
2 2 ayFy i=l i=l
G =
Consider the linear mapping
F G Hom {V, U).
Let
U).
it is
Goi?'
+
Goii'';
(ii)
{G
+ G')oF = GoF + G'oF;
o
LINEAR MAPPINGS
142
(Go(F + F'mv) = G{(F + F'){v)) = G{F(v) +
= Since
{G
(F
°
F'){v)
6
F'(v))
+ G{F'(v)) = {G'>F)(v) + {GoF')(v) = {G ° F + G o F'){v) = (G o F + G ° F'){v) for every vGV, Go {F + F') = G°F + G°F'. G(F{v))
&V,
For every v
(ii)
+
[CHAP.
+ G')°F)(v) = {G + G')(F{v)) = G{F{v)) + G'{F(v)) = (Go F)(v) + {G' °F){v) = (G ° F + G' F)(v) + G') ° F}(v) = {G ° F + G ° F')(v) for every v&V, (G + G')° F = G°F + {(G
Since
({G
GV,
For every v
(iii)
=
(k{G°F))(v)
k(G°F){v)
{k{G°Fmv) = k(GoF){v) =
and
G' °F.
=
k{G{F(v)))
=
k(G{F(v)))
=
=
{kG)(F{v))
=
G{kF{v))
(feG°F)(i;)
G{(kF){v))
=
{G°kF){v)
Accordingly, k{G°F) = (kG)oF = G°(kF). (We emphasize that two mappings are shown to be equal by showing that they assign the same image to each point in the domain.)
6.34.
F:V^V
Let (i)
G.U^W
rank {GoF)
rank (GoF)
By Theorem
(ii)
be linear.
rank (GoF)
(ii)
F{V) c U, we also have
Since
(i)
and ^ rank G,
= dim
Hence {GoF):V^W ^ rank F.
G(F{V)) c G(U)
and so
= dim
((GoF)(y))
is linear.
Show
that
^ dim G(V). Then G(?7) = rank G
dim G(F{V))
(G(F(y)))
^ dim
dim (G(F(y))) ^ dim F(y). Hence
6.4,
rank (GoF)
=
=
dim ((Go F)(y))
dim (G(F(y)))
=£
dim F(y)
=
rank
F
ALGEBRA OF LINEAR OPERATORS 6.35.
T be the linear operators on R^ defined by S{x, y) = {y, x) and T{x, y) Find formulas defining the operators S + T,2S- ZT, ST, TS, S^ and T^.
Let S and (0, x).
=
=
S(x,y) + T(x,y) = {y,x) + (0,x) = {y,2x). = 2S(x,y)-3T{x,y) = 2{y,x) - Z((i,x) = (2y,-x). (ST)(x,y) = S{.T(x,y)) = S(f),x) - (a;,0). (TS)(x,y) = T(S(x,y)) = T(y,x) = {0,y). SHx,y) = S{S{x,y)) = S{y,x) = (x,y). Note S^ = I, the identity mapping. THx, y) = T(T(x, y)) = 7(0, x) - (0, 0). Note T^ = 0, the zero mapping.
{S+T){x,y)
(2S-ZT)(x,y)
6.36.
T
Let
be the linear operator on R^ defined by
=
r(3, 1)
(By Theorem
(2,
-4)
and
T{1, 1)
such a linear operator exists and
6.2,
= is
(i)
(0, 2)
Find T{a,
unique.)
b).
In
particular, find r(7, 4). First write
(a, 6)
as a linear combination of (3,1) and (a, h)
Hence
(a, b)
=
{Sx, x)
+
(y, y)
=
=
a;(3, 1)
{Sx
+ y,
+
(1, 1)
using unknown scalars x and
y(l, 1)
(,2)
'Zx
x
+ y)
and so [^
Solving for x and y in terms of a and
X
Now
using
(2), {1)
and
T(a, b)
Thus
m, 4)
=
(7 - 4,
X
+ +
y y
= —
a b
b,
= ^o —
^6
and
y
= -|a + f 6
(3),
= =
+ yT(l, 1) = -4x) + (0, 2y) = = (3, -1).
xT{3, {2x,
20 - 21)
1)
y:
oo(2, (2a;,
+ 2/(0, 2) -4a; + 2y) = (a-b,5b- 3a) -4)
(5)
CHAP.
6.37.
LINEAR MAPPINGS
6]
Let
143
T be the operator on R^ defined by T{x, y, z) = T is invertible. (ii) Find a formula for T~^.
— y,2x + 3y-z).
{2x, 4a;
(i)
Show
that
W of T
The kernel
(i)
is
the set of all T(», y,
W
Thus
=
z)
2a;
Let
=
be the image of
(r, s, t)
T(x, y, z) = of r, s and t,
we
X
find
Let
V
be of if
i.e.,
(0, 0, 0),
(0, 0, 0)
=
(a;,
We
(x, y, z).
-
Sy
W — {0};
under T; then
s, t)
+
2x
0,
Thus
=
z
hence T
is
nonsingular and so by
s, t) under T^k y and z in terms
y, z) is the image of (r, will find the values of x,
and then substitute in the above formula for T~^. From
=
y
^r,
=
2r
T{x, y,
z)
— s,
=
z
= 7r
-y,2x + 3y-z) =
(2a;, 4a;
— Ss —
Thus T~^
t.
{r, s, t)
given by
is
= (^r,2r-s,lr-3s-t)
s, t)
dimension and let T be a linear operator on V. Recall that T if T is nonsingular or one-to-one. Show that T is invertible
finite
and only
invertible if
and only
=
y
(0, 0, 0).
(x, y, z)
and T-^r,
(r, s, f)
-
4x
0,
r-i(r,
6.38.
-y,2x + Sy-z) =
4x
(2a;,
=
y, z)
the solution space of the homogeneous system
is
which has only the trivial solution Theorem 6.9 is invertible. (ii)
such that T{x,
(x, y, z)
T
is if
onto.
is
By Theorem 6.4, dim V = dim (Im T) + dim (Ker T). Hence the following statements are (i) T is onto, (ii) Im T = V, (iii) dim (Im r) = dimV, (iv) dim (Ker T) = 0, Ker T = {0}, (vi) T is nonsingular, (vii) T is invertible.
equivalent: (v)
6.39.
Let
V
be of
dimension and let T be a linear operator on V for which TS = I, S on V. (We call S a right inverse of T.) (i) Show that T is Show that S = T~^. (iii) Give an example showing that the above
finite
for some operator invertible.
(ii)
need not hold (i)
Let
V is
if
V=
dim
of infinite dimension.
By
n.
(ii)
(iii)
Let
V
=
T(p{t))
+
ai
have
S(T(k))
V
= (r-ir)s =
=
=
S(0)
+
+ a2t+
n = rank
= r-»/=
r-i(rs)
over K; say,
= =
a„«""i
T(S{p{t)))
Oo
+
the identity mapping. 0¥'k. Accordingly,
I,
t
and only if T is onto; hence T / = rank TS — rank T — n.
invertible if
have
=
p(t)
ao
+ Oji + Ojf^ +
=
ajt
On
ST
+
and
= •
=
S{p(t))
+ Oit2 + + o„<« = p{t)
r(aot •
•
•
the other hand,
ii
k
a^t •
G
•
+
is
(TS){x,y) = T(S{x,y)) = T{0,x) the zero mapping: TS = 0.
it
does not assign
For any
{x,y)
= €
=
(0, 0)
R2,
=
{0,0).
Since
TS
=
T(T(x,y))
=
T{x,0)
=
assigns
S(x,0) = (0,x). For example, to every element of R^.
T^x.y)
+ a„t".
•
Let
T
+
a^t^
+
•
•
a„t«+i
K
and
fc
# 0,
then
(,ST){k)
=
¥= I.
T be the linear operators on R^ defined by S{x, y) = Show that TS = but ST # 0. Also show that T^ = T.
(ST){x,y)=S(T{x,y))
•
+ a„<»+i)
Let S and {x, 0).
r-i.
defined by
(rS)(p(«))
and so TS
6.40.
/s
be the space of polynomials in
and S be the operators on
We
=
Then s
/.
is
We
invertible if
rr-i = r-ir =
T
the preceding problem,
and only if rank T = n. Hence rank T = n and T is invertible.
is
=
(Sr)(4, 2)
{x,Q)
=
(0,0)
=
(0,
and T{x, y)
to every
(0, 4).
T{x,y).
x)
(a;,j/)GR2,
Thus ST
Hence T^
¥- 0,
=- T.
it
since
LINEAR MAPPINGS
144
[CHAP. 6
MISCELLANEOUS PROBLEMS 6.41.
Y
Let {ei, ei, 63} be a basis of Furthermore, suppose
= = -
T{ei) 2^(62)
T(e3)
Show that,
aifi &i/i Ci/i
GV,
for any v
and
+ 02/2 + b2/2 + C2/2
T-.V^U
Let
TJ.
A = ('''/
and
be linear.
''^
^
=
A[v]e
a basis of
{/i, /z)
where the vectors
[T{v)]f
in
K^ and K^ are written
as column vectors. Suppose V
=
fejei
+ fc2e2 + ^363; T(v)
then
=
kiT{ei)
-
kiiaJi
=
(Olfcl
Accordmgly,
+
[f]e
+
k^T^e^)
+ 0^2) +
+
h\ ^2
=
feifeg
ksTie^)
kiibJi
+ Cifcg)/! +
=
[Tiv)],
Also,
{ ^,\
+
62/2)
{a2ki
+
+
hi'ifi
+
"2/2)
+ C2fc3)/2
62^2
+
,[,1
+
:^,l
Computing, we obtain
^Me
6.42.
Let is
A;
=(„_._
Hence T T
Suppose
and so kT
is
is
singular.
that a linear singular if and only if
Then
T{v)
=
-
+ b,k2+c,kj
map T
—T
is
for some vector v ¥=
is
f^(^)l'
singular
if
and only
if
kT
=
&0
=
singular. 0.
Hence
ikT){v)
kT(v)
=
singular.
is
Now suppose kT is singular. Then = (kT)(w) = 0. But # and w
kT(w)
6.43.
[a,k,
Show
be a nonzero scalar.
singular.
-
)lfc2)
fc
{kT^(w) — implies
#^
w#
for some vector
kw
¥= 0;
thus
T
is
0;
hence
^(fcw)
=
also singular.
be a linear operator on V for which E^ = E. (Such an operator is termed a the kernel. Show that: (i) if m G C/, Let C/ be the image of E and then £'(m) - u, i.e. £7 is the identity map on U; (ii) if E ^I, then ^ is singular, i.e. E{v) = for some v^O; (iii) V =
Let
£•
W
projection.)
U®W.
(i)
If
u&TJ,
the image of S, then
E{v)
u ^ E(v) (ii)
U E ¥= I
then, for E'(v
(iii)
We
first
show that
some v £ F,
— m) =
S(v)
V - U+
definition,
m = E{v) S E(w)
and thus
We
v&W,
wG
=
J7,
=u — S(m) = Bl'y)
where v
m
v^ m.
—m =
By
(i),
where
using E^
E(u) v
-
—
E, we have
Thus
u.
— u¥=0
e y. Set m = E(v) and w = v- E{v). Then = £(l>) + — £'('U) = M + Of
W. Let v U
By
= u for some v GV. Hence = EHv) = E(E{v)) = E{u)
I)
We now show that w e TF, the kernel = E(v) - E^v) = E(v) - E(v) -
the image of E.
E{v
- E(v))
of E:
V = U + W.
W. Hence
UnW
next show that E{v) = 0. Thus V
=
-
E(v)
{0}.
=
Let v
and
The above two properties imply that
so
eUnW.
UnW
V = U®
-
W.
Since {0}.
vGU,
E(v)
=
v by
(i).
Since
CHAP.
6.44.
LINEAR MAPPINGS
6]
Show that a square matrix A with Theorem 6.9, page 130.)
is
145
and only
invertible if
if it is
nonsingular.
(Compare
Recall that A is invertible if and only if A is row equivalent to the identity matrix 7. Thus the following statements are equivalent: (i) A is invertible. (ii) A and 1 are row equivalent, (iii) The = and IX = have the same solution space, (iv) = has only the zero soluequations tion, (v) A is nonsingular.
AX
AX
Supplementary Problems MAPPINGS mapping from
6.45.
State whether each diagram defines a
6.46.
Define each of the following mappings / (i)
To each number
let
(ii)
To each number
let / assign its
(iii)
R -> R by
/ assign its square plus
To each number — 3 the number —2.
/:R^R
:
a formula:
3.
cube plus twice the number.
let / assign the
be defined by f(x)
{1, 2, 3} into {4, 5, 6}.
number squared, and
= x^-4x + 3.
Find
to each
number <
3 let / assign
- 2a;),
6.47.
Let
6.48.
Determine the number of different mappings from
6.49.
Let the mapping g assign to each name in the set {Betty, Martin, David, Alan, Rebecca} the of different letters needed to spell the name. Find (i) the graph of g, (ii) the image of g.
6.50.
Sketch the graph of each mapping:
6.51.
The mappings f:A-^B, g:B-^A, h:C-*B, diagram below.
(i)
f(x)
=
^x
(i)
/(4),
(ii)
/(-3),
(iii) /(j/
(iv)/(a!-2).
{o, 6} into {1, 2, 3}.
— 1,
(ii)
F-.B^C
g(x)
=
and
2x^
number
— 4x — 3.
GiA^C
are illustrated in the
Determine whether each of the following defines a composition mapping and, if it does, find domain and co-domain: {\)g°f, {n)h°f, (iii) Fo/, (iv)G°f, {y)g°h, (vi) h°G°g. 6.52.
Let
/:R^R
and
fir
:
R -^ R
defining the composition
6.53.
be defined by f(x)
mappings
(i)
f°g,
(ii)
For any mapping f:A->B, show that 1b° f
—
=
x^
g°f, f
—
+ Sx + l (iii)
f°'^A-
g°g,
and g(x) (iv)
f°f.
= 2x-3.
its
Find formulas
LINEAR MAPPINGS
146
6.54.
For each of the following mappings / Sx
- 7,
(ii)
fix)
=
x»
R -> R
:
[CHAP.
formula for the inverse mapping:
find a
(i)
f{x)
6
=
+ 2.
LINEAR MAPPINGS 6.55.
Show (ii) (iii)
jF
6.56.
Show
:
R2
^ R2
R3
-»
R2 defined by F{x,
:
defined
by F(x)
:
defined
by F(x,
=
(iii)
(iv)
F:R2->R
defined by Fix,y)
:
:
V
Let
S :V
->
defined
defined
by Fix)
r(A)
= MA,
Find Tia,
b)
6.60.
Find Tia,
b, c)
(ii)
TiA)
Suppose
W be
Let
defined by
where
d
a, 6, c,
e
R.
(x^, y^).
=
ix
+ l,y + z).
ix, 1).
=
\x-y\.
•
•
•
•
•
•
Show that
over K.
t
+ a^t") = + a„t") =
+
a^t
+
Q
a^t^
ax
+
+
•
+
•
+
a^t
T :V -*V and
mappings
the
a„t" + i
+
aj"--^
M
^
is
:
:
^V
T -.V
R3
^R 1) =
where T RS
F:V -*U
+ dy)
- MA
Til, 1,
6.62.
hy, ex
be an arbitrary matrix in V. matrices over K; and let are linear, but the third is not linear (unless -AM, (iii) TiA) =^ + A.
where T R2
6.59.
6.6L
+
nXn
Let V be the vector space ot that the first two mappings (i)
Zx).
be the vector space of polynomials in V defined below are linear:
+ ait + S(ao + ai« +
x).
+ y).
are not linear:
=
=
{z,x
[ax
by Fix, y,z)
Tiaa
6.58.
(2.x,
F
that the following mappings :
=
=
y)
- y,
{2x
y, z)
defined by F(x, y)
(ii)
are linear:
=
defined by F(x, y)
F R2 ^ R2 F R3 ^ R2 F R ^ R2
(i)
6.57.
:
R -> R2 F R2 ^ R2
(iv)
F
that the following mappings
F F
(i)
is linear.
M
defined
is
by
defined
=
r(l, 2)
-1,
(3,
5)
Show
M=
and
=
r(0, 1)
(2, 1,
0):
-1).
by
=
-2)
3,
r(0, 1,
Show
that, for
and
1
vGV,
any
a subspace of V. Show that the inclusion t(w) = w, is linear.
Fi-v)
map
of
= -2
^(O, 0, 1)
=
-Fiv).
W into
i:W cV
V, denoted by
and
KERNEL AND IMAGE OF LINEAR MAPPINGS 6.63.
For each of the following linear mappings F, find a basis and the dimension of and (6) its kernel W: (i) F R3 -> R8 defined by F(x, y, z) = ix + 2y,y-z,x + 2z). F R2 ^ R2 defined by Fix,y) = ix + y,x + y). (ii)
(a)
its
image
U
:
:
F
(iii)
6.64.
V
Let
:
R3
^ R2
defined
by Fix, y,z)
map defined by FiA) the image U of F.
6.65.
Find a linear mapping
F
6.66.
Find a linear mapping
F
6.67.
Let Dif)
6.68.
Let (ii)
ix
+ y,y + z).
be the vector space of 2 X 2 matrices over
linear (ii)
-
V be the = df/dt.
:
:
= MA.
R3
^ RS
R*
^ RS
R
and
let
M
=
f
j
Find a basis and the dimension of
whose image whose kernel
is
is
generated by generated by
vector space of polynomials in t over R. Find the kernel and image of D.
F:V-^U be linear. Show that the preimage of any subspace of U
(i)
is
Let
(i)
(1, 2, 3)
Let
F V -* V :
the kernel TF of
and
(1, 2, 3, 4)
D:V -*V
.
be the
F
and
(4, 5, 6).
and
(0, 1, 1, 1).
be the differential operator:
the image of any subspace of a subspace of V.
y
is
a subspace of
U
and
CHAP.
6.69.
LINEAR MAPPINGS
6]
Each of the following matrices determines a linear map from
'12 A =
(i)
(
2 ^1
r C
Let
C be
->
:
T(a +
or
space over
the conjugate
= a— bi
bi)
itself,
(ii)
where
Show
2
-1
2
-2/
U
field C. That is, T(z) = z where z G C, R. (i) Show that T is not linear if C is viewed as a vector is linear if C is viewed as a vector space over the real field R.
T
Find formulas defining the mappings
6.72.
H
=
by F{x, y, z) and SF — 20.
defined
:
:
W of each map.
and the kernel
e
OPERATIONS WITH LINEAR MAPPINGS 6.71. Let R3 -» R2 and G R^ ^ R2 be iJ'
B =
(ii) I
mapping on the complex
a, 6
that
into R^:
K,*
1^
-1 -3
Find a basis and the dimension of the image 6.70.
147
F+G
(y,x
+ z)
and G(x,
Let R2 - R2 be defined by H(x, y) — (y, 2x). Using the mappings F and problem, find formulas defining the mappings: (i) and °G, (ii) (in) Ho(F + G) and + H°G. :
H
H°F
=
y, z)
G
(2«,
x
- y).
in the preceding
F°H
G°H,
and
HoF
6.73.
Show
Horn
defined
(R2, R2)
G
and
H are linearly independent:
by
Fix, y) = {x, 2y), G{x, y) = {y,x + y), H{x, y) = (0, x). F,G,He Hom (R3, R) defined by F{x, y, z) = x + y + z, G(x, y,z) — y + z, H(x, y, z) =
(ii)
6.74.
that the following mappings F,
F,G,He
(i)
F,G & Rom
For
{V, U),
show
rank (F
that
+
G)
^ rank
x
i^
— z.
+
rank G.
V
(Here
has
finite
dimension.)
6.75.
F :V -^ U
Let
and G:U-*V be linear. Give an example where G°F
nonsingular.
6.76.
Hom (V,
Prove that
Theorem
U) does satisfy page 128.
6.6,
all
Show is
F
that if
G
and
nonsingular but
G
G°F
are nonsingular then
is
is not.
the required axioms of a vector space.
That
prove
is,
ALGEBRA OP LINEAR OPERATORS 6.77.
6.78.
6.79.
Let S and T be the linear operators on R2 defined by S{x, y) — {x + y, Find formulas defining the operators S + T, 5S - ST, ST, TS, S^ and T^. Let
T
p{t)
=
Show (i)
6.80.
that each of the following operators
Suppose
—
T{x, y)
{x
+ 2y,
3x
T{x, y)
+ Ay).
=
(—y,
x).
Find p(T) where
_ 5f _ 2. = (x-3y- 2z,
T{x, y, z)
sion.
6.81.
be the linear operator on R2 defined by t2
and
0)
y
- 4«,
z),
(ii)
S and T
Show
are linear operators on that rank (ST) = rank (TS)
T on
R^
T{x, y,z)
=
V and that S = rank T.
invertible,
is
{x
is
+ z,x- z,
and
find
a formula for
T~h
y).
nonsingular.
Assume
V
has
finite
dimen-
®
Suppose V = U W. Let Ei and E2 be the linear operators on V defined by Ei(v) = u, = w, where v = u + w, ue.U,w&W. Show that: (i) Bj = E^ and eI = E2, i.e. that Ei and E2E1 = 0. and £?2 are "projections"; (ii) Ei + E2 — I, the identity mapping; (iii) E1E2 =
E2(v)
6.82.
Let El and E2 be linear operators on is
6.83.
the direct
Show that
if
sum
of the image of
the linear operators
V
satisfying
(i),
(ii)
E^ and the image of £2-
S and T
and
(iii)
of Problem 6.81.
Show
that
V
^ — Im £?i © Im £2-
are invertible, then
ST
is
invertible
and (ST)-^
=
T-^S-^.
LINEAR MAPPINGS
148
6.84.
V
Let
Show
have
T be
dimension, and let
finite
TnlmT =
Ker
that
[CHAP.
V
a linear operator on
such that
rank
(T^)
=
rank
6
T.
{0}.
MISCELLANEOUS PROBLEMS 6.85.
T -.K^-^
Suppose
G
vector V 6.86.
6.87.
X"»
a linear mapping. Let {e^, e„} be the usual basis of Z" and let A be columns are the vectors r(ei), Show that, for every ., r(e„) respectively. Av, where v is written as a column vector.
is
.
.
mXn matrix whose
the
=
T(v)
R"-,
Suppose F -.V -* U is linear and same kernel and the same image.
Show
that
F:V -^ U
if
.
.
U-
dim
onto, then
is
.
Show
a nonzero scalar.
is
fc
,
dim V.
that the
Determine
maps
all
F
linear
and kF have the
maps
T:W-*R*
which are onto. 6.88.
Find those theorems of Chapter 3 which prove that the space of w-square matrices over associative algebra over K.
6.89.
Let
T :V ^ U
be linear and
let
W
he a subspace of V.
6.90.
The
wGW.
Tt^-.W^U defined by r^(w) = T{w), for every (iii) Im T^r = T(W). (ii) Ker T^ = Ker T n W.
6.45.
(i)
No,
6.46.
(i)
fix)
to
W
(i)
is
the
an
map
T^^ is linear.
Yes,
(ii)
=
+ 3,
x2
(iii)
/(»)
(ii)
Supplementary Problems
to
No.
=
+ 2a;,
a;3
{"^
=
fix)
(iii)
[-2
- 4xy + 4x2 ^^y + ^x + S,
(iv) a;^
if
»
if
a;
<
(i)
6.48.
Nine.
6.49.
(i)
{(Betty, 4), (Martin, 6), (David, 4), (Alan, 3), (Rebecca, 5)}.
(ii)
Image of g {g o f)
(i)
24,
(ii)
3,
:
A
(iii) j/2
-*
(i)
(/ ° g){x)
(ii)
(f^°/)(a;)
f-Hx)
=
-
A,
No,
(ii)
+ 7)/3,
6.54.
(i)
6.59.
T{a, b)
6.60.
T{a, b,
6.61.
F(v)
+
6.63.
(i)
(a) {(1, 0, 1), (0, 1,
= c)
{-a
=
F{-v)
(iii)
(F o /)
/-!(«)
(ii)
-
=
36
F(v
dim
(ii)
(a) {(1, 1)},
(iii)
(a) {(1, 0), (0, 1)},
o
No,
(iv)
(g o g)(x) /)(a;)
(v)
(goh) :C
a;*
+ Ga;* + 14a;2 + 15x + 5
2c.
-2)},
=
1;
dim
t/
dim
U=
(6) {(1,
=
=
2;
2;
0;
hence F(-v)
(6) {(2,
-1)},
(6) {(1,
-
-F{v).
-1, -1)}, dim
dim T^ -1,
^ A,
=Ax-9 =
= V^^^^.
F(0)
?7
-» C,
(/
+ (-u)) =
-
A
(iii)
- 6).
6,
:
(iv)
7a
+ 26, -3a +
8o
3
{3, 4, 5, 6}.
= 4a;2 - 6a; + 1 = 2a;2 + 6a;-l
(x
3
- 8a! + 15.
6.47.
6.52.
T
is
Two operators S, T G A(V) are said to be similar if there exists an invertible operator P G A{V) for which S = P-^TP. Prove the following, (i) Similarity of operators is an equivalence relation. (ii) Similar operators have the same rank (when V has finite dimension).
Answers
6.51.
restriction of
Prove the following,
K
1)},
=
1.
dim
W=
\.
W=
\.
(yi)
{h°G°g) iB
^ B.
CHAP.
6.64.
LINEAR MAPPINGS
6]
U^
(i)
(")
")'
{(_2
F(x, y,
=
{x
6.66.
F(x, y,z,w)
=
6.67.
The kernel of
6.69.
(i)
(ii)
6.71.
(F
6.72.
(i)
+
+ 4y,
D
2x
+ 5y,
+ y - z,
(x
(a)
{(1,2,1), (0,1,1)}
(6)
{(4,
(a)
ImB =
-2, -5,
G)(,x, y, z)
2x
Sx
0), (1,
+ y - w,
R3;
=
(6)
(y
basis of
-3,
2.
0).
D
is
0, 5)}
Im A; dim(ImA) = 2. basis of KerA; dim(KerA) =
2.
basis of
{(-1,2/3,1,1)}
+ 2z, 2x-y + z),
The image of
(3F
-
KerB; dim(KerB)
2G)(x, y,
z)
=
(3j/
the entire space V.
=
l.
-Az,x + 2y + Zz).
+ z,2y), (H°G)(x,y,z) - {x-y.iz). (il) Not defined. (Ho(F + G)){x, y,z) = {HoF + Ho G)(x, y, z) = {2x-y + z, 2y + 4«). =
{x
+ T)(x, y) = (x, x) (5S - 3r)(a;, y) = (5a; + 8y, (S
SHx,
2.
+ 6y).
the set of constant polynomials.
is
(H°F){x,y,z)
(iii)
6.77.
z)
ImF; dim(ImF) =
^^sisof
l)' (I _2)|
6.65.
KerF; dim(KerF) =
I'asisof
i)|
(o
149
v)
Ti(x^ y)
=
= =
(x
+ y,
0);
-3x)
note that S^
{-X, -y); note that
=
(ST){x, y)
=z
{x- y,
(TS){x, y)
=
(0,
a;
0)
+ y)
S.
T^-\-I
=
Q,
hence
T
is
a zero of
6.78.
v{T)
6.79.
(i)
6.87.
There are no linear maps from RS into R* which are onto.
x'^
+
1.
0.
T-Hr,
s, t)
=
(14*
+ 3s + r,
4t
+ s,
t),
(ii)
T-^r,
s, t)
=
(^r
+ ^s,
t,
^r
- |s).
chapter 7
Matrices and Linear Operators INTRODUCTION
K
and, for v GV, suppose a basis of a vector space V over a field the coordinate vector of v relative to {ei}, which we write as a column vector unless otherwise specified or implied, is
Suppose
V
=
ttiei
{ei,
.
.
+ 0.262 +
•
•
.
•
,
e„} is
+ Omen. Then
\an
Recall that the
mapping v
l^ [v]e,
determined by the basis
{Ci}, is
an isomorphism from
V
onto the space K". In this chapter we show that there is also an isomorphism, determined by the basis from the algebra A{V) of linear operators on V onto the algebra cA of n-square matrices
{ei},
over K.
A
similar result also holds for linear
mappings F:V-^U, from one space
into another.
MATRIX REPRESENTATION OF A LINEAR OPERATOR
K
en} is and suppose (ei, a linear operator on a vector space V over a field of combination is linear each a so r(e„) vectors in V and are Now T{ei), the elements of the basis {e,}:
Let
r be
a basis of V.
.
.
.
r(e2)
= =
T{en)
=
T(ei)
The following Definition:
Example
.
.
.
,
,
01262
02161
+ +
Oniei
+
an2e2
anCi
+ 02262 +
•
•
•
•
•
•
+
•
•
+ + •
+
oi^en a2n6n
o„„e„
definition applies.
The transpose of the above matrix of called the
matrix representation of
matrix of
T
7.1
in the basis
T
coefficients,
denoted by
[T]e or [T], is
relative to the basis {ei} or simply the
{et}:
(On
021
...
Onl
012
022
.
.
fln2
Om
a2n
.
...
'
0,
-* V Let V be the vector space of polynomials in t over R of degree ^ 3, and let D V be the differential operator defined by D{p(t)) = d{p(t))/dt. We compute the matrix of D in the basis {1, t, t^, fi}. We have: :
:
D(l) D(t) D(fi) D{fi)
= = = =
150
= + Of + 0*2 + 0*3 1 = 1 + Ot + 0(2 + 0*3 + 2t + 0f2 + 0«3 It = + Ot + 3t2 + 0t3 3t2 =
CHAP.
MATRICES AND LINEAR OPERATORS
7]
151
Accordingly,
=
[D]
Example
Let T be the linear operator on R2 defined by T(x, y) — (ix — 2y, 2x + y pute the matrix of T in the basis {/i = (1, 1), /a = (-1, 0)}. We have
7.2:
T(Ji)
=
Tifz)
= n-1,
=
r(l, 1)
(2,3)
=
0)
3(1, 1)
=
(-4, -2)
-2(1,
1)
+
3/1
2(-l,
+
/2
=
0)
we
v
its
K
use the usual basis of K".
first theorem tells us that the "action" of an operator T on a vector v matrix representation:
Theorem
7.1:
Let
(ei,
.
.
That
we
e„}
.,
any vector
then
2/2
Av
t^
Our by
+
-2/1
A over defines a linear operator on K" by (where v is written as a column vector). We show (Problem that the matrix representation of this operator is precisely the matrix A
map
7.7) if
=
(-1, 0)
2
Recall that any n-square matrix
the
+
com-
-2
3
(3
Remark:
=
We
V
be a basis of
vGV,
=
[T]e [v]e
and
T
let
is
be any operator on V.
preserved
Then, for
[Tiv)]e.
is, if we multiply the coordinate vector of v by the matrix representation of T, obtain the coordinate vector of T{v).
Example
7.3:
D:V -^V
Consider the differential operator p{t)
= a+bt +
cfi
Hence, relative to the basis
show that Theorem
{1,
t, t^,
7.1
D{p{t))
=
/j
=
=
:
=
(5,7)
T{v)
=
(6,
17)
and
fz
=
[T]f in
m,M,
3dt^
= [D(pm
=
7(1, 1)
=
in
+
17(1, 1)
(-1, 0).
Example
^
7.2,
Example
7.2:
T{x, y)
2(-l, 0)
=
7/1
+
+
0)
=
17/i
11(-1,
=
we
[T(v)]f
— 2y,
2x
+
11/2 {/i, /a),
= 11
verify that Theorem 7.1 holds here:
(r^G) <2y
(4a;
2/2
Hence, relative to the basis
and
Using the matrix
+
o/\
T R2 ^ R2
V
(1, 1)
2ct
o\
1
2
Consider the linear operator Let V = (5, 7). Then
where
+
does hold here:
iO
7.4:
b
=
[D(p{t))]
3,,
Example
Let
7.1.
t^,
and
'0
[D][Pit)]
and so
dt^
=
[p(t)]
We
+
Example
in
=
(ID = i^<*
+
j
MATRICES AND LINEAR OPERATORS
152
[CHAP. 7
Now we have associated By our
on V.
a matrix [T]e to each T in A{V), the algebra of linear operators theorem the action of an individual operator T is preserved by this The next two theorems tell us that the three basic operations with these
first
representation.
operators addition,
(i)
scalar multiplication,
(ii)
composition
(iii)
are also preserved.
Theorem
7.2:
Let {ei, ...,e„} be a basis of V over K, and let oA be the algebra of «-square matrices over K. Then the mapping T h* [T]e is a vector space isomorphism from A{V) onto cA. That is, the mapping is one-one and onto and, for any
S,T G A{V) and any
+ S]e =
[T
Theorem
We V, and
7.3:
T{ei) 7(62)
{T
(T
= =
+ a^ez + 6262
aiei biei
+ S){ei) -
'tti J
l^ttz
Also, for k
EK, we
T{ei)
+ S){e2) =
'^^"®
+ +
r(e2)
ci
&i
k[T]e
[S]e [T]e.
=
ca
bz
+
— =
S{ei) '
8(62)
S(ei)
Suppose
2.
{ei, ez} is
a basis of
+ 0262 diCi + did CiCi
i^i-
-
aiCi
=
(tti
+ 5(62) = =
+ di\ + dzj
+ Ci)ei +
bid (&i
= (:;*; + 0262 + ciCi +
+
6262
+ di)ei +
/«!
&i\
[az
bzj
+ 62)62
(a2
+
6262
+
did
(62
d^ez
+ ^2)62
/ci
di
\cz
dzj
kaiCi
+
ka^ez
kbiei
+
kbzez
,
have (A;r)(ei)
=
fcr(ei)
{kT){ez)
=
kTiez)
+ azBz) = = k(biei + bzez) =
=
k{aiei
kbi\
fkai
Finally,
=
[kT]e
V for which
[^i-Cy
^
=
[ST]e
above theorems in the case dim V
are operators on
Now we have
and
[T]e+[S]e
For any operators S,T G A{V),
illustrate the
T and S
keK,
/ai
,
bi\
,
.-m.
we have
+ a2e2) = aiS(ei) + = ai{ciei + CzCz) + azidiBi + dzCz) = (ttiCi + a2di)ei + (aiCz + azdz)ez
(Sr)(ei) == S(r(ei))
=
S(aiei
=
=
S(biei
(Sr)(e2)
= =
S{T{ez)) bi{ciei {biCi
+ CzCz) +
+ bzCz) =
bzidiBi
+ bzdi)ei +
{biCz
biS{ei)
+
a2S(e2)
b^Siez)
+ dzBz)
+ bzdz)ez
Accordingly, ^
J'
_ ~
/aiCi
[aicz
+ azdi + azdz
biCi
biCz
+ +
bzdi\
bzdz)
_ ~
/ci
[cz
dA
/ai
dz) \az
bi\
bzj
_
,
._
L>IJ«
CHAP.
MATRICES AND LINEAR OPERATORS
7]
153
CHANGE OF BASIS We have shown that we can represent vectors by n-tuples (column vectors) and linear operators by matrices once we have selected a basis. We ask the following natural question: How does our representation change if we select another basis? In order to answer this we
question,
first
Definition:
Let
need a definition. [ei,
.
.
.,e„}
be a basis of
V
and
be another basis.
let {/i, ...,/«}
/i
=
anei
+
ai2C2
+
•
•
•
+
ai„e„
/2
=
aziBi
+
022^2
+
•
•
•
+
a2T.e„
fn
=
ttnlCi
+
a„2e2
+
+
UnnCn
•
•
•
Then the transpose P of the above matrix of coeflScients tion matrix from the "old" basis {d} to the "new" basis
is
Suppose
termed the transi-
{/{}:
'
_
p
We comment
I
fflll
ft21
.
*12
^'22
.
.
ftnl
.
ffin2
.
.
that since the vectors fi, .,fn are linearly independent, the matrix P is In fact, its inverse P^Ms the transition matrix from the basis back to the basis {Ci}. .
.
invertible (Problem 5.47). {/,}
Example
7.5:
Consider the following two bases of R^: {ei
=
(1, 0),
A =
Then
=
/2
62
=
(1,1)
=
(-1,0)
Hence the transition matrix
and
(0, 1)}
(1,0)
=
P from
+
(0,1)
-(1,0)
+
also
have
=
D,
(1,
+
e^
e^
=
(1, 0)
=
0(1, 1)
62
=
(0,1)
=
(1,1)
Observe that
P
-ei
+
0e2
-
+
=
is
-V =
(-1, 0)
(-1,0)
O/i
-
/g
= /1+/2
Hence the transition matrix Q from the basis {/J back
Q
(-1. 0)}
e^
=
0(0,1)
A=
the basis {ej} to the basis {/J '1
We
=
ih
to the basis {e^} is
IN ,
and Q are inverses:
We now show how coordinate vectors are affected by a change of basis. Theorem
We {Ci} to
basis
7.4:
Let P be the transition matrix from a basis space V. Then, for any vector v G V, P[v]f
emphasize that even though the
{fi}
new
basis
P
{/i}, its effect is
{Ci}
=
to a basis {fi} in a vector
[v]e.
Hence
[v]f
=
P~'^[v]e.
is called the transition matrix from the old basis to transform the coordinates of a vector in the new
back to the coordinates in the old basis
{ei}.
:
.
MATRICES AND LINEAR OPERATORS
154
We
illustrate the
matrix from a basis
Now
GF
suppose V
above,
we
+ 0262 + 0363 biBi + 6262 + 6363 C161 + 6262 + 6363
ttiCi
= =
fa
P
above theorem in the case dim F = 3. Suppose {61,62,63} of F to a basis {fufzifa} of V; say,
A = /2
[CHAP. 7
and, say,
v
=
+ /i;2/2 + fcs/s.
fei/i
1 0.1
bi
ci\
02
&2
C2
\aa
ba
Caj
P =
Hence
is
the transition
Then, substituting for the
/i
from
obtain
V
= =
+ a262 + a363) + fc2(biei + 6262 + 6363) + kaiciei + 0262 + 6363) (aifci + bife + Cika)ei. + (azki + bzki + C2ka)e2 + {aaki + bakz + 63^3)63 /i;i(aiei
Thus
+ a^ki + \a3ki +
[v],
=
and
Ikz]
-
[v]e
\lc3l
Accordingly,
/
P[v]f
-
^,^\
^
^^
a2
&2
62^2
\a3
ba
Caj\kaj
I
=
P-'[v]e 7.6:
Let v
-
(a, b)
e
R2.
V
= ^
V
jaikr
=
ttzifci
\a3k1
+ + +
bife
+ cM + 62^3 bakz + Cakaj
bik2
Hence
=
P-'P[v]f
=
I[V],
[V]e
[V],
Then, for the bases of R* in the preceding example, (a, b) (a, 6)
= =
+
b{l,l)
+ (b-a)(-l,0)
6(0, 1)
+ =
ae^
Mf =
and
=-
[v]^
=
a(l, 0)
(J^j
By the preceding example, p-i are given by
verify the result of
-M.
P
the transition matrix
-
I'D We
=
62^2
by P~S we have
Also, multiplying the above equation
Example
+ Cifc3^ 62^2 + dka b3k2 + cakal
jaiki
/jcA
Theorem
be^ bfi
(ft
+
ib-a)/^
_ „)
from {ej
'- =
to {/J
and
its
inverse
(-:
7.4:
= (_:
DC)
= (.!.) =
..
The next theorem shows how matrix representations of linear operators are affected by a change of basis.
Theorem
7.5:
Example
Let P be the transition matrix from a basis {Ci} to a basis {/i} in a vector space V. Then for any linear operator T on F, [T]t = P-i[T]eP. 7.7:
Let T be the linear operator on E^ defined by the bases of R^ in Example 7.5, we have r(ei)
ne^)
= =
r(l, 0)
:=
(4, 2)
r(0,l)
=
(-2,1)
=
4(1, 0)
=
+
-2(1,0)
/4 -2 Accordingly,
[T]e
V2
=
(4a;
2(0, 1)
=
T(x, y)
1
+
(0,1)
- 2j/, 4ei
=
2a;
+
+
j/).
2e2
-2ei
+
e2
Then for
CHAP.
MATRICES AND LINEAR OPERATORS
7]
We
compute
[T]f
using Theorem
7.5:
m, - p-i... = (_: Note that
Remark:
Suppose
P-
{ei, .... en} is
this agrees
155
ixi -ixi
with the derivation of
[T]f in
-I)
Example
-
(i -I
7.2.
is any «-square invertible matrix over a field K. a basis of a vector space V over K, then the n vectors
(Oij)
/i
=
aiiCi
+
02162
+
•
•
•
+
i=l,
a„ie„,
.
.
Now
if
.,n
are linearly independent (Problem 5.47) and so form another basis of V. Furthermore, P is the transition matrix from the basis {«{} to the basis {/{}. Accordingly, if A is any matrix representation of a linear operator T on V, then the matrix B = P~^AP is also a matrix representation of T.
SIMILARITY Suppose A and B are square matrices for vs^hich there exists an invertible matrix P such that B = P~^AP. Then B is said to be similar to A or is said to be obtained from A by a similarity transformation. We show (Problem 7.22) that similarity of matrices is an equivalence relation. Thus by Theorem 7.5 and the above remark, we have the following basic result.
Theorem That
matrices A and B represent the same linear operator they are similar to each other.
Two
7.6:
is, all
the matrix representations of the linear operator
T
if
T form an
and only
if
equivalence
class of similar matrices.
A
linear operator
T
is
if for some basis (Ci} it is represented then said to diagonalize T. The preceding theorem
said to be diagonalizable
by a diagonal matrix; the basis
{«{} is
gives us the following result.
Theorem
7.7:
A
be a matrix representation of a linear operator T. diagonalizable if and only if there exists an invertible matrix Let
P~^AP
is
Then T
P
is
such that
a diagonal matrix.
That is, T is diagonalizable by a similarity transformation.
if
and only
if its
matrix representation can be diagonalized
We emphasize that not every operator is diagonalizable. However, we will show (Chapter 10) that every operator T can be represented by certain "standard" matrices called its normal or canonical forms. We comment now that that discussion will require some theory of fields, polynomials and determinants. Now
a function on square matrices which assigns the same value to similar matrices; that is, f{A) = f{B) whenever A is similar to B. Then / induces a function, also denoted by /, on linear operators T in the following natural way: f{T) = f{[T]e), where {d} The function is well-defined by the preceding theorem. is any basis. suppose /
is
The determinant
perhaps the most important example of the above type of functions. Another important example follows. Example
7.8:
is
The its
trace of a square matrix diagonal elements:
A =
tr (A)
(oy),
= an +
We
written tr (A), 022
+
•
•
•
+
is
defined to be the
sum
of
a„„
show (Problem 7.22) that similar matrices have the same trace. Thus we can speak of the trace of a linear operator T; it is the trace of any one of its matrix representations:
tr {T)
=
tr ([T]g).
MATRICES AND LINEAR OPERATORS
156
[CHAP.
7
MATRICES AND LINEAR MAPPINGS We now consider the general case of linear mappings from one space into another. Let V and U be vector spaces over the same field K and, say, dim V = m and dim U = n. Furthermore, let {ei, em} and {/i, ...,/«} be arbitrary but fixed bases of V and U .
.
.
,
respectively.
F:V^U
Suppose so each
U and
is
Then the vectors F{ei), is a linear mapping. a linear combination of the fc
F{e2)
= =
F{em)
—
F{ei)
The transpose
022/2
+ +
ttml/l
+
dmifl
+
•
•
•
•
•
•
'
•
..,
F{em) belong to
+ ai„fn + aznfn +
Ctmn/n
of the above matrix of coefficients, denoted by [F]l is called the matrix relative to the bases {ei} and {ft}, or the matrix of F in the bases {ec}
representation of
and
ai2/2
ttzi/i
+ +
aii/i
.
F
{/i}: / ftll
ft21
.
.
.
ami
^^12
CI22
.
.
•
ttm2
din
0/2n
_ -
rmf L^Je
\
The following theorems
Theorem
7.8:
.
.
•
dmn
apply.
For any vector v GV,
[F]l [v]e
=
[F{v)],.
multiplying the coordinate vector of v in the basis obtain the coordinate vector of F{v) in the basis {fi).
That
Theorem
is,
7.9:
(ei}
by the matrix
we
The mapping F ^ [F]f is an isomorphism from Hem (V, U) onto the vector space of % X m matrices over K. That is, the mapping is one-one and onto
G G Horn {V, U) and any e K, [kF]f = and [F + G]f = [F]i + [G]/
and, for any F,
Remark:
[F]l,
fc
k[F]i
nxm matrix A
over K has been identified with the linear mapZ" given by v M' Av. Now suppose V and U are vector and n respectively, and suppose {e;} is a basis spaces over K of dimensions of V and {fi} is a basis of U. Then in view of the preceding theorem, we shall given by [F{v)]f = A[v]e. We also identify A with the linear mapping then A is identified with given, are comment that if other bases of V and U Recall that any ping from K'" into
m
F:V^U
another linear mapping from
Theorem
7.10:
Let
and
{ei}, {fi}
into U.
be bases of V, U and be linear mappings. Then
and
G:U-*W
V
{Qi}
[GoFYe
=
W respectively.
Let
F:V-*U
[GYfWVe
That is, relative to the appropriate bases, the matrix representation of the composition of the of two linear mappings is equal to the product of the matrix representations individual mappings.
We lastly show how the matrix when new bases are Theorem
7.11:
representation of a linear
mapping
F:V-*U
is
affected
selected.
P be the transition matrix from a basis {ei} to a basis (e,'} in V, and let Qbe the transition matrix from a basis {/i} to a basis {//} in [/. Then for any linear mapping F:V ^ U, Let
[Ft =
Q-'inp
CHAP.
MATRICES AND LINEAR OPERATORS
7]
Thus i.e.
in particular,
when
the change of basis only takes place in
=
[F]l.
when
i.e.
157
Note that Theorems and 7.11 respectively.
JJ;
and
[F]iP
the change of basis only takes place in V. 7.1, 7.2, 7.3
and
7.5 are special cases of
Theorems
7.8, 7.9, 7.10
The next theorem shows that every linear mapping from one space into another can be represented by a very simple matrix.
Theorem
7.12:
F:V-*U be linear and, say, rankF = r. Then there V and of V such that the matrix representation of F has the
Let
exist bases of
form
I
A =
We
where / is the r-square identity matrix. form of F.
call
A
the normal or canonical
WARNING As noted previously, some texts write the operator symbol V on which it acts, that is, vT instead of T{v)
T
to the right of the vector
In such texts, vectors and operators are represented by n-tuples and matrices which are the transposes of those appearing here. That is, if felCl
+
feez
+
•
+
•
•
knCn
then they write [v]e
And
if
=
(A;i,
fe,
.
.
.,
instead of
kn)
[v]e
r(ei)
=
aiei
+ aid +
•
•
•
+ a„en
T{e2)
=
6iei
+
6262
+
•
•
•
+
r(e„)
=
ciei
+
0262
+
•
•
•
+
=
&ne„
c„e„
then they write
[T]e
=
'tti
Oi
bi
b2
lCi
instead of
[T]e
=
C2
is also true for the transition matrix from one basis to another and for matrix repcomment that such texts have theorems resentations of linear mappings F:V ^ U. which are analogous to the ones appearing here.
This
We
MATRICES AND LINEAR OPERATORS
158
[CHAP.
7
Solved Problems
MATRIX REPRESENTATIONS OF LINEAR OPERATORS 7.1.
Find the matrix representation of each of the following operators the usual (i)
T{x, y) Note
(i)
first
r(ei) T{e^)
(ii)
r(ei) Tie^)
7.2.
R^ relative to
that
r(l,0)
= =
r(l,0)
S
(a, b)
if
= =
= =
T{Q,1)
(2,-1)
= =
r(0,l)
then
R2,
(0,3)
(3,1)
(-4,5)
=
(a, b)
= =
0ei
+
2ei-
e^
= =
3ei+
62
+
-461
ae^
3e2
+
be^.
/O
and^
.^, rri.
=
(
and
[rig
=
(
2
\S„ -1 /3 -4
\1
562
5
Find the matrix representation of each operator T in the preceding problem relative to the basis {A = (1,3), /a = (2, 5)}. We must first find the coordinates of an arbitrary vector (a, b) G K^ with respect to the basis {/J.
We
have
=
(a, b)
or
X
or
a;
Thus (i)
We
have
T{x, y)
= =
r(/2)
(ii)
+
x(l, 3)
2y
=
We
have T(x, r(/i)
T(h)
Suppose that T
=
Sx
= =
r(l,3) r(2, 5)
- y).
is
=
+ 2y,
=
and
Sx
+
5y
—
5a
and
y
~
3a
=
(26
(3a
- 6)/2
- 5a)/i +
+ 5y)
Zx
—
b 6
Hence
(6,0) (10, 1)
= =
-3O/1
-48/i
+ +
I8/2
and
=
[T]f
29/2
30
-48
18
29
(3x
and
[T]f
77
124
-43
-69
the linear operator on R^ defined by z)
=
that the matrix of
(ttiic
T
+ a2.y + aaz,
-
the rows of [T]e are obtained ponents of T{x, y, z). T{ei) 7(62) 7(63)
= = =
bix
+ h^y + bsz,
cix
in the usual basis (ei} is given
is,
T{1, 0, 0) T(0, 1, 0)
r(0, 0, 1)
= = =
lT]e
ai
a2
as
61
&2
bs
Ci
C2
Cs
from the (ai, 61, Ci) (02, 62, C2) (aa, 63, C3)
=
+ dy + Cs^)
by
coefficients of x,
= = =
a^ei 0361 agei
/ai
03
aaX
(h
62
63
Accordingly,
Remark:
(x
a
[T]e
That
=
2/(2, 5)
— 4y,x + 5y). Hence = r(l,3) = (-9,16) = llfi-ASf^ = r(2,5) = (-14,27) = 124/1-69/2 y)
T{x, y,
Show
(2y,
+
=
26
(a, 6)
r(/i)
7.3.
T on
basis {ei = (1, 0), 62 = (0, 1)}: = {2y, Sx - y), (ii) T{x, y) = (3x -4y,x + 5y).
+ + +
b^ez 6262
6363
+ + +
y and
z in the
c^e^ 6363 6363
1
This property holds for any space K'^ but only relative to the usual basis {ei
=
(l, 0,
...,0), 62
=
(0,1,0, ...,0),
..., e„
=
(0,
...,0,1)}
com-
CHAP.
7.4.
MATRICES AND LINEAR OPERATORS
7]
Find the matrix representation of each of the following linear operators relative to the usual basis (i)
T{x,y,z)
(ii)
T{x,y,z)
By Problem
7.5.
159
=
(ei
=
(1, 0, 0), 62
=
(0, 1, 0), 63
(i)
(ii)
on R^
- {2x-Zy + Az,5x-y + 2z,Ax + ly), = {2y + z,x-4:y,Zx). 7.3:
=
[T]^
(i)
|
-1
5
2
=
[T]^
(ii)
,
)
-4
1
\
+ z,x — Ay, 3a;). (1, 1, 0), fa = (1, 0, 0)}
= Find the matrix of T in the basis {/i = (1, 1, 1), /a = Verify that [T], [v]f = [T{v)]s for any vector v G R^.
T be
Let
T
(0, 0, 1)}:
by
the linear operator on R^ defined
T(x, y, z)
{2y
We
must first find the coordinates of an arbitrary vector (a, h, c) G R^ with respect to the basis {fvfz'fai- Write {a,b,c) as a linear combination of the /j using unknown scalars x, y and z:
=
x{l, 1, 1)
=
(x
(a, b, c)
+
+
y{l, 1, 0)
z(l, 0, 0)
+ y + z,x + y,x)
Set corresponding components equal to each other to obtain the system of equations
X
+
y
+
z
=
Solve the system for x, y and z in terms of (a, 6, c)
Since
(i)
r(/i) r(/2)
T{fa)
(ii)
=
T(x,y,z)
Suppose v
= = = ~ V
a, b
c/i
+
and
y
=
x
b,
c to find
+ (6 - c)/2 +
x
=
=
c
c,
y
=
b
~e,
z
= a — b.
Thus
(a - 6)/3
+ z, - 4j/, 3a;) = (3,-3,3) = 3/1-6/2 + 6/3 = (2,-3,3) = 3/1-6/2 + 5/3 = (0,1,3) = 3/1-2/2- /a a;
(2j/
r(l,l,l) r(l,l,0) r(l,0,0)
and
[T]f
and so
[v]
then
(a, b, c);
=
=
X
a,
(a,b,c)
=
c/i
+
(6
— c)/2 +
(a
—
6)/3
Also,
= =
T{v)
T(a,
b, c)
3a/i
+
=
(-2a
(26 + c, a - 46, 3(i) - 46)/2 + (-0 + 66 + c)/3
and so
[T{v)]f
Thus [T]f[v]f
7.6.
=
Let
A -
V
written as a column vector).
is
( j
= (1, 0), = (l,3),
(i)
{ei
(ii)
{/i
(i)
Tie,)
and
let
T be
= (0, 1)}, /2 = (2,5)}. 62
6-c
-6 -6 -2
-2a-45
=
lT(v)]f
the linear operator on R^ defined by T{v)
i.e.
Find the matrix of
T
in
-
3e,-
=(:) =2ex + 4e2
= Av
(where
each of the following bases:
the usual basis;
2yi\ ^ =(l (I) = u, + \3 4/\0/ ^3/
^(^^)={3 4)(l)
=
/I
2
MATRICES AND LINEAR OPERATORS
160
[CHAP.
7
Observe that the matrix of T in the usual basis is precisely the original matrix A which This is not unusual. In fact, we show in the next problem that this is true for any matrix A when using the usual basis. defined T.
By Problem
(ii)
7.2,
=
(o, 5)
(26
X)
^<« = (a
- 5o)/i +
©
=
- h)!^.
(3o
Hence
-'-'
=
and thus
/-5
-8N
6
loy
[r]/
V
7.7.
any «-square matrix A = {an) may be viewed as the linear operator T on K" by T{v) = Av, where v is written as a column vector. Show that the matrix representation of T relative to the usual basis {et} of K" is the matrix A, that Recall that defined
[T]e
is,
= A. I Oil
T(ei)
=
Ol2
•
••
«lJl
fail
\/ 1
—
=
Aei
OiiBi
+
021^2
+
+
a„ie„
+
+
«n2
+
Onnen
<»12
=
Aea
T(ei)
aji2j
r(e„)
= Ae„ =
Oil
<*12
«2i
«22
«ln«l
I
(That
is,
r(e,)
=
Ae,
is
the ith column of A.)
me
= ,
7.8.
+
02n«2
+
•••
Accordingly,
0.11
%2
•
•
•
''itl
021
*22
•
•
•
<*2n
Onl
<'^n2
a.nn
\
= A J
of the sets (i) {l,t,e\te*) and (ii) {f^\t^*,t^e^*} is a basis of a vector space V be the differential operator on V, that is, D{f) — df/dt. of functions / R -» R, Let in the given basis. Find the matrix of
Each
D
:
D
(i)
(ii)
= = 1 I?(t) £>(et) = e* i)(«e«) = e« +
I>(1)
te*
= = = =
0(1) 1(1) 0(1) 0(1)
+ + + +
= = 3e»« = D(
+ + 0(t) + 0(t) + 0(t)
0(e«)
0(t)
0(et)
3(e30 l(e30 0(e3«)
l(eO l(e«)
+ + +
+ + + +
O(te')
O(te')
and
0(«e3') 3(«e3t)
2(«e3')
[D]
=
0(
+ 0(t263t) + 0(t2e3t) + 3(t2e3')
and
[D]
=
CHAP.
7.9.
MATRICES AND LINEAR OPERATORS
7]
Prove Theorem 7.1: Suppose {ei, on F. Then for any vGV, [T]e [v]e .
Suppose, for
—
i
1,
.
.
V
.,e„} is a basis of
=
161
and T
is
a linear operator
[T{v)]e.
.,n,
.
n
=
Tiei)
Then
+
Bjiei
+
ffljaea
matrix whose ith row
[r]e is the n-square
•
•
+
•
2
=
Oi„e„
%«}
is
n
Now
suppose
=
V
+
k^ei
+
kzBz
•
•
+
•
2
=
fc„e„
K^i
Writing a column vector as the transpose of a row vector,
=
[v],
...,fe„)t
fcj,
(&i,
(2)
Furthermore, using the linearity of T, T{v)
2 he^ = 2
=
T
=
2(2
(^
n
n
/
j— 1 \ i=l
Thus
[r(i;)]g is
On
the column vector whose jth entry
+
7.10.
=
a«e,-
(oij-fci
+ a2jfc2 ^
h a„^fe„)ej
is
+
a^jk^
the other hand, the ith entry of [r]e[^]e by (2). But the product of (1) and (2) [T], [v],
2
i—
/
•
+
•
a„j&„
(^)
obtained by multiplying the ;th row of [T\g by [v]^, hence [r]c[v]e and [T(v)\g have the same entries.
is
i.e. (1)
Thus
2
=
) «j
aijfci
fci (^
n
\
««*>
2
hned =
is (3);
[T(v%.
V over X^, and let cA be the algebra Then the mapping T ^ [T]e is a vector space isomorphism from A{V) onto cA. That is, the mapping is one-one and onto and, for any S,T& A{V) and any kGK, [T + S\e = {T]e + [S\e and [kT]e = k[T]e. Prove Theorem
Let
7.2:
{ei,
.
.
.
be a basis of
e„}
,
of %-square matrices over K.
The mapping
is
values on a basis. operator
its
one-one since, by Theorem 8.1, a linear mapping is completely determined by The mapping is onto since each matrix & cA is the image of the linear
M
^
2
-
F(e^
»»«e^
=
i
l,...,n
i=l
where (wy)
Now
is
the transpose of the matrix
suppose, for
i
=
1,
.
.
. ,
M.
w,
n
n
T{eO
Let
A
for
i
and
=
1,
B
be the matrices
2 i=i
=
A=
B=
Observe that
=
S{ei)
Then
(6y).
...,%,
2 i=i [r]^
A + J?
is
the matrix (ay
[T also have, for
i
=
1, ..
=
T(ei)
+ 6y).
+
=
S{ei)
2K +
A'
+
fit
=
[r],+
A* and
is
the matrix
=
(fcay).
k
is
proved.
T(et)
=
fc
2
[S]e
n ayej
=
Accordingly,
[kT],
Thus the theorem
6«)ej
.,n,
n
kA
=
Accordingly,
+ S], = (A+B)t =
(fcrXej)
Observe that
SijCj
„
(r + SKej)
We
and
(ay)
and
cmej
=
(kA)t
=
kAt
=
k[T],
2
ikaij)ej
[5]^
= B*. We
have,
1
MATRICES AND LINEAR OPERATORS
162
7.11.
1
Prove Theorem
S,Te A{V),
Let
7.3:
=
[ST]e
{ei,
.
.
.
,
B=
=
r(ej)
Then
(bjk).
linear operators
n
2 1=1
[T]^
7
[S]e [T]e.
n
Suppose
Then for any
be a basis of V.
e„}
[CHAP.
=
and
"ij^j
A* and
[S]^
=
(ST)iei)
-
2 fc=l
=
S(ej)
i
a«
\i— 1
be the matrices
2
/
n
\
2 =
(
\fc
l
B
= sCSoije,) =
n/«
2 =
and
A=
(ay)
and
We have
B*.
S(7'(ei))
=
A
Let
6jk«/c.
6ifc6fc
IC=1
/
l
/
2(2 \3 =
=
)
a«S(e,)
i— n
\
«k
aijftjic
1
/
n
AB
Recall that
AB =
the matrix
is
where
(cjfc)
=
Cj^
J
=
[ST],
CHANGE OF 7.12.
"iibjk-
=
B*At
Accordingly,
[S]AT]e
MATRICES
BASIS, SIMILAR
Consider these bases of R^: {ei = (1,0), cz = (0,1)} and {/i ^ (1,3), /2 = (2,5)}. (ii) Find the transition matrix Q (i) Find the transition matrix P from {ei} to {/i}. = that [vy = P-^[v]e for any Show (iv) P'K that Q from {/i} to {ei}. (iii) Verify T on R^ defined by = operator for the P-'[T]eP that [T]f vector V eR^ (v) Show T{x, y)
=
{2y,
Sx
- y).
(See Problems 7.1 and 7.2.)
(i)
= =
/i
/2
(ii)
By Problem
7.2,
(a, 6)
61
62
(iv)
= = =
If
i;
=
= =
(1,3) (2,5)
(26
lei 261
- 5a)/i +
and
-5 3
/O
By Problems
7.1
and
7.2;
+
- 6)/2.
3/2
M/ =
Consider the following bases of R«:
{ei
5
V
1)
"
2
-1
3
'
/26-5a\
g^-^)-
(
2\/a\
2\
^
\3
Q^/-5
^^^
and
=
Hence
_ /-5a +26
-1A6/ ~
['ne=(g_^)
/^
Thus
5)(~3 -1) = Co
(3
P-'\vl
(v)
(3a
(0,1)
-5/1
p =
^^^ 562
=2/1-/2
Me=(j,)
then
362
=
/a\ (a,6),
+ +
(1,0)
^« =
(-)
7.13.
=
(AB)t
2 —
3a
I
[T]f
=
(1,0,0), 62
-6 /-30 -48 \ 29 j" (^13
=
(0,1,0), 63
^""^
=
(0,0,1)}
and
= (1,0,0)}. (i) Find the transition matrix P f rom {ei} {/i = (1,1,1), /2 = (1,1,0), /3 that Q = P \ to {/i}. (ii) Find the transition matrix Q from {A} to {ei}. (iii) Verify (iv)
Show
for the (i)
T
that [v]/ = P-^[v]e for any vector v G R^ (v) Show that [T]f = P ^[T]eP defined by T{x, y, z) = {2y + z,x- Ay, 3a;). (See Problems 7.4 and 7.5.) /l
/a fs
= = =
(1,1,1) (1, 1, 0)
(1,0,0)
= Iei+l62+l63 = lei + 1^2 + Oeg = lei + 062 + 063
and
P =
CHAP.
MATRICES AND LINEAR OPERATORS
7]
By Problem
(ii)
(a, b, c)
7.5,
= = =
ei
62
63
=
cf^
(b
(0,1,0) (0,0,1)
'1
(v)
It
=
v
(a, 6, c),
By Problems
then
7.4(ii)
[v],
and
7.5,
[7]^
and
=
[v]f
/O
2
h
-4
l\/o
-1
Prove Theorem 7.4: in a vector space V.
2
0^
=
Let
P
U_
^
\a-
bi
)
Thus
.
1\
/
and
[T\f
l\/l
1
l\
1
1
=
3
3
0/
P[v]f
=
Also,
[v]e.
.
{cj}
[v]f
Thus
3\
= m,
-6 -6 -2 6 5 -1/ \
be the transition matrix from a basis
vGV,
3\
3
3
-6 -6 -2 6 5 -1/ \ /
=
o/\l
0/\3
Then for any
0;
=
I
0/
=0 1-11-4 \l -1
1^
1-1 \l -1
0/\l -1
/O
7.14.
/o
Q =
and
1
= \b\
Thus
1^
1
\3
P-^[T\eP
- b)fs.
(a
iWo
1
\l ,1
(iv)
- c)/2 +
= 0/1 + 0/2 + 1/3 = 0/1 + 1/2-1/3 = 1/1-1/2 + 0/3
(1,0,0)
PQ ^
(iii)
+
163
to a basis {h)
= P-^[v]e.
n
Suppose, for i=l,...,n,
matrix whose jth row
A =
ajiei
+
+
04262
•
•
+
•
aj„e„
is
=
2 ^0^. j=i
Then
P
is
the «-square
(oij, a2j, .... a„j)
(i)
n
+ kj„ =
Also suppose V - kj^ + k2f2+ transpose of a row vector,
[V]f
Substituting for
/j
2
=
(fci,
Then writing a column vector as the
Vj-
*-i
^2, ...,fc„)t
(2)
in the equation for v,
V
=
2v,
=
2^i(i««e,) = i(|«iA)«i
n
= Accordingly,
[11]^
is
2
(aijfci
+ a2jk2 +
(1)
•
•
+ a„jkn)ej
the column vector whose jth entry aijfci
On
•
+
a2jfc2
+
•
•
is •
+
a„jk„
PMf =
row of P by [vh, i.e. same entries and thus
Me-
Furthermore, multiplying the above by P-i gives P~^[v]e
7.15.
(s)
the other hand, the yth entry of Plv]f is obtained by multiplying the ith by (2). But the product of (1) and (2) is (5); hence P[v]f and [v]^ have the
=
P-iP[v]f
=
[v]f.
Prove Theorem 7.5: Let P be the transition matrix from a basis {d} to a basis a vector space F. Then, for any linear operator T on V, [T]t = P-i [T]eP. For any vector
vGV,
P-HT]^P[v]f
=
P-^[T],[v],
=
p-i[T(v)]^
=
[T(v)]f.
{/i}
in
MATRICES AND LINEAR OPERATORS
164
But
[T]f[v]f
=
mapping v
Since the
l-»
Accordingly, P-i[r],P
7.16.
=
hence P-^[T],P[v]f
[T{v)]f;
is
[v]f
=
7
[T],[v]f.
P-i[T]^PX = [T]fX
onto K»,
[CHAP.
X £ iC«.
for every
[7]^.
that similarity of matrices is an equivalence relation, that is: (i) A is similar to A; (ii) if A is similar to B, then B is similar to A; (iii) if A is similar to B and B is similar to C then A is similar to C.
Show
(i)
(11)
(iii)
A = I-^AI, A is similar to A. Since A is similar to B there exists an invertible matrix P such that A = P-^BP. Hence B = PAP-i = (P-i)-»AP-i and P^^ is invertible. Thus B is similar to A. p-iPP, and since Since A is similar to B there exists an invertible matrix P such that A = Hence A = = Q-^CQ. that B such matrix invertible an Q exists similar there to C B is to C. similar A p-iBP = P-^(Q-^CQ)P = (QP)->C(QP) and QP is invertible. Thus is identity matrix / is invertible
The
=
and /
/"i.
Since
TRACE 7.17.
The
A=
trace of a square matrix
= an +
^+
elements: tr (A) is similar to B then tr (A)
=
•
•
+
•
written tr (A), is the sum of its diagonal Show that (i) tr (AB) = tr (BA), (ii) if A
(oij),
a„„.
tr (B). n
(1)
A=
Suppose
B=
and
(a„)
(fty).
Then
AP =
where
(ci^)
n
tr(AP)
=
n
2 i=l
^
aij&jfc-
Thus
n
2 2 i=l }=1
=
Cii
=
Cj^
ttyfeji
n
On
BA =
the other hand,
(d^^)
=
tr(PA)
j
(ii)
If
A
2 =
dji
2 i=l
=
2 2
=
6ii««
=
tr(P-iPP)
Find the trace of the following operator on
We
and
Let
first
tr (T)
must
=
z)
find a
tr ([T],)
=
(aiflj
+ a2y + a^z,
tr
We
must
V
>=1
P
=
2aa6;i 5
=1
such that
=
(PPP-i)
tr(AP)
A = P-^BP.
Using
(i),
tr (P)
R^:
bix
+ h^y + hsz,
Cix
+ Czy + csz)
matrix representation of T. Choosing the usual basis {ej,
=
defined
first find
2
=
di
+
63
+
h
62
&3
Cl
C2
C3 /
C3-
V be the space of 2 x 2 matrices over R, and let
operator on
Thus
6ji«Hic-
similar to B, there exists an invertible matrix
is
T{x, y,
7.19.
=
dj^
3=1 i=l
l
tr(A)
7.18.
where
by T{A) = MA.
Af
=
f
g
^
j
.
Let
Find the trace of T.
a matrix representation of T.
Choose the usual basis of V:
T be the
linear
.
CHAP.
MATRICES AND LINEAR OPERATORS
7]
165
Then
= ME, =
nS,)
^)(J
(^l
=
°)
°)
(^^
=
nE,)
= ME, =
T(E,)
= ME, = (I ^Y° I) ^ (^ I) V4 3 4/Vl 0/
T(E,)
= ME^ =
^g ^^(^J J^
/I
2\/0
ON
(^3
J(^^
J
^J J)
/O
2\
(^^
J
=
Hence
2
/l
tr (T)
=
1
+
1
+
4
+
4
=
+
OE,
+
3^3
+
0^4
=
0^1
+
IE,
+
OE,
+
SE,
::.
2E,
+
OE,
+
4E,
+
OE,
=
OE,
+
2E,
+
OE,
+
4E^
2
4
3
3
\0
and
IE,
0\
10
=
[T]e
=
4^
10.
MATRIX REPRESENTATIONS OF LINEAR MAPPINGS 7:20.
Let (i)
(ii)
if
:
R3
^ R2
be the linear mapping defined by F{x,
F in the following bases of R* and R^: {ft = (1, 1, 1), h = (1, 1, 0), fs = (1, 0, 0)}, {9i = (1, 3), g, = (2, 5)} Verify that the action of F is preserved by its matrix representation; that any vGR^ [F]nv]f = [F{v)],
F(/i)
F(fa)
(ii)
If
7.2,
=
(a, b)
F(l,l,l)
F(1,0,0)
v-(x,y,z)
then,
=
F(v)
(Sx
- 5a)ffi + (3a - b)g2. = (1,-1) = -7g,+ 4g, =
=
=
(3,1)
by Problem
7
r„,
F{Xi, X2,
Show is
.
.
. ,
zf,
+
-
QQ
19
1
Q\
8/
{y
( 8,
- 2O3/ + 26z \ + 11/- 15. )•
^
T^^
- 20j/ + 26«\ + ll.-15. ) =
/-13a;
SJU-H
19
7 4 »ff2
/-13a;
,
[i^(-)]a
for
- z)/, + {x - y)/,. Also, = (-13a; - 20y + 26z)gi + (8a; + lly - 15z)g2,
-
-33 -13 \/
4
F:R^-^K^
v
+ 2y-4z,x-5y + 3z) ,
Let
7.5,
-135^1+
is,
Hence
(26
and
7.21.
+ 2y-4z,x-5y + Sz).
{Sx
Find the matrix of
By Problem
(i)
=
y, z)
==
(8a;
,„, ^, t^(^>^'
be the linear mapping defined by
Xn)
=
{anXi
+
•
•
•
+ amXn,
that the matrix representation of given by
021X1
+
•
•
•
+ aanXn,
F relative to the
'
ai2
...
ttln
0,21
CI22
...
(l2n
flml
(tm2
•
dmnl
.
.
.
,
OmlXi
+
usual bases of
(ttu
.
.
•
•
•
+ amnXn)
K" and
of
K"
\
MATRICES AND LINEAR OPERATORS
166
That
is,
of F{xi,
the rows of [F] are obtained from the coefficients of the .
.
F{0,
7
the components
Xi in
x„), respectively.
.,
^(1,0
0)
1,
.
F{0,0,
7.22.
[CHAP.
.
.,
.
.
0)
,
.
1)
= =
(ail, aai, ...,(i„i) (ai2, a22'
=
••> "m2)
(ai„, (l2r»
•
•>
•
,
rpi
/«n
«12
••
"21
«22
•
•
•
«a
y^ml
«m2
•
•
•
«tr
-
«in
I
"rnn)
Find the matrix representation of each of the following linear mappings relative to the usual bases of R": (i)
(ii)
(iii)
F F F
:
:
:
^ R3 R* ^ R2 R3 ^ R* R2
By Problem
=
defined
by F{x,
y)
defined
by F{x,
y, s, t)
defined
by F{x,
y, z)
we need
7.21,
-y,2x + 4y, 5x - ey)
{Zx
=
+ 2s-U,hx + ly-s- 2t) = {2x + Zy-%z,x + y + z. Ax - 5z, &y) (3a;
-4:y
only look at the coefficients of the unknowns in F{x, y, 3
g\
6
0/
.
.).
.
Thus
(2
7.23.
T:R2^R2
Let
the bases
{ei
(We can view T
own
= =
Tie^)
A =
Let fined
/
(0, 1)}
mapping from one space
Show
W
r(l,0)
r(0,l)
—3
A
Recall that
.
\1 -4
=
= (26-5o)/i + (3a-6)/2. = (2,1) = -8/1+ 5/2 = (-3,4) ^ 23/1-13/2
{a,b)
5
2
(
by F(v)
of (ii)
(1, 0), 62
as a linear
7.2,
r(ei)
(i)
=
the matrix of
7/ Av where v
in
into another, each having its
U=
(1, 1, 1),
f '
^^^ ^'^
^ /-8 \
5
F-.W^B?
de-
F
F
relative to the usual basis of R^
and
relative to the following bases of R^
(1, 1, 0),
h = (1, 0, 0)},
(1 _4
^)
(j _J
^)
{^1
=
(1, 3),
g^
=
(i)
F(1,0,0)
=
F(0,1,0)
=
/
from which
By Problem
W\ = 7.2,
{
(a, 6)
2
5
_.
-,
=
(26
23
"13
written as a column vector.
is
Find the matrix representation of
=
Then
determines a linear mapping
that the matrix representation of is the matrix A itself: [F] = A.
{/i
(ii)
T
of R^ respectively.
basis.)
By Problem
7.24.
= (2x-Zy,x + Ay). Find and {A = (1,3), ^ = (2,5)}
be defined by T{x,y)
=
3\
„
)
=
-A-
- 5a)flri +
1
= (1)
=
=
- 561-462
(_J)
(Compare with Problem
(Za-V^g^.
Then
261
+
7.7.)
162
(2,
5)}
and
R*.
CHAP.
MATRICES AND LINEAR OPERATORS
7]
F(h)
=
F{f2)
=
F(h)
=
5
Prove Theorem basis of
where
/
C7 is
^( 4)
il -4
==
-41flri
+
24fir2
-
-SfiTi
'-
I)
+
5fr2
5
Let F:y-*J] be linear.
7.12:
V-
Then there
F
Ml that {mi,
.
. ,
.
W
mJ
is
=
M2
F('Ui),
a basis of
J7',
{%,
=
...,
^(va),
Mr
=
A =
form
and a
(
\
V
the image of F.
m — r.
r\
Y
exists a basis of
JJ = n. Let be the kernel of F and hence the dimension of the kernel of F is and extend this to a basis of V:
and dim
m,
Set
J7.
8^2
such that the matrix representation A of i^ has the the r-square identity matrix and r is the rank ofF.
Suppose dim
of
+
5
il -4
24
are given that rank F = be a basis of the kernel of
We note
-12flri
:)
-12 -41 ~^^
[F]«
8
7.25.
-
--
(I -4 -?)(;
5
and
167
Let {wi,
.
.
We
.,«)„_ J
F(t)^)
the image of F. Extend this to a basis .
.
.,
M„
Mr+l,
.
.
.,
M„}
Observe that F(t;i)
=
Ml
=
^(va)
=
M2
—
F(i;,)
=
M,
F(Wi)
=0
= =
F(w^_r)
=
Thus the matrix of
F in the
0^2
+
+
1^2
+
+ +
0^2 OM2
+ +
= 0% +
OM2
+
1mi 0*
Omi Omi
+
•
•
•
+ +
Om,
+
Om^
+
1m^
+
OWr
Om^+i
+
•
+
Om^+i
+
•
+
OMy+ 1
+
+
Om^+i
+
+ +
0m„
+ +
0m„
+
0m„
0m„
Om„
above bases has the required form.
Supplementary Problems MATRIX REPRESENTATIONS OF LINEAR OPERATORS 7.26.
Find the matrix of each of the following linear operators {ei
7.27.
(1, 0), 62
=
(0, 1)}:
(i)
=
(2, 3)}.
=
r(», y)
Find the matrix of each operator /2
7.28.
=
T
T
-3y,x + y),
(ii)
T
on R2 with respect to the usual basis
T{x, y)
=
(5x
+ y,Zx- 2y).
in the preceding problem with respect to the basis {T(v)]f for any v e R2. {T\f{v\f
In each case, verify that
Find the matrix of each operator
(2x
in
-
Problem 7.26
in the basis
{g^
=
(1, 3),
g^
=
{/i
(1, 4)}.
=
(1, 2),
MATRICES AND LINEAR OPERATORS
168
7.29.
7.30.
Find the matrix representation of each of the following linear operators usual basis: (i)
T(x,y,z)
(ii)
T{x, y, z)
(iii)
T(,x,
y,z)
= = =
-7y - Az, 3x + y + 4z, (z,y + z, x + y + z)
Let D be the differential operator, vector space V of functions / E (ii)
7.32.
{sin
t,
cos
t},
(ii)
V be T{A)
7.36.
7.37.
7.38.
(iv) {1, t,
the vector space of 2
X
D
= AM,
(iii)
T(A)
T on V
sets is
each basis:
in
(i)
a basis of a {e', e^t, te^'^},
sin St, cos 3t}.
2 matrices over
R
and
=
Af
let
in the usual basis (see
Problem
be the conjugation {1
+ i, 1 + 2i}.
Find the matrix of each
.
)
(
T (ii)
7.19) of V:
(i)
T{A)
= MA,
=^MA- AM.
BASIS, SIMILAR
MATRICES
Consider the following bases of R^: (i)
7.35.
Each of the following
dfldt.
Let ly and Oy denote the identity and zero operators, respectively, on a vector space V. Show that, for any basis {ej of V, (i) [1^]^ = I, the identity matrix, (ii) [Oy]^ = 0, the zero matrix.
CHANGE OF 7.34.
—
Find the matrix of
Consider the complex field C as a vector space over the real field E. Let operator on C, i.e. T(z) = z. Find the matrix of T in each basis: (i) {1, i),
Let
relative to the
- 83/ + z)
6a;
D(f)
i.e.
^ R.
(iii) {e5«, te^*, t^e^t},
of the following linear operators
7.33.
T on R3
(x,y,0) (2x
:
7.31.
[CHAP. 7
Find the transition matrices Verify Q = P-i.
(ii)
Show
that
[v]^
(iii)
Show
that
[T]f
= -
-
(1, 0), eg
and
Q from
{e^
P
for any vector
P[v]f
v
G
=
{/i
=
to
{gj}
{/i
=
^=
(1, 2),
{/J and from
(2, 3)}.
{/j} to {ej, respectively.
fP.
P-'^[T]^P for each operator
Repeat Problem 7.34 for the bases
and
(0, 1)}
(1,2), /a
T
=
in
Problem
(2,3)}
and
7.26.
{g^
=
(1,3),
g^
=
(1.4)}.
Suppose {e^, e^} is a basis of V and T :V -^V is the linear operator for which T^e^) = Se^ — 2e2 and T{e2) = ej + 4e2. Suppose {/i, /a} is the basis of V for which /i = ei + e^ and /z = 2ei + 3e2. Find the matrix of T in the basis {/i, /j}. Consider the bases B — {1, i} and B' = {1 + i, 1 + 2i} of the complex field C over the real field R. (i) Find the transition matrices P and Q from B to B' and from B' to B, respectively. Verify that Q = P-\ (ii) Show that [T]^, = P-'^[T]bP for the conjugation operator T in Problem 7.31.
Suppose {ej, {/J and {flrj} are bases of V, and that P and Q are the transition matrices from {ej and from {/J to {ffj, respectively. Show that PQ is the transition matrix from {ej to {fTj}.
to {/j}
7.39.
Let
A
be a 2 by 2 matrix such that only
A
is
similar to
A = Generalize to
7.40.
wXw
matrices.
c
itself.
Show
that
A
has the form
:)
Show that all the matrices similar to an invertible matrix are invertible. similar matrices have the same rank.
More
generally,
MATRIX REPRESENTATIONS OF LINEAR MAPPINGS 7.41.
Find the matrix representation of the linear mappings relative to the usual bases for R":
(ii)
F F
(iii)
F:R*-*B,
defined
by
(iv)
i^
R ^ R2
defined
by
(i)
R3
-*
:
Ri!
^ R*
:
defined by
=
- 4j/ + 9s,
+ Sy- 2z) F{x, y) = (3a; + 4j/, 5x -2y,x + ly, ix) F(x, y, s,t) = 2x + 3y-7s-t F(x) = (3x, 5x)
R2 defined by F{x,
:
y, z)
(2x
5x
show that
CHAP.
7.42.
MATRICES AND LINEAR OPERATORS
7]
Let (i)
i<'
:
^ R2
R3
be the linear mapping defined by F{x,
Find the matrix of {/i
(ii)
7.43.
=
F
y, z)
=
(2x
169
+ y — z,
Sx
— 2y + iz).
in the following bases of RS and R^:
(l,l,l), /2
=
Verify that, for any vector v
(1,1,0), /a
G
=
(1,0,0)}
[F]^ [v]f
R3,
=
and
=
{jti
(1, 3),
fir^
=
(1,
4)}
[F{v)]g.
Let {ej and {/J be bases of V, and let ly be the identity mapping on V. Show that the matrix of ly in the bases {ej and {/;} is the inverse of the transition matrix P from {e^} to {f^}^, that is,
See Problem
7.44.
Prove Theorem
7.7,
7.45.
Prove Theorem
7.8.
{Hint.
See Problem
7.10.)
7.46.
Prove Theorem
7.9.
{Hint.
See Problem
7.11.)
7.47.
Prove Theorem
7.10.
{Hint.
See Problem
7.15.)
page 155.
{Hint.
page
7.9,
161.)
MISCELLANEOUS PROBLEMS 7.48.
Let
r
be a linear operator on
V
W
— m. T{W) C W. Suppose dim where A is an •m X to submatrix. 7.49.
Let
y =
dim U A and 7.50.
©
W, and
let
U
and
let
Show
W
that
T
be a subspace of
has a matrix representation of the form
G
Two
mX n matrices A
that
if
an operator
and
B
T V
is,
/A B\ .
I
)
y
are said to be similar
F is
diagonalizable, then
over
K
if,
if
if
ON
B
(
I
,
where
there exists an invertible
for any basis {ej of V, the
any similar operator
are said to be equivalent
Suppose
V.
/A ''
respectively.
on
->
:
has a matrix representation of the form
Show that linear operators F and G are similar if and only matrix representations [F]^ and [G]g are similar matrices. Show
invariant under T, that
W each be invariant under a linear operator
= m and dim V = n. Show that T B are mXm and nX n submatrices,
(ii)
V
^
Recall that two linear operators F and operator T on V such that G = T-^FT. (i)
7.51.
1/
and
G
is also
diagonalizable.
there exists an m-square invertible
Q and an n-square invertible matrix P such that B — QAP. Show that equivalence of matrices is an equivalence relation.
matrix (i)
(ii)
Show
that
and only (iii)
Show
if
A and B A and B
can be matrix representations of the same linear operator
that every matrix A is equivalent to a matrix of the form V and r = rank A. (
identity matrix
7.52.
Two
F :V -> U
if
are equivalent. j '
where /
is
the r-square
A and B over a field K are said to be isomorphic (as algebras) if there exists a bijective A -* B such that for u,v S A and k G K, f{u + v) = f(u) + f(v), (ii) /(few) = fe/(w),
algebras
mapping
/
:
(i)
=
f{uv) f{u)f{v). (That is, / preserves the three operations of an algebra: vector addition, scalar onto multiplication, and vector multiplication.) The mapping / is then called an isomorphism of B. Show that the relation of algebra isomorphism is an equivalence relation. (iii)
A
7.53.
Let cA be the algebra of *i-square matrices over K, and let P be an invertible matrix in cA. that the map A \-^ P~^AP, where A G c/f, is an algebra isomorphism of a4 onto itself.
Show
c
MATRICES AND LINEAR OPERATORS
170
Answers /2 -3 7.26.
(i)
7.27.
Here
7.28.
Here
7.29.
(i)
1
=
(a, 6)
(26
=
(a, 6)
1
-2
(4a
- 3a)/i +
- h)gi +
10 10
(6
Supplementary Problems
to
(2a
- b)!^.
- Za)g2.
14
3
6-8
1
2
(i)
^..^
/-23 -39
^"^
(
(")
V-27 -32
-32 -45
35/
15
26
35
41
(iii)
,0
'0
101 5
(iii)
(ii)
2,
,0
25 \
-11 -15 j
« (25
5
1
,18
«
(
0'
1
^.,
-7 -4I
'2
0,
,0
7.30.
6 3
(ii)
\1
[CHAP.
2
o\
1
(iv)
0-3
5, 3
iO
1 7.31.
(i)
(ii)
-1
-3 4 -2 -3 ^c
-6 7.32.
(i)
(iii)
(ii)
P =
7.35.
P =
8 7.36.
7.37.
7.41.
-3
Q =
Q =
2
2
5
-1 -3
11
-2 -1
P =
-4
9
3
-2
(i)
3 7.42.
5
3
-1 -2
(i)
11
-1 -8
2
-1
-1
1
(iii)
(2,3,-7,-1)
6
d—a
c
7.34.
h
a-d
(iv)
—
0/
7
chapter 8
Determinants INTRODUCTION
K
To every square matrix A over a field determinant of A; it is usually denoted by
there
det(A)
or
assigned a specific scalar called the
is
|A|
This determinant function was first discovered in the investigation of systems of linear We shall see in the succeeding chapters that the determinant is an indispensable tool in investigating and obtaining properties of a linear operator. equations.
We comment that the definition in the case
We
of the determinant and most of its properties where the entries of a matrix come from a ring (see Appendix B).
with a discussion of permutations, which
shall begin the chapter
is
also apply
necessary for
the definition of the determinant.
PERMUTATIONS A one-to-one mapping denote the permutation
1
2
.
.
h H
of the set {1,2,
<7
.
.
.,«} onto itself is called a permutation.
We
by
n\
...
or
. .
•
.
—
.
.
3i32
.... =
,
where
3n,
3i
cr(i)
JnJ
.
Observe that since o- is one-to-one and onto, the sequence /i J2 jn is simply a rearrange«. We remark that the number of such permutations is n !, ment of the numbers 1, 2, and that the set of them is usually denoted by S„. We also remark that if <7 € /S„, then the inverse mapping cr"^ G S„; and if a.rGSn, then the composition mapping o-orGSn. In particular, the identity mapping .
.
belongs to Sn.
(In fact,
Example
8.1
Example
8.2:
:
e
—
.
.
.
.
,
12...n.)
There are 2
!
There are 3!
=
2
•
1
=
2 permutations in Sg: 12
= 3'2'1 =
and
6 permutations in S3: 123, 132, 213, 231, 312, 321.
Consider an arbitrary permutation a in Sn. a = ji jz jn. according as to whether there is an even or odd number of pairs .
i>
We then
k
define the sign or parity of
but a,
i
precedes
written sgn
=
A;
.
.
in
by
1
if
<7
is
even
[—1
if
a
IS
odd
r
sgna
21.
J.
171
er
We (i,
say a is even or odd for which
k)
(*)
DETERMINANTS
172
Example
Consider the permutation a
8.3:
3 3,
and 5
—
35142
and are greater than
5 precede
and
is
greater than
identity permutation
e
= 12
even, and 21
is
odd.
Example
8.4:
The
Example
8.5
In S2, 12
is
a
(*),
.
n
.
(3, 1)
hence
2;
and
(5, 4) satisfies
=
r{i)
We
call T
There are
2{j
— i — l) +
Thus the transposition
(Oij)
If
i
(*).
satisfy
(*).
=
1.
(*).
(h^)!
i^t
_
k^
k,
and
i
i,
;
and leaves the
j
then
j,
pairs satisfying
l
ij-l)i{j+l) ...n
...
(*):
=
where x
i),
i+1,
.
.
.,j—l
t is odd.
an «-square matrix over a
A
<
=
T(fc)
i,
= 12 ...{i-l)}{i+l)
(j,i),
DETERMINANT Let A — be
=
r(j)
j,
a transposition. T
(4, 2)
even since no pair can satisfy
is
Let t he the permutation which interchanges two numbers other numbers fixed:
8.6:
satisfy
and
(*).
even and sgn
is
(5, 1)
(3, 2), (5, 2)
and 312 are even, and 132, 213 and 321 are odd.
123, 231
In S3,
.
hence
1;
hence
4;
Since exactly six pairs satisfy
Example
in S5.
and 4 precede and are greater than
5 precedes
:
[CHAP. 8
K:
field
jdll
(tl2
...
0-21
0.22
.
\a„i
CLn2
I
O-ln^ 0'2n
Consider a product of n elements of A such that one and only one element comes from each row and one and only one element comes from each column. Such a product can be written in the form ftlil 0.212
^^'n
where the factors come from successive rows and so the first subscripts are in the .,n. Now since the factors come from different columns, the sequence natural order 1,2, = ji 32 Conversely, each permutajn in Sn. of second subscripts form a permutation matrix A contains n\ such tion in Sn determines a product of the above form. Thus the that
is,
.
.
.
.
.
products. Definition:
The determinant of the w-square matrix is
the following
sum which
is
A=
summed over
denoted by det(A) or permutations o- = ji jz
|A|,
(Odi),
all
.
.
.
h
in Sn. \A\
That
=
2
is,
The determinant denoted by
2/ (sgn
of the w-square matrix
o-)a'UiO,2j^
(sgn
A
is
.
.
.
a„j„
C7)aicr(l) tt2(T(2)
•
•
•
Ona-in)
said to be of order
ail
ai2
ttln
0-21
0.22
0.2n
Onl
0,n2
n and
is
frequently
CHAP.
DETERMINANTS
8]
173
We
emphasize that a square array of scalars enclosed by straight lines is not a matrix but rather the scalar that the determinant assigns to the matrix formed by the array of scalars. Example
Example
8.7:
8.8:
The determinant of a 1 X 1 matrix A (We note that the one permutation in Sj In
1S2,
the permutation 12
is
—
(an)
is
even.)
is
even and the permutation 21
"H
«12
0-1^
a^i
4 -5 -1 -2
Example
8.9:
=
-
4(--2)
-13
(-5
is
itself:
\A\
=
On.
ad
—
be.
Hence
odd.
'*12'*21
''ll'*22
Thus
on
the scalar
and
a
b
e
d
=
In 1S3, the permutations 123, 231 and 312 are even, and the permutations 321, 213 and Hence 132 are odd.
may
This
On
ai2
ai3
0-21
'*22
"•23
<*ll'*22'*33
131
*32
<*33
—
be written Oll(
+
+
<*i2(l23a3l
—
di^a^ia^i
ttl2<*2l'>33
"" <*ll''23*32
as:
~ ''23'*32)
~ ®23*3l) +
*12("21<'33
a22
«23
O32
'''33
~
«21
"^23
«31
"33
+
«12
'*13('''2l'''32
021
022
O3I
032
~ <*22'*3l)
«13
which
is a linear combination of three determinants of order two whose coefficients (with alternating signs) form the first row of the given matrix. Note that each matrix can be obtained by deleting, in the original matrix, the row and column
2X2
containing
its coefficient:
ttll
«12
Oi3 "23
«ii
8.10:
(i)
2
3
4
6
6
7
8
9
"12
<*33
«.l1
Example
-
=
6
7
9
1
-
2
»I2
Oi3
"21
'h.i
«2:i
031
"gi
'*33
5
7
8
1
3
+
+
«13 a31
5
6
8
9
Itj.,
033
4
1
2(6-63) 2
3
-4 2
1
-4 -1
(ii)
<»11
=
—4
-
v
2
-1
5
2(-20
+ 2)
3(5-56)
-
2 3 1
5
+
4(45-48)
+
(-4) 1
27
-4 -1
5
As n increases, the number of terms in the determinant becomes astronomical. Accordwe use indirect methods to evaluate determinants rather than its definition. In fact we prove a number of properties about determinants which will permit us to shorten the computation considerably. In particular, we show that a determinant of order n is equal to a linear combination of determinants of order m. — 1 as in case n = 3 above. ingly,
PROPERTIES OF DETERMINANTS We now list basic properties of the Theorem
8.1:
The determinant
determinant.
of a matrix
A
and
its
transpose A* are equal:
\A\
=
\A*\.
DETERMINANTS
174
By
[CHAP. 8
any theorem about the determinant of a matrix A which concerns the have an analogous theorem concerning the columns of A.
this theorem,
rows of
A
will
The next theorem gives certain cases for which the determinant can be obtained immediately.
Theorem
8.2:
Let
A
be a square matrix.
A has a row (column) of zeros, then \A\ = 0. A has two identical rows (columns), then |A| = 0. If A is triangular, i.e. A has zeros above or below If
(i)
If
(ii)
(iii)
=
\A\
/
is
product of diagonal elements. the identity matrix.
Thus
The next theorem shows how the determinant of a matrix
is
the diagonal, then
in particular,
affected
|/|
=
1
where
by the "elementary"
operations.
Theorem
8.3:
Let (i)
(ii)
(iii)
B be the matrix obtained from a matrix A by multiplying a row (column) of A by a scalar
We now state two Theorem
8.4:
fc;
then
|B|
=
interchanging two rows (columns) of |A|; then |Z?j = — |A|. adding a multiiJle of a row (column) of A to another; then
fe|A|.
jB]
=
|A|.
of the most important and useful theorems on determinants.
A be any n-square matrix. Then the following are equivalent: A is invertible, i.e. A has an inverse A~^. A is nonsingular, i.e. AX - has only the zero solution, or rank A = n, or the rows (columns) of A are linearly independent, |A| ¥= 0. (iii) The determinant of A is not zero:
Let (i)
(ii)
Theorem
8.5:
The determinant is a multiplicative a product of two matrices A and B minants: \A B\
=
is, the determinant of product of their deterequal to the
function. is
That
\A\ \B\
above two theorems using the theory of elementary matrices page 56) and the following lemma.
We
Lemma
We
shall prove the
8.6:
Let
E
Then, for any matrix A,
be an elementary matrix.
comment that one can
also prove the preceding
two theorems
\E A\
-
(see
\E\\A\.
directly without
resorting to the theory of elementary matrices.
MINORS AND COFACTORS Consider an w-square matrix A = (ay). Let M« denote the (w- l)-square submatrix of A obtained by deleting its ith row and .7th column. The determinant \Mij\ is called the mmor "signed" of the element ay of A, and we define the cofactor of Oni, denoted by A«, to be the ^^""^=
A« = i-iy^^m
Note that the "signs" (-1)*+^' accompanying the minors form a chessboard pattern with +'s on the main diagonal:
-
; : ; +
We emphasize that My
-
+
denotes a matrix whereas
::-\
Ay denotes
a scalar.
CHAP.
DETERMINANTS
8]
Example
8.11:
A =
Let
=
M23
Then
175
3
4
5
6
7
{
2
8
= The following theorem
Theorem
8.7
2
3
8
9
9
3
8
9
and
1
=
(-1)2
2
=
-(18-24)
6
applies
The determinant of the matrix A = (Odj) is equal to the sum of the products obtained by multiplying the elements of any row (column) by their respective cofactors: \A\
„
=
+
OiiAii
+
ai2Ai2
•
•
+
•
=
ttinAin
2
o^a-^a
^
OijAij
j=i
and
+
CLljAij
+
Ct2i-'4.2j
•
•
+
•
—
anjAnj
The above formulas, called the Laplace expansions of the determinant of A by the ith. row and the yth column respectively, offer a method of simplifying the computation of \A\. That is, by adding a multiple of a row (column) to another row (column) we can reduce A to a matrix containing a row or column with one entry 1 and the others 0. Expanding by this row or column reduces the computation of \A\ to the computation of a determinant of order one less than that of
Example
8.12:
\A\.
A =
Compute the determinant of
'
5
4
2
2
3
1
l\
—2
-5 -7 -3 1 -2 -1 Perform the following
Note that a 1 appears in the second row, third column, operations on A, where fij denotes the ith row: (i)
add -2R2 to
By Theorem tions; that
(ii)
jRi,
8.3(iii),
add 3^2 to Ra,
li?2 to R^.
the value of the determinant does not change by these opera4
2
1
1-2 -7-3 9
2
=
\A\
3
-5
1-2-1 if
add
is.
5
Now
(iii)
we expand by
4
the third column,
we may
1
-2
2
3
1
2
3
3
1
2
5
-2
1
neglect all terms which contain
Thus
\A\
=
1
_2
2
3
1
2
3
3
1
2
5 1
2
3
1
2
-
1
3
3
2
(an) '
A = 1
over a
field
K:
an
ai2
ttin
0,21
ffl22
tt2re
fflnl
ffln2
+
5
-123
12
3
(-2)
CLASSICAL ADJOINT
A=
=
(-1)2
{
Consider an n-square matrix
1-2
-2
1
2
3
1
=
5
}
38
0.
DETERMINANTS
176
[CHAP. 8
The transpose of the matrix of cofactors of the elements
Oij
of A, denoted by adj A,
is
called
the classical adjoint of A:
'An
A
adj
'^^^
-
A21
...
A„i
^^^
•••
^"^
^nn
\
I
We say "classical adjoint" in
instead of simply "adjoint" because the term adjoint will be used Chapter 13 for an entirely different concept.
Example
8.13:
A
Let
-
The cofactors
An = + ^21
A31
— —
-4 -1
5
2
3
-4
-1
5
3
-4
= + -4
= = =
-18,
-11,
- +
A22
-10,
2
—
A12
~ —
A32
A
of the nine elements of
1
5
2
-4
1
5
2
-4
2
=
Ai3
2,
are
-4
= +
-l|
1
= =
A23
14,
-4,
A33
= -
= +
2
3
1
-1
2
2
3
=
4
=
5
= -8
-4
We form the transpose of the above matrix of cofactors to obtain the classical adjoint of A: I
^
adj
-18 -11 -10
=
2
4
\
Theorem
8.8:
\
14-4 5-8/
For any square matrix A,
A -(adj A) = where
(adj
Thus,
/ is the identity matrix.
if
A) -A
=
\A\I
\A\¥' 0,
Observe that the above theorem gives us an important method of obtaining the inverse of a given matrix.
Example
8.14:
Consider the matrix
/2
A (adj A) =
3
-4 \l -1
A
of the preceding example for which
-4\ /-18 -11 -10\ = 2 14 -4 2 4 5 -8/ 5/\
\A\
We
—46.
0\
/-46
0-46 \
—
/l
= -46 -46/
also have,
by Theorem
\0
-18/-46 -11/-46 -10/-46\
2/-46
14/-46
4/-46
5/-46
-4/-46 -8/-46/
APPLICATIONS TO LINEAR EQUATIONS n
n unknowns:
linear equations in
a2\X\
+ ai2a;2 + + a22a;2 +
anliCi
+
anX\
an%X2
1^
|A|/
8.8,
A-i = r4T(adjA)
Consider a system of
0^ 1
= -46/ =
We
have
+
•
•
•
•
•
•
+ ai„a;» = + ainXn — "T
annXn
bi
&2
bn
/
=
9/23
11/46 5/23 \
-1/23 -7/23 2/23
\-2/23 -5/46 4/23/
CHAP.
DETERMINANTS
8]
177
Let A denote the determinant of the matrix A — (oij) of coefficients: A = \A\. Also, let As denote the determinant of the matrix obtained by replacing the ith column of A by the column of constant terms. The fundamental relationship between determinants and the solution of the above system follows.
Theorem
8.9:
The above system has a unique is given by
solution if
and only
A
if
?^ 0.
In this case
the unique solution
_
^
_
Al
_
An
The above theorem
is known as "Cramer's rule" for solving systems of linear equations. emphasize that the theorem only refers to a system with the same number of equations as unknowns, and that it only gives the solution when A ^ 0. In fact, if A = the theorem does not tell whether or not the system has a solution. However, in the case of a homogeneous system we have the following useful result.
We
Theorem
8.10:
Example
The homogeneous system Ax — A = |A| = 0.
— +
\2x 8.15:
Solve, using determinants:
has a nonzero solution
< 3a;
First compute the determinant
Since
A
52/
A
2
-3
3
5
=
A
Zy
= =
A.
=
7
-3
1
5
=
^x
=
38
=
T"i9
if
1
of the matrix of coefficients:
=
10
+
9
We
=
19
also
2
7
3
1
have -19
38,
Accordingly, the unique solution of the system *'
and only
1
the system has a unique solution.
^^ 0,
if
2,
^
=
is
-19
^y
T=l9-
= -i
We remark that the preceding theorem is of interest more for theoretical and historical reasons than for practical reasons. The previous method of solving systems of linear equations, i.e. by reducing a system to echelon form, is usually much more efficient than by using determinants.
DETERMINANT OF A LINEAR OPERATOR Using the multiplicative property of the determinant (Theorem
Theorem
8.11:
Suppose
A
and
B
are similar matrices.
Then
|A|
8.5),
=
we
\B\.
Now suppose T is an arbitrary linear operator on a vector space V. determinant of T, written det (T), by det(r)
=
obtain
We
define the
|[r]e|
By the above theorem this definition [T]e is the matrix of T in a basis {et}. dependent of the particular basis that is chosen.
where
The next theorem follows from the analogous theorems on matrices.
Theorem
8.12:
Let
T and
(i)
det (S
(ii)
r
is
iS
o
be linear operators on a vector space V.
T)
=
det {S) det {T),
invertible if
•
and only
if
det
(7)^0.
Then
is in-
DETERMINANTS
178
We also remark that det(r)-i
if
T
where Iv
1
the identity mapping, and that det (T~^)
is
-
invertible.
is
Example
=
det (Iv)
[CHAP. 8
Let
8.16:
T
be the linear operator on R3 defined by T(x, y, z)
—
— 4y + z, X — 2y + 3z,
(2x
'2
The matrix
T
of
in the usual basis of
R3
is
=
[T]
X
\
5x
+ y — z)
-4 -2
l\
,5
2-4 1 1-2 3 1-1 5
=
det(r)
2(2
- 3) +
A
over a
4(-l
-
15)
3
.
Then
1-1/ +
1(1
+ 10) = -55
MULTILINEARITY AND DETERMINANTS Let cA denote the set of n-tuple consisting of
its
n-square matrices
all
row vectors
Ai, A2,
A = Hence cA may be viewed as the
The following Definition:
.
.,
.
{Ai, A2,
field
K.
We may
view
A
as an
A„: .
.
.,
An)
set of n-tuples of w-tuples in K:
definitions apply.
A
function
D.cA
components; that if
(i)
if
K
said to be multilinear if
is
it is
linear in each of the
is:
row Ai = B + C, then D{A)
(ii)
-^
= D(...,B + C,...) =
D{...,B,
...)
+
D(...,C,
...);
row Ai = kB where k G K, then D{A)
=
= kD{...,B,
D(...,kB, ...)
...).
We also say n-linear for multilinear if there are n components. Definition:
A function D-.cA^K two
have the following basic
Theorem
8.13:
said to be alternating
if
D{A) =
whenever
A
has
is,
for
identical rows:
D{Ai, A2,
We
is
.
.
An)
.,
result;
=
whenever
D
is
multilinear,
-
Aj, i¥^ j
here / denotes the identity matrix.
There exists a unique function D-.oA -*K such (i)
A,
(ii)
D
is
alternating,
that:
(iii)
D{I)
=
1.
This function D is none other than the determinant function; that any matrix A^cA, D(A) = jA|.
CHAP.
DETERMINANTS
8]
179
Solved Problems
COMPUTATION OF DETERMINANTS 8.1.
a
Evaluate the determinant of each matrix:
(i)
3
-2
4
5
= 3-5-
8.2.
a 23.
=
k
k
4
2k
=
2A;2
-
the determinant
2,
=
4fc
2
'1 (i)
3\
1
2
3
4
-2
3
2
(4
(ii)
2
=
-2
3
5
-1
1
1
=
-3
2
2
2 3
3
5 2
1
3
2-3
10 2-4 13 3
b
=
0.
-
Hence k
0.
= (a—b)(a+b) — a'a
+
and k
Q;
=
2.
That
-62.
is, if
fe
=
or
/
)
(iii)
,
(
2
0\
1^
3
-3
2
, I
3
(iv) I
2 -4
. |
-3
-
4
3
2
-1
2
- 2(-4-6) + -
1
4
-3
5
1
+
+
4
-2
2
5
3
3(20
+ 4) =
|4
2
6
3
ll
=
79
24
1
-1 -3
(iv)
2k
b
\-l -3
-1
5
4
(iii)
4
l\
2 -3
1(2-15)
(ii)
k
+
of each matrix:
/2
3
- 2) =
a
k
5 -ly
^2
(i)
2fc(fe
a
b
a
a
is zero.
Compute the determinant
[4-2
or
0,
—
(ii)
Determine those values of k for which
fe
8.3.
=
(-2) -4
a
b
a
5
^4 (i)
—
(ii)
=
2(10-9)
+
l(-9
+ 2) = -5
5
=
1(6
+ 4) =
10
4
8.4.
Consider the 3-square matrix
A —
\a2
&2
C2
\a3
bs
cs
Show
that the diagrams below can
be used to obtain the determinant of A:
Form the product of each of the three numbers joined by an arrow in the diagram on the and precede each product by a plus sign as follows:
left,
DETERMINANTS
180
[CHAP. 8
Now form right,
the product of each of the three numbers joined by an arrow and precede each product by a minus sign as follows:
— Then the determinant of
\A\
=
A
«!
61
Ci
a2
62
"2
«3
&3
"3
The above method of computing
8.5.
—
asftgCi
Cgagfei
two expressions:
of the above
does not hold for determinants of order greater than
\A\
3.
Evaluate the determinant of each matrix:
(i)
/2
-l\
3
2
A (i)
-3
/a
b
(ii)
(iii)
2 -
3
/
10-2
(iii)
\-2
7/
Expand the determinant by
Use the method
Add
c>
(ii)
3
3/
the second column, neglecting terms containing a
2
0-1
3
2
4-3
=
2
-1
3
2
—
abc
-(-3)
+
3(4
0:
21
3)
7
of the preceding problem:
a
b
c
a
b
b
c
a
c
=
a^
+
63
+
—
c*
abc
—
abc
=
63
+
3a6c
c3
twice the first column to the third column, and then expand by the second row: 3
2-4 + 2(3)
1
-2 +
-2
3
3
Evaluate the determinant of
First multiply the first
row by
A =
+ 2(-2)
-
24|A|
3
= +
3
-1
2
3
-1
=
8
4
2-4 1-4 1
=
-2
2
-1
3-6-2 6-4|A|
2
10
and the second row by
6
2
3
2(1)
fi 8.6.
—
bgC^ai
sum
precisely the
is
diagram on the
in the
3
=
1
1
6 28,
4.
Then
-2 - (3) + 4(3) -4 -(3) -4 + 4(1) 1~(1)
-6 + 4(3)
3
2
and
\A\
=
28/24
1
6-5 14
-7
10 =
7/6.
14
8.7.
Evaluate the determinant of
Note that a (where
i2j
1
A
appears in the third row, first column. Apply the following operations on Thus (iii) add IB3 to R^. (i) add —2R3 to iSj, (ii) add 2R3 to i?2,
denotes the ith row):
A
CHAP.
DETERMINANTS
8]
=
\A\
2
5
2
-3
1
3
1
-6
-3 -2 2 -5 -2 2 4
-1
1 -6 -2 -1 -2 2
3 3
1
-3
3
181
2
-1
= +
-6 + 6(1) 1 +1 3-2 -2 -1 + 6(-2)
8.8.
+2
2
5
1 -6 -2 -1
-3
2
5
5
-1
-3
3
1
+ 6(2)
— —
-2 -13
1
-13 17
= -4
17
2
1
1
-1
A
Evaluate the determinant of
First reduce A to a matrix which has 1 as an entry, such as adding twice the second row, and then proceed as in the preceding problem.
-2 -5
3 |A|
5
=
7
2
-3 -5
8
3
-2 -5 -2 -2
2
8
4
-5 + 2(3) -2
1
7
-3
-2
-3 -5
8
2
4
3
1
2
+ 2(-2)
4
+ 2(-2)
1
-5
3
3
-1
2
4
2 4
3 1
-1
1
1-6 4
-6
1
3
/t
A =
I
\
Add
4-3(3) 3-3(1)
+ 2(-2) -3-3(-2)
-5 + 2(2) 4
8-3(2)
-5-(l)
1
= —
3 1
-3(12
1-13
Evaluate the determinant of
-5 + 2(4)
-1
3
- (3)
2-(-l)
2
3
8.9.
7
-3 + 2(2)
-5
3
4
+ 2(-5)
-5 + 2(3) -2 + 2(1)
1
2
8
4
-2 + 2(3) -2 + 2(1)
3
4
+
-1
S
5
t-B
6
-6
+ 6) = -54
1 1 t
+ 4/
the second column to the first column, and then add the third column to the second column
to obtain t
\A\
Now
to the
8
3
2
-5
row
2-3-57-3
4
1
2
3-2
4
-5 -3
2
first
factor
t
+2
from the
first
=
t
column and
|A|
=
(t
+ +
t — 2
2
1
t-2 t-2
2
1 t
+
4
from the second column
10 11
+ 2)(t-2)
1
to get
1 1 «
+
4
Finally subtract the first column from the third column to obtain
\A\
=
(<
+ 2)(t-2)
10 11 1
= t
+
4
(t+2)(t-2)(t +
4)
DETERMINANTS
182
[CHAP. 8
COFACTORS 8.10.
Find the cofactor of the 7 in the matrix
2
1
-3
1
5
-4
7
-2
4 3
-2
The exponent 2 +
8.11.
6
-3
5
2
•
(i)
=
(adj
A)
\A\
=
3
(iv)
adj
is
adj
A
2
3
4
\l
5
7/
4
7
7
is
3
4
5
7
2
3
5
7
2
3
3
4
+
-
(i)
,
(iv)
8.12.
+ 2
4
1
7
1
3
1
7
1
3
2
4
2
3
1
sl
(X)
10
A-i
=
4
-3
\
7
10
|A|
,
(ii)
Find adj A.
=
2
(iii)
Verify
21
+
+ Observe that the "signs" in the
the transpose of the matrix of cof actors.
u
- +\ — + — \+ - +/
.
=
|A|/
|]4i(adjA)
Find adj A. adj
/
61
A '(adj A) =
(ii)
A = (^_|^|
(ii)
7
= 1-20 +
3
Consider an arbitrary 2 by 2 matrix (i)
-3
Compute
matrix of cofactors form the chessboard pattern
(iii)
4
1
4
A-\
Find
1
+ That
3\
2
A
—
rz
2
2
4
+ (ii)
-2
/I
A =
1
5
2
-3
from the fact that 7 appears in the second row, third column.
3 comes
\A\ I.
4
1
4 3
Consider the matrix
A
2
^
(-1)2
adj (adj A)
=
Show ^1^1 j
adj
A
that adj (adj A)
-
d
-b
-e
a
\^_^
-
a
b
c
d
•
„;
= -
A.
(^_,
/ +la|
-l-ciy
\^_|_6|
+|di;
„;
^ /a v''
c
^
CD-=
A
CHAP.
DETERMINANTS
8]
183
DETERMINANTS AND SYSTEMS OF LINEAR EQUATIONS 8.13.
Solve for x and y, using determinants:
+ y = 3x — 5y —
1
2a; (i)
=
y
(ii)
2
1
3
-5
A„/A
=
3a
—
—c/a, y
— 2hy — Sax — 5by — ax
__
4 -13,
=
A:,
7
1
4
-5
=
Aj,/A
=
a6,
2c
3a;
First arrange the system in standard
2x 3x
X
Compute the determinant A
=
+ 3y + 5y + -2y -
z
3z
A
-4
=
-13.
Then x
= — ac.
Then
=
=
3,
A^-M
=
A^/A
Aj,
a.
c
3a
2c
=
a;
=
S
+1 - 5y
a;
-
z
.
2i/ in columns:
of coefficients:
3
5
2
-2
-3
=
2(-15
+ 4) - 3(-9-2) - l(-6-5)
=
22
in the
8
-
7
3
3-1
5
2
=
=
A„
66,
-1 -2 -3 and X
2
= 1 - 8 = -1
2z
13-1 =
A,
=
the system has a unique solution. To obtain A^., A,, and A^, replace the coefficients of matrix A by the column of constants. Thus
t^ 0,
unknown
0.
form with the unknowns appearing
-
of the matrix
2
A
-be,
+ 2x = + 22; -1 =
3z
Since A
¥^
—c/b
Solve using determinants:
the
where ab
,
2c
-39, Ay
-26 -56
c
=
Ax
Sy 8.14.
=
c
1.
-26 -56
a
A =
,
(ii)
Aj./A
=
3,
2/
=
2
1-1
3
8
=
-1, z
=
|A|.
=
A^/A
A,
=
-1 -3
1
Aj,/A
-22,
2
=
2
3
1
3
5
8
1
=
44
-2 -1
2.
PROOF OF THEOREMS 8.15.
Prove Theorem
8.1:
|A*|
A=
(tty).
Then A*
Suppose
= =
l^'l
(6jj)
=
By Problem
sgn r
8.36,
Hence
=
|A«|
However, as a runs through
8.16.
Thus
\At\
=
all
a^;.
Hence
(Sgn
2
(sgn ff)a„(i).ia
=
2
sgn
S„.
=
2
aes„
= Let T
where 6y
a) 6io.(i) 62,^(2)
a,
(sgn
.
.
6„o-(n)
.
and
t)
ai^d) a2T(2)
the elements of S„,
t
=
•
•
•
«nT(n)
also runs through all the elements of
|A|.
Prove Theorem 8.3(ii): Let B be obtained from a square matrix two rows (columns) of A. Then |B| = — |A|.
We
A
by interchanging
prove the theorem for the case that two columns are interchanged. Let t be the transtwo numbers corresponding to the two columns of A that are interchanged. If A = (oy) and B — (6jj), then 6y — Ojtcj)- Hence, for any permutation a. position which interchanges the
DETERMINANTS
184
Thus
\B\
Since the transposition t
—
sgn
But as
8.17.
=
=
2
(sgn
ff)6io.(i) 620.(2) ... &no-(n)
=
2
(sgn
a) OiTirCl) "2to-(2)
=
an odd permutation, sgn ra
is
•
•
sgn r
8
"-nraCn)
•
sgn
•
= —
sgn
Thus sgn a
a.
=
and so
TO,
a runs through -\A\.
2
= —
\B\
\B\
[CHAP.
(sgn
TOr)
a.iTO-(l)a2T
•
"^nrcrCn)
•
•
the elements of S„, to also runs through
all
the elements of S^; hence
all
Prove Theorem 8.2: (i) If A has a row (column) of zeros, then \A\ = 0. (ii) If A has two identical rows (columns), then \A\ - 0. (iii) If A is triangular, then \A\ = product of diagonal elements. Thus in particular, |/| = 1 where / is the identity matrix. (i)
(ii)
Each term in |A| contains a factor from every row and so from the row of term of |A| is zero and so \A\ = 0.
Thus each
zeros.
in K. If we interchange the two identical rows of A, we Suppose 1 + 1 # the preceding problem, 1A| = — |A| and so \A\ = 0. Hence by matrix A.
obtain the
still
in K. Then sgn
Now
(iii)
A =
Suppose a^j
=
(ay)
whenever
Suppose ii 7^ 1
lower triangular, that is, the entries above the diagonal are Consider a term t of the determinant of A:
is
<
j.
t
=
i
1<
Then
ii ¥- 1.
(sgn
aiij a2i2
•
•
where
•
i^H ...in
'*"*n'
=
and so a^^
ii
=
hence
0;
t
=
That
0.
is,
Now
ti
=
1
^
1
or
suppose ij
but 12
Then 2 <
iz ¥- 2.
^
and
ig
so
a^ =
hence
0;
.
Prove Theorem
Let
8.3:
B
.
A
be obtained from
(iii) (i)
If the jth \B\
row
A
of
k\A\. That
is
(iii)
or
...
Proved in Problem
by
multiplied
fc,
then every term in
|A|
fe
=
= The
first
=
c'0
+
or)
sum 1A|
is zero.
|B1
is
=
\A\.
by
multiplied
fc
and so
=
2
=
fc
an
(sgn o)
2a (sgn
a2t2
a) aii
•
•
a2i2
•
.
C^^jiP .
.
•
Oni„
•
•
=
«ni„
^
1^1
8.16.
2 (sgn c 2 (sgn
hence by Theorem |B|
n
|A|
Suppose c times the feth row is added to the jth row of A. Using the symbol yth position in a determinant term, we have \B\
t„ 9^
is,
|B|
(ii)
Thus each
0.
by
A
multiplying a row (column) of
=
=
.
by a scalar fe; then |B| = interchanging two rows (columns) of A; then |B| = - |A|. adding a multiple of a row (column) of A to another; then
(ii)
t
is zero.
2
Similarly we obtain that each term for which ij 7^ 1 or % # 2 or a^n = product of diagonal elements. Accordingly, 1A| = a^^a^^
(i)
each term for which
is zero.
term for which
8.18.
all zero:
=
is
a„j
.
agi^
.
{ca^i^
.
•
.
•
«fci^
+ ajj.) •
•
.
««!„
.
.
a„i^
+
2 (sgn
c) a^i^ a^i^.
.
.
a^.
.
. .
a„i^
the determinant of a matrix whose feth and ;th rows are identical; the sum is zero. The second sum is the determinant of A. Thus
8.2(ii)
A.
aii aji^
to denote the
/\
CHAP.
8.19.
DETERMINANTS
8]
Prove
Lemma
For any elementary matrix
8.6:
185
£",
=
l^'A]
IE"!
|A|,
Consider the following elementary row operations: (i) multiply a row by a constant A; # 0; interchange two rows; (iii) add a multiple of one row to another. Let E^, JS?2 and E^ be the corresponding elementary matrices. That is, Sj, E^ and E^ are obtained by applying the above (ii)
operations, respectively, to the identity matrix
=
l^il
=
k\I\
k,
=
\E^\
By
/.
the preceding problem,
=
-\I\
-1,
=
\E,\
=
|/|
1
Recall (page 56) that SjA is identical to the matrix obtained by applying the corresponding operation to A. Thus by the preceding problem,
=
\E^A\
k\A\
and the lemma
8.20.
is
=
=
\B\
By
(i)
By
\En\ \Er,-i\
.
(i)
A
.
\E2\ \Ei\ \A\,
.
the preceding problem,
=
8.4:
is invertible,
By Problem
^ =
^
Let
A
=
\A\
1|A|
=
I^g] 1A|
nonsingular, (ii)
E2E1A where \A\
if
the E, are
^ 0.
\B\ ¥=
and only
if
if
\A\ ¥- 0.
Then the following are
equivalent:
|A| 9^ 0.
(iii)
are equivalent.
.
\E^\\E,_,\...\E2\\E,\\A\
Hence
i.
.
induction,
fey
A be an w-square matrix. is
and
(i)
for each
.
and only
if
Hence
|Bi| 1A|.
\E„\\E„_,...E2E,A\
(ii)
6.44,
B = EnEn-i
\B\
(ii)
=
|J7iA|
the preceding problem, ^i
Prove Theorem
=
\E,A\
l^^l lA],
proved.
\B\
(ii)
=
-\A\
Suppose B is row equivalent to A; say elementary matrices. Show that: (i)
8.21.
=
lE^A]
\Ei\\A\,
Hence
it suffices to
show that
(i)
and
(iii)
are
equivalent.
Suppose A is invertible. Then A is row equivalent to the identity matrix /. But |/| ?* 0; hence by the preceding problem, |A| ^ 0. On the other hand, suppose A is not invertible. Then A is row equivalent to a matrix B which has a zero row. By Theorem 8.2(i), \B\ = 0; then by the preceding problem, \A\ = 0. Thus (i) and (iii) are equivalent.
8.22.
Prove Theorem If is
A
is
A =
=
\AB\
singular, then
nonsingular, then |A|
8.23.
8.5:
=^
AB E^
.
is .
.
\A\\B\. also singular and so \AB\ = = |A| |B|. On the other hand E2E1, a product of elementary matrices. Thus, by Problem
=
\E^...E^E,I\
and so
|AJ5|
=
Prove Theorem
8.7:
Let
\E„\...\E2\\E,\\I\
\E„...E2E,B\
A=
(a«);
=
then
\EJ
\A\
.
.
.
=
A
8.20,
\EJ...\E2\\E,\
=
lE^WE^WB]
= anAn +
if
ttizAia
|A| |B|
+
•
•
•
+ ai„Ai„, where
Aij is the cofactor of an.
Each term in \A\ contains one and we can write \A in the form
only one entry of the ith row
(aij.Ojg,
.
.
.,
a,„)
of A.
Hence
\
|A|
=
ajiAfi
+
ai2A*2
+
+
ai„Af„
(Note Ay is a sum of terms involving no entry of the ith row of A.) we can show that At. = A;,. = (-l)«+i|M„I
where Afy
Thus the theorem
is
proved
if
the matrix obtained by deleting the row and column containing the entry ay. (HisAy was defined as the cofactor of Oy, and so the theorem reduces to showing that the two definitions of the cofactor are equivalent.) is
torically, the expression
.
DETERMINANTS
186
First
we
consider the case that
=
Orm'^nn
=
i
n, j
2 a
«nn
=
(sgn
Then the sum of terms
n. a)
[CHAP. 8
02
ffli
•
•
in \A\ containing a„„ is
<»n-l,cr(n-l)
•
where we sum over all permutations aSS„ for which ain) = n. However, this is equivalent (Prob.,n-l}. Thus A*„ = |M„„| = (-!)»+« 1M„„| lem 8.63) to summing over all permutations of {1, .
.
consider any i and }. We interchange the ith. row with each succeeding row until it is and we interchange the jth column with each succeeding column until it is last. Note that the determinant |Afy| is not affected since the relative positions of the other rows and columns are not affected by these interchanges. However, the "sign" of |A| and of Ay is changed n — i and then
Now we
last,
n—j
Accordingly,
times.
A% =
8.24.
A = (an) and by the row vector
Let
let
B
(bn,
.
.
and By
(6y).
=
By
biiAn
j ¥=
+
+
bt2Aa
by replacing the
+
(lizAti
+
aijAii
+
023^2!
+
=
does not depend upon the ith
•
•
4-
•
|A'|
8.25.
—
we
\A\,
Prove Theorem
of
A
&i„Ai„
•
•
•
•
•
+
ttjnAin
—
4-
a„iA„i
=
•
ftiiBa
+
+
6i2^i2
= Ay
row of B, By
=
ftjiAji
+
+
•
•
+
6i2Ai2
•
•
6i„Bi„
for
+
•
=
ajiAji
also obtain that
+
aj2-^i2
Oti-^ii
+
+
•
•
«2j-^2i
+
•
+
n.
1
Hence
6i„Ai„
A
O-jn^ln
•"
A-(adjA) = (adjA)-A =
8.8:
=
j
Now let A' be obtained from A by replacing the ith row of has two identical rows, |A'| = 0. Thus by the above result,
|A*|
row
the preceding problem,
\B\
Using
ith
i,
ajiAn
\B\
Since
A
. ,
Furthermore, show that, for
B=
(-l)* + MMy|
be the matrix obtained from &m). Show that |B|
Let
=
(-l)»-i + »-i |M„|
+
•
by the
row
of A.
|AI
^
Since A'
=
«ni-Ani
|A|/.
jth
=
Thus
0.
if
0,
A-* =
(l/|Al)(ad3A).
A=
Let
and
(ay)
let
A
(adj
•
A)
=
The
(fty).
ffliz.
(«ii.
ith
row
of
A
is '^)
•
•
•
.
<*»tl)
Since adj A is the transpose of the matrix of cofactors, the ith column of adj the cofactors of the jth row of A: (A,i,Aj2, ...,A,„)«
Now
by,
the ij-entry in
A
•
(adj A), is obtained
6y
Thus by Theorem
8.7
=
a^Aji
A
•
(adj
A)
=
A -(adj A) \A]
I.
Uj^Ajz
+
•
•
•
+
(1)
and
is
the transpose of (2)
(2):
ttin'Ajn
and the preceding problem, "
Accordingly,
+
by multiplying
A
is
\
if
i
^i
the diagonal matrix with each diagonal element \A\.
Similarly,
(adj
A)-A =
\A\
1.
In other words,
CHAP.
8.26.
DETERMINANTS
8]
187
Prove Theorem 8.9: The system of linear equations Ax = b has a unique solution if and only if A = ]A| ^ 0. In this case the unique solution is given by Xi = Ai/A, X2
=
A2/A,
By
.
.
.
Xn
,
=
An/A.
Ax = b = |A| #
preceding results, and only if A
invertible if
Now
A#0. By
suppose
has a unique solution
and only
if
A
if
is invertible,
and
A
is
0.
Problem
A-i =
8.25,
Ax =
Multiplying
(1/A)(adj A).
b
by
A -1, we
obtain
= A-^Ax =
X
Notethattheithrowof (l/A)(adjA)
=
Xi
However, as in Problem
is
A^,
(l/A)(Aii,
(1/A)(6i Aii
+
A)b
(1/A)(adj .,
.
.
b^A^i
A„j).
++
(1)
If
6
6„A
J
=
(61, 63,
+
+
62^21
•
•
+
•
=
6„A„j
P
P-^P =
8.28.
is I.
Show
invertible.
Hence
Prove Theorem
.,
&„)'
then,
by
(i),
A;
A
the determinant of the matrix obtained by replacing the ith column of Thus Xi = (l/A)Aj, as required.
Suppose
•
8.24;
6iAii
8.27.
•
=
1
=
|/|
|p-ip|
Suppose
8.11:
that |P-i]
A
=
=
S
6.
|P|-i.
|p-i| |p|,
and
by the column vector
and so
|P-i|
=
|P|-i.
Then |A| = |B|, B = P-^AP. Then
are similar matrices.
Since A and B are similar, there exists an invertible matrix P such that by the preceding problem, |P| = |P-iAP| = |P-i| \A\ \P\ = \A\ lP-i| \P\ - \A\
We and
8.29.
remark that although the matrices P-i and A may not commute, their determinants |P-i| do commute since they are scalars in the field K.
\A\
Prove Theorem
8.13:
multilinear,
D
(ii)
There exists a unique function D.cA^K such that (i) D is (iii) D(/) = 1. This function D is the determinant
alternating,
is
i.e. D{A) = |A|. D be the determinant function: D(A) = \A\. We must that D is the only function satisfying (ii) and
function, Let
and
By
A=
D
show that
satisfies
(i), (ii)
and
(iii),
(iii).
(1),
preceding results, D satisfies (ii) and (iii); hence we need show that it is multilinear. Suppose = (Ai, Ag, ., A„) where A^ is the fcth row of A. Furthermore, suppose for a fixed i,
(««)
.
Aj
.
=
Accordingly,
Expanding D(A)
D(A)
\A\
where Bi
Cj,
=
a^
-
+
Bi
+
61
by the
= D(A„ (61
=
(fciAii
+
ajj
(b^,
.
=63 +
02,
A„)
=
.
.
6„)
and
...,
«;„
,
Q= =
(ci,
+
6„
.
.
. ,
c„)
c„
ith row, .
.
=
Cj,
=
.,
ci)Aii
+
+
Bi
+
Ci,
(62
62Ai2
+
•
.
.
.,
+ C2)A;2 + •
•
.
.
.
+ 6„Ai„) +
aaA^,
+
+
+ c„)Ai„
(6„
(ciA(i
at^A^^
+ C2A,.2 +
+
•
+
•
Ui^A^,
+ c^Ai^)
However, by Problem 8.24, the two sums above are the determinants of the matrices obtained from by replacing the ith row by Bj and respectively. That is,
A
Q
D(A)
Furthermore, by Theorem
=
I>(A„...,Bi
=
I>(Ai, ...,Bi, ...,A„)
B
is
multilinear,
i.e.
Ci,
...,A„)
+
Z)(Ai, ...,Ci, ...,A„)
8.3(i),
Z)(Ai, ...,fcAj
Thus
+
D
satisfies
(iii).
A„)
=
fcD(A„...,Ai, ...,A„)
DETERMINANTS
188
We next must prove the uniqueness of D. the usual basis of K", then by (iii), Z>(ei, eg, 8.73) that is
•'%) =
Z)(ejj,ei2.
Now
A =
suppose
(ay).
Observe that the
^k = Thus
=
I'(A)
>
(«ki. «fc2.
I>(aiiei
+
•
•
D
Suppose .
=
e„)
.,
.
row
•
•
•
A
=
+
in^
and
(ii)
(i),
Using
1.
.
.
If {e^,
(iii).
we
(ii)
also
.
.
.
e„}
(D
i^
.
,
have (Problem
is
+
«fci«i
+
02161
=
where a A^^ of
=
"fcn)
satisfies
D{I)
sgna,
fcth
+ ai„e„,
•
[CHAP. 8
+
"k2«2
a2„e„,
.
.
•
.
,
+
•
o„iei
afc„e„
+
•
+
•
a„„e„)
Using the multilinearity of D, we can write D(A) as a sum of terms of the form
= 2 ("Hj a2i2 where the sum
is
are equal, say
ij
summed
=
i^
over
but
•
sequences 11^2 then by (ii),
all
j ¥' k,
=
D
is
2 "212 «ni„) 2 (sgn a„^ a2i^ ••
(«iii
= Hence
%•
in
•
where
i^
D{ei^,
•
a)
.
' K^
^'•\'
'^"iJ
need only be summed over
(2)
D{A)
•
G
{1,
.
.
. ,
n}.
If
=
i^i^
two of the indices
•'\) =
^(%. %, Accordingly, the sum in we finally have that
•
.
.
permutations
all
Bj^,
.
.
.
,
the determinant function and so the theorem
is
.
J„.
Using
(1),
are:
ejj
where a
a„i^,
a
-
i^i2
...in
proved.
PERMUTATIONS 8.30.
Determine the parity of a Method
542163.
1.
We need
to obtain the
3 numbers
(5,
2 numbers
(5
3
numbers
1
number
(5,
(5)
number
of pairs
(i, j)
for which
4 and 2) greater than and preceding
and
4)
4 and
greater than and preceding 6)
greater than and preceding 5,
numbers greater than and preceding
6.
+ 2+3 +
1
+
=
+
9
is
i>
j
and
i
in
a.
There
sgn a
=
-1.
precedes
3
1,
2,
greater than and preceding
numbers greater than and preceding
Since 3
Method
=
3,
4,
odd, a is an odd permutation and so
2.
Transpose
1 to the first position as follows:
^42163
to
154263
to
125463
to
123546
to
123456
Transpose 2 to the second position:
154263 Transpose 3 to the third position:
125463 Transpose 4 to the fourth position: 1
2
sT^
6
Note that 5 and 6 are in the "correct" positions. Count the number of numbers "jumped": 3 + 2 + 3 + 1 = 9. Since 9 is odd, a is an odd permutation. (Remark: This method is essentially the same as the preceding method.)
CHAP.
DETERMINANTS
8]
Method
189
3.
An
interchange of two. numbers in a permutation is equivalent to multiplying the permutation by a transposition. Hence transform a to the identity permutation using transpositions; such as.
X
2^1
6
3
2
5
6
3
X4
4
5
14
Since an odd number,
8.31.
5,
1
2
5
6/3
1
2
S^
6^4
X
12
3
4
6
12
3
4
5
was
of transpositions
5
X
6
used, a is
Let a = 24513 and t = 41352 be permutations in mutations to„ and oroT, (ii) a~^. Recall that a
=
24513 and t
_
/I
2
=
an odd permutation.
Find
Ss.
the composition per-
(i)
41352 are short ways of writing
3
5\
4
""(^24513/
and
T
_ ~
/I
2
3
4
5N
V4
1
3
5
2/
which means <,(1)
=
2,
=
4,
=
4,
=
5,
(4)
=
1
and
=
r(2)
=
1,
r(3)
=
3,
T (4)
=
5
and
t(5)
=2
3
and r(l)
12
(i)
»
T
to
=
15243
4
5
i
1
4,
J,
j
2
4
5
13
j
4
1
1
1
2
4
3
15 Thus
3
and a°T
=
12 ^/
and ..
That
8.32.
is,
<7
=
i> is
a pair
^r
^^
^^
13
5
2
i
i
i
i
i
5
3
4
^/
2
3
1
3\
4
5/
=
/I
2
3
4
5\
V4
1
5
2
3/
41523.
Consider any permutation
there
5
12534.
Vl
=
4
4
12
(ii)
ff-i
3
(i*,
jiji
.
.
.
and
k
Show
jn.
i
that for each pair
(i,
k)
precedes kin a
k*) such that i*
<
k*
and
cr(i*)
>
(i)
and vice versa. Thus cr is even or odd according as to whether there odd number of pairs satisfying (1). i* and A;* so that a(i*) = i and precedes k in a it and only if i* < k*.
Choose
and
i
such that
a{k*)
=
fc.
Then
i
>
fe
if
and only
is
if
an even or
>
a(fe*),
.
DETERMINANTS
190
8.33.
Consider the polynomial g - g{xi, ...,Xn) polynomial g — g{xi, X2, Xs, xt).
sum
=
[CHAP.
Write out
y[{Xi-Xj).
8
explicitly the
'"^^
The symbol 11 is used for a product of terms in the same way that the symbol 2 of terms. That is, Yl (a'i ~ ""j) means the product of all terms (Kj — Xj) for which
is
used for a
<
i
j.
Hence
i
9
8.34.
Let
-
=
g{Xl,...,Xi)
— X2){Xi — Xa)(Xi — Xi)(X2-X3)(X2-Xi)(X3 — X^)
iXi
For the polynomial g
be an arbitrary permutation.
(T
define
~
= Yl (^-^c" 'i<3
Show
^(tw)-
that
g
if
o-
is
even
[—g
it
a
IS
odd
{
Accordingly, aig) Since a
=
(sgn
one-one and onto,
is
n
=
a{g)
-
(a'.ra)
=
a'.ru))
—
a{g)
of the
form
g or a(g)
(«;
- Xj)
= —g
where
i
>
,{xi-Xj)
.11. KjorOj .
to"
Thus
in the preceding problem,
according as to whether there is an even or an odd number of terms Note that for each pair (i, j) for which j.
<
i
and
}
>
17(1)
(1)
is a term (ajo-fj, - x^^j-^) in a(g) for which a(i) > a(j). Since a is even even number of pairs satisfying (1), we have a(g) = g it and only if a if and only if a is odd.
there
8.35.
if is
and only
if
there
even; hence
is
an
= -g
Let u,tG Sn. Show that sgn (to a) = (sgn T)(sgn
odd.
Using the preceding problem, we have sgXi-{r
Accordingly, sgn(TO
8.36.
°a)g
=
=
(sgn T)(sgn
=
Consider the permutation a " ttj^i
aj^2
We have a-^°
a
-
JiJz
.
=
.
c,
in
ttj^n
=
{T°c)(g)
=
T(CT(flr))
J1J2
.
.
jn.
.
aik^chk^
is
sgn
^1^2
kn-
i
=
=
1,
.
1,
.
. ,
(
Thus a°T
—
e,
(sgn T)(sgn
where
Omk^
•
Since
is
e
=
(7-1
<7~^
=
sgn
.
.
.
kn
even, cri and a are both even or
a.
a permutation,
Then for
=
Show that sgn
the identity permutation.
=
ff(fci)
=
T((sgn
aj^i Oj^a
have the property that
Let T
=
^(fea)
=
2,
•
•
•
.
.
a,j^n
.,
=
"(K)
"iki "zkz
=n
w,
=
a{T(i))
the identity permutation; hence t
=
=
a(fcj)
ct~i.
=
i
»nk„-
Then
fej,
k^,
..,k„
CHAP.
DETERMINANTS
8]
191
MISCELLANEOUS PROBLEMS 8.37.
Find det (T) for each linear operator (i)
T
is
the operator on R^ defined by T{x, y,
(ii)
T
T:
—
z)
{2x
— z,x + 2y- 4:Z, Sx-3y + z)
the operator on the vector space
is
T(A)
=
MA
M
where
V
of 2-square matrices over
K
d,
c
1
(i)
Find the matrix representation of T relative
say, the usual basis:
to,
=
[r]
12-4
=
3-3 (ii)
Find a matrix representation of T 1
2(2-12)
- l(-3-6) =
some basis of V,
in
1
E. =
T(E^)
b\/l dj\0
a
=
T(Ei)
c
c
nE^)
=
T{Ei)
=
c
e
'a
Thus
=
aEi
+
0^2
+
cE^
+
OE4
=
0iS7i
+
aE^
+
OS3
+
cE^
=
bEi
+
OE2
+
dEs
+
OEi
=
0^1
+
bE2
+
OE3
+
dEi
c
'b
^d b
d
1
dj c
a
a
=
1,
and
b
a det (T)
VO
c
d
b ^0
=
0\
c
a
=
[T]e
a
1
b\/0 djio
a
E^ "*
0/'
c
b\/0 d)\i
a
-11
say,
£, = a
6\/0 dj\0
a
=
2—4 -3
1
^1
Then
3
0-1'
0-1
2 det (T)
2 1
I
1
Then
c
=
a
c
+
d
a
d
6
c
e
-
6
d
b
a^d^
+
2abcd
bH"^
d
b
d
6
/111 8.38.
Find the inverse of
A =
1
(
1
.0
The inverse of Set
AA-'^
Ai4-i
—
A
l,
is
of the
1,
form (Problem
8.53):
A-» =
1
the identity matrix:
*
z '
=
=
Set corresponding entries equal to each other to obtain the system a;-|-l
=
0,
by
defined
2/-t-z-|-l
=
0,
z
+
l
=
/
1/
.
DETERMINANTS
192
[CHAP. 8
/l-l The
solution of the system
is
a;
=
=
—1, y
=
z
0,
A
Hence
—1.
^
—
0\
—1
1
\
A~^
8.39.
D
Let
(adj i4.)/|A|
= - D{B,
Show that D{A, B) multilinear and alternating, then
be a 2-linear, alternating function.
show that
generally,
if
D
is
D{...,A, ...,B, that
A-> =
could also be found by the formula
the sign
is,
Since
D
is
= -D{...,B,
...)
is
D{A + B,A+B) =
alternating,
=
and
I>(B,
= Similarly,
=
B)
Let
V
D(.
=
D(,...,A
D(...,A
...,A, ...)
Furthermore, since
0.
.
.,
(i)
A + B,
.
.,
.
A+ B, +
A,...)
D(...,A, ...,B,
D{A, B)
or
D(...,B,...,A,
D
multilinear,
is
D{M) =
D.V^R
.
is 2-linear
A=
Let
Yes.
A=
(1, 1)
= = d(\ 3)
(tti,
D(A,C)
a^,
=
B=
dT'
Hence for any scalars
D(sA
s, <
G
(61, 63)
"^)
is,
D(B, A)
.)
+ D{...,B....,B,
D(...,B, ...,A,
...)
...)
A,...).
M
=
(
,
over R.
j
(with respect to the rows)
if
(i)
Determine
D{M) = a + d,
D
is
B-
and
D(2A.B)
and
4
and
=
Then
(3, 3).
C=
3)
=
5
#
2D(A, B)
then
(cj, Ca);
D(B,C)
and
01C2
- C^^
= d(^'
^')
= V2
R,
+ tB,C) = =
That
= -
ad.
For example, suppose
No.
D{B, B)
D(....,A, ...,B, ...)
+
...)
.
...)
= -D(...,B
B,...)
D(A,B)
(ii)
D(B, A)
be the vector space of 2 by 2 matrices
whether or not (ii)
+
=
=
8.40.
i?/«»i s(aiC2)
+ +
**^
sa,+
=
t(6iC2)
tb,^^
^
8D(A, C)
^sa,
+
+
th,)c,
tD{B, C)
linear with respect to the first row.
Furthermore,
=
D(C,A) = d(^;
D(C, B)
and
c,a,
= D^^^
I'J
Hence for any scalars
s, t
S
D(C,sA +
That
is,
D
is
More
Hence
0.
D(A, B)
+
and thus
A).
changed whenever two components are interchanged.
= D{A+B,A+B) = D(A,A + B) + D(B,A + B) = D{A, A) + D(A, B) + D(B, A) + But D(A, A)
.
1/
\0
^=
R, tB)
^
d(
"J-.,
=
s(cia2)
+
„?..)=
t(ci62)
linear with respect to the second row.
Both linearity conditions imply that
D
is 2-linear.
=
sI>(C,A)
Ci(sa2+t62)
+ «D(C,B)
c,b.
CHAP.
DETERMINANTS
8]
193
Supplementary Problems COMPUTATION OF DETERMINANTS 8.41.
8.42.
Evaluate the determinant of each matrix:
Compute the determinant
2
5
4
1
(i)
t-2
of each matrix:
(i)
For each matrix
8.44.
Compute the determinant /2
5
\1
8.45.
4/
/-2 -1
-2 -4\ 5 -1
2
(ii)
,
\
/t-5
....
<">
7
-1
(
+
3
which the determinant
4
(iii)
,
Evaluate the determinant of each matrix:
-
f
For each matrix
-3 -3
3
1
-3 -6
(ii) I
+
t
5
t-4,1
6
in the preceding problem,
determine those values of
for which the determinant
t
is zero.
/l 8.47.
2
2
2
3
10-20
Evaluate the determinant of each matrix:
(i)
3-1
3
1
^"^ I
2)
1
l2
-1
4
2 -1
COFACTORS, CLASSICAL ADJOINTS, INVERSES
2-2
'l
For the matrix
I
4 ,1 (i)
the entry
8.49.
Let
A
8.50.
Let
A =
4,
(ii)
'1
2
3
1
a
1
f
3^ I,
„
„ 2
7
2 -3^
find the cofactor of:
1
the entry
5,
(iii)
the entry
(i)
adj A,
(ii)
A-i.
Find
(i)
adj A,
(ii)
A-i.
2'
1,
8.51.
Find the
8.52.
Determine the general 2 by 2 matrix
8.53.
Suppose
classical adjoint of each
A
is
7.
Find
B
diagonal and '
is
for which
A =
adj A.
...
02
,0
A
8.47.
triangular; say,
di
A =
matrix in Problem
...
and
62
B lO
2\
1-2
3
1-21'
\4 -3
8.48.
is zero.
1/
6
\0
(i)
8.46.
-2
of each matrix:
/3
-2
-3
1
3
in the preceding problem, find those values of t for
l\
1
(i)
-3
t-l)'
4
8.43.
6 (ii)
K
j
3
1/
DETERMINANTS
194
A
(i)
Show
(ii)
Show
that
(iii)
Show
that the inverses of
that adj
B
A
B
and
t-i
=
hence
lO
iff all
^
5_,
'
a-
«{
^ 0.
form
/&r* I
...
That is, the diagonal elements of elements of A and B.
invertible
is
\
••
ar^
A
either exists) are of the
(if
...
|0
8
is triangular.
6; ¥= 0;
is invertible iff all
A-1
B
diagonal and adj
is
[CHAP.
di2
••
dm
62
...
d2n
\o
A-i and B-^
are the inverses of the corresponding diagonal
DETERMINANT OF A LINEAR OPERATOR 8.54.
Let
T
be the linear operator on R^ defined by T{x, y,
Find det 8.55.
Let
DiV^V
erated by
8.56.
8.57.
(i)
det (S°T)
Show
(3a;
be the differential operator,
{l,t, .... t"},
Prove Theorem (i)
=
z)
- 2z,
+ 7z,x + y + z)
5y
(T).
=
=
l
{sin
(iii)
dv/dt.
cos
V
if
the space gen-
is
Then:
be linear operators on V. is
invertible if
where ly
is
the identity operator;
(ii)
Find det (D)
t}.
T
det(S)'det(r);
(i)det(lv)
that:
T and S
Let
8.12:
= t,
D{v)
i.e.
{e*, e^*, e^t},
(ii)
and only
det (7) 9^
if
0.
det (T-i)
(ii)
=
det(r)-i
T
if
is
invertible.
DETERMINANTS AND LINEAR EQUATIONS 8.58.
8.59.
8.60.
Prove Theorem
i
(1)
Solve by determinants:
A =
+ [ix — (Sx
,..
Solve by determinants:
(i)
5y „
,
2y
= =
8
,..,
,
(")
l
= = [sx-iy-Gz = (2x-5y + x + 2y -
}
2z
7
Az
3
.
.
,
+3 = - Sz = [sy + z = (2z
Ax =
•
+ Sx + 1. - 2x
y
x
<
(ii)
5
The homogeneous system
8.10:
f2x-Sy = -1 „ I4x + 7y = —1
i
2y 2
has a nonzero solution
if
and only
if
IA| == 0.
PERMUTATIONS 8.61.
Determine the parity of these permutations
8.62.
For the permutations
8.63.
Let T
8.64.
e
5„.
Show
t
and v
in
in Sg:
Problem
(i)
a
8.61, find
=
3 2 1 5 4,
(i)
r°c,
(ii)
r
=
(ii) Tr°
1 3 5
S„
is,
have the property that
aeS„
4,
(iii)
cr-i,
(iii)
that t°
2
=
n,
(iv)
= =
y
=
4253
1.
t-i.
{t « a
:
a
e
(i)
iSf„}.
Show
c* runs through
:
MULTILINEARITY Let V = (K"")"", 8.65. D:V-^K. (i)
Show
i.e.
V
is
the space of m-square matrices viewed as m-tuples of
that the following weaker statement is equivalent to
i?(Ai,A2, ...,A„)
whenever Aj (ii)
=
Ai+i for some
D
row
vectors.
Let
being alternating:
=
i.
Suppose D is m-linear and alternating. then D(Ai,...,AJ = 0.
Show
that
if
A^.Az,
,A^
are linearly dependent,
.
CHAP.
8.66.
DETERMINANTS
8]
V
Let
D:V->^K
8.67.
=
(
j
2-linear (with respect to the rows) if
is
D{M) =
(iii)
M
be the space of 2 by 2 matrices
(iv)
0,
D{M) =
195
(i)
over R.
Determine whether or not
D{M) = ac-hd,
(ii)
D{M)
Let V be the space of M-square matrices over K. Suppose B B V is invertible and so Define Z> V -* by Z?(A) = det (Afi)/det (B) where A G V. Hence
X
:
fl(Ai, Ag,
where Aj
row
the tth
is
.
is
=
A„)
. ,
.
A and !>(/) = 1. of
and that det (A) det (B). This method alternating,
= ah- ed,
1.
A^B
so
det (B)
¥= 0.
det (AiB, A^B, .... A„B)/det (B)
is
the ith
row
of
AB. Show that
(Thus by Theorem 8.13, D{A) used by some texts to prove Theorem
-
det (A)
8.5, i.e.
Z> is
multilinear and
and so
|Ai5|
=
det (AB)
=
\A\ |B|.)
MISCELLANEOUS PROBLEMS
A
8.68.
Let
8.69.
Prove:
be an w-square matrix.
The above
8.70.
is
called the
Prove
1
x^
x\
1
X2
xf '2
1
a:„
xl
IfeAl
=
fc" \A\
x\~^ ~"-l X^
...
«»-»
...
Vandermonde determinant
M
Consider the block matrix
=
f
j
where
of order n.
A
and
C
are square matrices. Prove \M\
M
More
generally, prove that if is a triangular block matrix with square matrices Aj, the diagonal, then \M\ = |Ai| [Agl "l^ml•
8.71.
Let A, B,
^ 8.72.
8.73.
"
A
is ortfeoflroMttZ,
Consider a permutation tth
row
W
is
that
=
a
A=
e^., i.e.
is,
Jiij
=
•
.,
A^
on
Consider
the
2»-square
block
matrix
\A\\D\-\B\\C\.
A^A = •
.
\A\ \C\.
•
commuting m-square matrices.
be
P'^^^^tl'**
(c d)-
Suppose
whose
8.74.
C and D
.
=
•
(e,-^, e^-^,
.
Show
Let
in.
I.
.,
e^^).
{e<}
that
|A|
=
±1.
be the usual basis of X», and that |A| = sgn a.
let
A
be the matrix
Show
Let A be an M-square matrix. The determinantal rank of A is the order of the largest submatrix of A (obtained by deleting rows and columns of A) whose determinant is not zero. Show that the determinantal rank of A is equal to its rank, i.e. the maximum number of linearly independent rows (or columns).
Answers 8.41.
(i)
-18,
8.42.
(i)
t2
8.43.
(i)
t
8.44.
(i)
21,
-
=
(ii)
3t
5,
(ii)
t-
to
Supplementary Problems
-15.
10,
-2;
-11,
(ii)
t^
(ii)
(iii)
t
2t
-
100,
-
8.
i, t
-
(iv)
0.
-2.
DETERMINANTS
196
+ 2)(t-3)(t-4),
8.45.
(i)
(<
8.46.
(i)
3, 4,
8.47.
(i)
-131,
(ii)
-55.
8.48.
(i)
-135,
(ii)
-103,
8.49.
adj
-2;
(ii)
A =
4,
-2;
-1
1
2
-2
\
adj
A =
|
4,
(iii)
(iii)
+ 2)2(t-4),
1
(t
+ 2)2(f-4).
-31.
-1
=
A-i
,
i -^ \-l 1
=
(adj A)/|A1
0/
-3 -1 2
(iii)
-2.
-2\
1 8.50.
(t
(li)
[CHAP. 8
6
A-i
,
=31 /-I
2\
\-2 -1
5/
-5/
-29 -26 -2\ (-16 -30 -38 -16 29 -8 51 -13 -1 -13 1 28 -18/
-14 -17 -19^
21
/ ,...
1
-44
H
33
11
-29
1
13
21
17
7
I
(") \
| 0,
-19 -18;
/k
8.52.
A =
8.54.
det(r)
8.55.
(i)
0,
8.58.
(i)
X
=
21/26,
8.59.
(i)
X
=
5,
8.61.
agn a
8.62.
(i)
T°v =
8.66.
(i)
Yes,
(
„
,
=
4.
(ii)
6,
=
y
1,
(iii)
—
y
=
1, z
agn t
53142,
(ii)
1.
No,
29/26;
=
= (ii)
(iii)
1.
=
(ii)
x
(ii)
Since
—1, sgn v
ir°(r
Yes,
=
-5/13, y
=
1/13.
the system cannot be solved by determinants.
—1.
52413,
(iv)
A = 0,
=
No.
(iii)
=
32154,
(iv)
t-i
=
14253.
chapter 9
Eigenvalues and Eigenvectors INTRODUCTION In this chapter we investigate the theory of a single linear operator T on a vector space V of finite dimension. In particular, we find conditions under which T is diagonalizable. As was seen in Chapter 7, this question is closely related to the theory of similarity transformations for matrices.
We
with an operator T: its characteristic These polynomials and their roots play a major role in the investigation of T. We comment that the particular field K also plays an important part in the theory since the existence of roots of a polynomial depends on K. shall also associate certain polynomials
polynomial and
its
minimum
polynomial.
POLYNOMIALS OF MATRICES AND LINEAR OPERATORS Consider a polynomial f{t) over a field K: f(t) = Ont" + + ait + Oo. •
-
If
•
A
is
a square
matrix over K, then we define
=
/(A)
where
/ is the identity matrix.
nomial
fit)
if /(A)
Example
9.1:
=
a„A"
+
•
•
•
+
we say
In particular,
+
ttiA
ao/
that
A
is
a root or zero of the poly-
0.
V
A =
Let
and
/(t)
let
=
2t2
- 3t + 7,
g(t)
=
t^
- 5t - 2.
Then
(
«^)-i:!r-<3:)-a;) ^
A
The following theorem
Theorem
9.1:
=
.(A)
Thus
is
1
^3
a zero of
2V
J
-/I
2\
„/l
-5^3 ^^-2^^
=
a^;:: /O ^^
0\ ^^
^
g(t).
applies.
Let / and g be polynomials over K, and
let
A
be an w-square matrix over K.
Then (i)
(ii)
+ flr)(A) = /(A)+flr(A) {fg)(A) = f(A)giA) (/
and, for any scalar k (iii)
Furthermore, since
=
{kf){A)
f{t) g{t)
=
G K, kf{A)
g{t) f{t)
for any polynomials
f{A)g{A)
That
is,
any two polynomials
in the
matrix
=
A
giA)f{A)
commute.
197
f{t)
and
g{t),
EIGENVALUES AND EIGENVECTORS
198
Now +
cut"
•
T :V -^ V is a linear operator on a vector space V over K. If then we define f{T) in the same way as we did for matrices:
suppose •
+
+
ait
[CHAP.
9
—
f{t)
do,
=
f{T)
a^T"
+
• +aiT +
aoI
where / is now the identity mapping. We also say that T is a zero or root of f(t) if f{T) = 0. remark that the relations in Theorem 9.1 hold for operators as they do for matrices; hence any two polynomials in T commute.
We
Furthermore, of f(T).
if
A
is
In particular,
a matrix representation of T, then /(A) if and only if /(A) = 0. f{T) =
is
the matrix representation
EIGENVALUES AND EIGENVECTORS Let is called
T -.V -*V
be a linear operator on a vector space V over a if there exists a nonzero vector v
field
eV
an eigenvalue of T
T{v)
=
K.
A
scalar X
G
K
for which
\v
Every vector satisfying this relation is then called an eigenvector of T belonging to the eigenvalue A. Note that each scalar multiple kv is such an eigenvector:
=
T{kv)
The
kT{v)
set of all such vectors is a subspace of
The terms
V
=
k(\v)
=
(Problem
\{kv)
9.6) called the
eigenspace of
\.
and characteristic vector (or: proper value and proper and eigenvector.
characteristic value
vector) are frequently used instead of eigenvalue
Example
9.2:
I.V^V
Let
Hence Example
9.3
:
1 is
Then, for every vGV, I{v) = v = Iv. be the identity mapping. /, and every vector in V is an eigenvector belonging to 1.
an eigenvalue of
Let 7" R^ ^ R2 be the linear operator which v GB? by an angle rotates each vector = 90°. Note that no nonzero vector is a Hence T has no eigenmultiple of itself. :
values and so no eigenvectors.
Example
9.4:
Let D be the differential operator on the vector space V of diflferentiable functions. We have Hence 5 is an eigenvalue of D £)(e*') = 5e^*. with eigenvector e^'.
If A is an w-square matrix over K, then an eigenvalue of A means an eigenvalue of A is an eigenvalue of A if, for some nonzero viewed as an operator on K". That is,
\gK
(column) vector v
In this case v Example
is
G X",
an eigenvector of
9.5:
Av =
A
\v
belonging to
A.
A = AX = tX:
Find eigenvalues and associated nonzero eigenvectors of the matrix
We
seek a scalar
t
and a nonzero vector
i
The above matrix equation X ,
[3x
+ +
2y 2y
is
= =
DO
X = =
(
\
such that
1
2
3
2
•(:
equivalent to the homogeneous system tx
(
or ty
-s
{t-l)x- 2y =
\-Zx + (t-2)y
=
(i)
CHAP.
EIGENVALUES AND EIGENVECTORS
9]
199
Recall that the homogeneous system has a nonzero solution terminant of the matrix of coefficients is 0:
t-1
-2
Thus
t is
Thus V =
=
t
4 in
=
)
{
A
-
4
and only
if
t
if
3x
~
2y
+
2y
„
is
)
= =
= -l
-3x
w =
=
t
I
4
or
t
=
—1.
,
.
3x
—
=
2y
a nonzero eigenvector belonging to the eigenvalue
—
)
to
<
=
4
is
a multiple of
{
1
—
= =
2y 3y
is
Proof. X
is
x
—1, and every other eigenvector belonging to
+
=
y
t
= —1
is
a multiple of w.
T{v) i.e.
Xl
—T
is
T:V -^V
=
singular.
XV
We
or also
T
if
and only
Theorem
sta+2 a very useful
9.3:
Example
have that v
is
eigenvectors independent. Consider the functions
D
fre-
XGK The
there exists a nonzero vector v such that
=
{Xl-T){v)
or
in the eigenspace of X if
=
and only
if
the above
— T.
theorem which we prove (Problem
Nonzero
9.6:
if
{Xl){v)-T{v)
relations hold; hence v is in the kernel of XI
We now
is
an eigenvalue of T
an eigenvalue of
4,
a nonzero eigenvector belonging to the eigenvalue
is
)
,
.
or simply
be a linear operator on a vector space over K. Then if and only if the operator Xl — T is singular. eigenspace of A is then the kernel of XI — T.
Let
9.2:
=
v.
The next theorem gives an important characterization of eigenvalues which quently used as its definition.
Theorem
t
in (1),
-2x
Thus
=
or simply
and every other eigenvector belonging t
= {t-4){t+l) =
\3/
\V/
Setting
the de-
if
(1),
-Sx
(
3t
and only
2
an eigenvalue of
Setting
-
t2
«-
-3
if
belonging
e°i', 6°^',
.
.
to
., e''n'
the differential operator then
=
aj,
by induction: are
eigenvalues
distinct
where
9.14)
.
.
linearly
.,a„ are distinct real numbers.
Accordingly, e^i', ..., e°»' and so, by are eigenvectors of D belonging to the distinct eigenvalues ai, , a„, Theorem 9.3, are linearly independent. If
is
D(e°i'')
a^e'^''*.
.
We remark that independent eigenvectors can belong Problem 9.7). DIAGONALIZATION AND EIGENVECTORS Let T:V -^ V be a linear operator on a vector that
T can be represented by a
space
diagonal matrix ...
'A;i
k2
,0
...
...
kni
V
to the
with
.
.
same eigenvalue
finite
dimension
n.
(see
Note
EIGENVALUES AND EIGENVECTORS
200
if
and only
if
there exists a basis
[vi,
.
T{v2)
= —
T{V„)
-
T{vi)
that
is,
values
Theorem
.
9
for which
kivi kiVi
knVn
such that the vectors vi, .,Vn are eigenvectors of In other words: ., fen.
ki,
We
V
.,v„} of
.
[CHAP.
T belonging
respectively to eigen-
.
A
linear operator
if
and only
T V -» V can be represented by a diagonal matrix B has a basis consisting of eigenvectors of T. In this case the diagonal elements of B are the corresponding eigenvalues.
9.4:
if
:
V
have the following equivalent statement.
Alternate
Form
of
Theorem
In the above theorem, eigenvectors of A, then B Example
9.7:
if
w-square matrix A is similar to a diagonal matrix and only if A has n linearly independent eigenvectors. In this case the diagonal elements of B are the corresponding eigenvalues.
An
9.4:
we
let
B
if
P
be the matrix whose columns are the
A =
Consider the matrix
eigenvectors
/2\
and
A
is
(
1\
/
By Example
.
j
P =
Set
(^_J.
(^^j
Then
n independent
= P~^AP.
/2
1\
{^^
_^j
,
9.5,
A
and so
has two independent
P„_,i -
/1/5
1/5
(^^^^
_^^^
similar to the diagonal matrix
B = P-^AP =
/1/5
l/5\/l
2\/2
1\
1^3/5
-2/5/^3 and —1 of
2/1^3
-1/
the diagonal As expected, the diagonal elements 4 values corresponding to the given eigenvectors.
^ /4 0\ ~ \0 -l) matrix B are the
eigen-
CHARACTERISTIC POLYNOMIAL, CAYLEY-HAMILTON THEOREM Consider an %-square matrix
A
over a
field
ai2
(0,11
The matrix
tin
K:
Chl
ffl22
ttnl
ttre2
~tt21
ain fflZn
.
flnn
•
determinant a polynomial in
AA{t) t,
is
—
det
— — a22 ffli2
t
— ttn2
— ftnl
is
.
.
/„ is the n-square identity matrix and matrix of A:
(t — an
which
.
.
— A, where
called the characteristic
Its
.
.
{tin
—
* is
.
.
.
—ain
.
.
.
—a2n
...
t
aim
A)
called the characteristic polynomial of A. AA{t)
the characteristic equation of A.
=
det
(tin
an indeterminant,
-A) =
We
also call
is
CHAP.
EIGENVALUES AND EIGENVECTORS
9]
Now
201
each term in the determinant contains one and only one entry from each row and characteristic polynomial is of the form
from each column; hence the above
-
AA{t)
{t- an){t
- 022) ••(*- ann) terms with at most n — 2 factors of the form
+
t
— an
Accordingly,
=
AaC*)
—
t"
(au
+ a22+
•
•
•
+
+aTO)t"~^
terms of lower degree
A is the sum of its diagonal elements. Thus the characteristic polynomial Aa(*) = det (i/„ — A) of A is a monic polynomial of degree n, and the coefficient of i"~^ is the negative of the trace of A. (A polynomial is monic if its leading coefficient is 1.)
Recall that the trace of
Furthermore,
if
we
set
i
=
=
Aa(0)
But
Aa(0) is the constant
we
in Aa(<),
obtain
=
\-A\
(-1)"[A|
term of the polynomial AaC*). Thus the constant term of the charmatrix A is (—1)" \A\ where n is the order of A.
acteristic polynomial of the
13 Example
The
9.8:
characteristic polynomial of the matrix
A =
—2
(
0^
2—1 -21
4 *
=
^(t)
-
=
\tI-A\
1
2
-3
t-2
-4 As
We now
expected, A(t)
state one of the
Cayley-Hamilton Theorem Example
The
9.9:
t
+
Every matrix
characteristic polynomial of the matrix
As
=
\tI-A\
=
2«
+
28
characteristic polynomial.
its
t-3
1
I
-2
o
=
t-2 A
2^
/I
A =
expected from the Cayley-Hamilton theorem,
is
'<^'^G iJ-'Q i)-'Q The next theorem shows the intimate
+
3.
V
A(t)
<2
in linear algebra.
a zero of
is
-
«3
2
a monic polynomial of degree
is
most important theorems
9.5:
=
1
Ji
t2
-
3t
a zero of
°)
A{t):
= (:
:)
relationship between characteristic polynomials
and eigenvalues.
Theorem
9.6:
Let
A
\GK
is an eigenbe an n-square matrix over a field K. A scalar A if and only if A is a root of the characteristic polynomial A(t) of A.
value of Proof.
By Theorem
9.2,
A
an eigenvalue of A if and only if XI — A is singular. XI — A is singular if and only if |a7 — A| = 0, i.e. A is a root is
Furthermore, by Theorem 8.4, of A(t). Thus the theorem is proved.
Using Theorems Corollary
9.7:
9.3, 9.4
and
9.6,
If the characteristic
we
obtain
polynomial
of distinct linear factors:
A(t) of
an ^-square matrix
A
is
a product
EIGENVALUES AND EIGENVECTORS
202
=
A{t) i.e. if ai,
.
.
.
{t- ai){t -
aa)
On are distinct roots of A{t),
,
matrix whose diagonal elements are the
•
•
•
(t
then
[CHAP.
- an)
A
is
similar to a diagonal
ok.
Furthermore, using the Fundamental Theorem of Algebra (every polynomial over has a root) and the above theorem, we obtain Corollary
A
Let
9.8:
least
Example
9.10:
Let
be an w-square matrix over the complex one eigenvalue.
A =
2
—5
Its characteristic
.
t
A(t)
-
Then
field C.
polynomial
A
C
has at
is
3
=
«-
2
=
5
-1
We
9
t
+
(«-3)(t2
+
l)
2
consider two cases:
(i)
A
(ii)
A
Then A has only the one eigenvalue is a matrix over the real field R. Since 3 has only one independent eigenvector, A is not diagonalizable.
3.
a matrix over the complex field C. Then A has three distinct eigenvalues: and —i. Thus there exists an invertible matrix P over the complex field C for which is
3, i
/3
P-iAP =
0^ i
\0
A
i.e.
Now
suppose
A
show that
and
is
-i,
diagonalizable.
A and B are similar matrices, say B = P^^AP where P is invertible. We B have the same characteristic polynomial. Using tl — P~^tIP, \tI-B\
=
\tI-P-'AP\
=
\P-mi-A)P\
=
\P-HIP - P-'AP\
=
\P-^\\tI-A\\P\
Since determinants are scalars and commute, and since
\tI-B\
|P~i| |P|
=
1,
we
finally obtain
= \tI-A\
Thus we have proved
Theorem
9.9:
Similar matrices have the same characteristic polynomial.
MINIMUM POLYNOMIAL Let A be an w-square matrix over a field K. f{t)
for which
polynomials
we
coefficient is 1, 9.25);
we
Theorem
i.e.
which
is
monic.
Such a polynomial m{t) exists and
minimum polynomial
the
9.10:
The minimum polynomial m{f) of
is
is
Among
these
whose leading
unique (Problem
of A.
call it
as a zero.
There
Observe that there are nonzero polynomials
f{A) — 0; for example, the characteristic polynomial of A. consider those of lowest degree and from them we select one
A divides every polynomial which has A In particular, m{t) divides the characteristic polynomial A(t) of A.
an even stronger relationship between m{t) and
A(i).
CHAP.
EIGENVALUES AND EIGENVECTORS
9]
Theorem
9.11:
The
characteristic
203
and minimum polynomials of a matrix
A
have the same
irreducible factors.
This theorem does not say that w(i) = A(t); only that any irreducible factor of one must In particular, since a linear factor is irreducible, m(t) and A(t) have the same linear factors; hence they have the same roots. Thus from Theorem 9.6 we obtain
divide the other.
Theorem
9.12:
A
scalar A
minimum
Example
an eigenvalue for a matrix
is
Find the minimum polynomial
9.11:
A
if
and only
if
2
1
A
is
a root of the
poljTiomial of A.
>«(*)
of the matrix
2
A
2 5
The characteristic polynomial of A is A(t) = |t/- A] = (t- 2)»(t- 5). By Theorem 9.11, both t — 2 and t — 5 must be factors of m(t). But by Theorem 9.10, vnif) must divide A(t); hence m(i) must be one of the following three polynomials: TOi(t)
=
= (t-2)3(t-5) Cayley-Hamilton theorem that in^iA) — A(A) = 0. The reader # but WgCA) = 0. Accordingly, m^it) = {t — 2)^ (t — 5) is
(t-2)(t-5),
We know from the
=
W2(t)
(t-2)2(t-5),
m^fy
verify that »ni(A) minimum polynomial of A.
Example
Let
9.12:
A
be a 3 by 3 matrix over the real
of the polynomial f{t) = t^ + 1. its characteristic poljmomial A(t).
We
field R.
show that
A
By
can the
cannot be a zero A is a zero of it has at least
the Cayley-Hamilton theorem, Note that A(t) is of degree 3; hence
one real root.
Now suppose A is a zero of f{t). Since f(t) is irreducible over R, f{t) must be the minimal polynomial of A. But /(t) has no real root. This contradicts the fact that the characteristic and minimal polynomials have the same roots. Thus A is not a zero of f{t). The reader can verify that the following field
C
is
a zero of
3
by 3 matrix over the complex
f(t):
fo
-1
0^
10 lO
CHARACTERISTIC AND MINIMUM POLYNOMIALS OF LINEAR OPERATORS Now suppose T:V^V is a linear operator on a vector space V with finite dimension. We define the characteristic polynomial A(<) of T to be the characteristic polynomial of any matrix representation of T. By Theorem 9.9, A{t) is independent of the particular basis in which the matrix representation is computed. Note that the degree of A{t) is equal to the dimension of V. We have theorems for T which are similar to the ones we had for matrices:
Theorem
9.5':
Theorem
9.6':
T
is
a zero of
The scalar
its
characteristic polynomial.
\GK
is
an eigenvalue of T
if
and only
if
A
is
a root of the
characteristic polynomial of T.
\G K
of T is defined to be the multiplicity The algebraic multiplicity of an eigenvalue of A as a root of the characteristic polynomial of T. The geometric multiplicity of the eigenvalue A is defined to be the dimension of its eigenspace.
Theorem
9.13:
The geometric multiplicity.
multiplicity of
an eigenvalue A does not exceed
its
algebraic
EIGENVALUES AND EIGENVECTORS
204 Example
[CHAP.
Let V be the vector space of functions which has {sin be the differential operator on V. Then
9.13:
A
The matrix
of
D
=
D{sme)
=
Z>(cos e)
= — sin 9
in the
cosfl
above basis
0(sin e)
=
+
and
D
let
l(cos9)
— l(sin e) +
therefore
is
cos 9} as a basis,
e,
9
A =
O(cos e)
V
=
[D]
Thus
(
det (tl
t
-1
1
t
=
-A)
and the characteristic polynomial of
D
is
=
A(t)
=
t^
t2
+
l.
On the other hand, the minimum polynomial m{t) of the operator T is defined independently of the theory of matrices, as the polynomial of lowest degree and leading coefficient 1 which has T as a zero. However, for any polynomial f{t), f{T)
=
if
and only
=
f{A)
if
A is any matrix representation of T. Accordingly, T and A have the same minimum polynomial. We remark that all the theorems in this chapter on the minimum polynomial of a matrix also hold for the minimum polynomial of the operator T. where
Solved Problems
POLYNOMIALS OF MATRICES AND LINEAR OPERATORS 9.1.
Find f{A) where
9.2.
Show
that
^ = =
'«-'
9.3.
T
A =
(
«
o
)
~z\ and
^^
^ ^^^^ ^^
f{t)
^
f(^)
f-M +
=
*^
-
4t
1.
-
5.
--"— a;)'-G :)-<;:)
=
a:
Let V be the vector space of functions which has {sin 6, cos ^} as a basis, and D be the differential operator on V. Show that D is a zero of f{t) = t^ + 1. Apply f(D)
to each basis vector:
/(D)(sin9) /(D)(cos
e)
= =
Since each basis vector
fm =
(Z>2
{D^ is
+ 7)(sin e) + /)(cos e)
mapped
= r»2(sin e) + = DHcos e) +
into 0, every vector
/(sine) /(cos e) i;
SV
= =
is also
-sin
«
-cos
9
+ +
mapped
sin 9
cos e into
= = by
f(D).
0.
This result
let
is
expected since, by Example 9.13,
/(<) is
the characteristic polynomial of D.
Thus
CHAP.
9.4.
EIGENVALUES AND EIGENVECTORS
9]
A be a matrix representation of an operator T. representation of f{T), for any polynomial f{t).
Show
Let
Let be the mapping T h* A, i.e. which sends the operator need to prove that (f(T)) - f(A). Suppose fit) = a„t" -\duction on n, the degree of fit).
T
We
Suppose matrix.
=
TO
Recall that
0.
=
{f(T))
=
and so the theorem holds for n
=
0(/')
Thus
where
/
=
(Hr)
205
into its
+
•
V
=
ao0(/')
=
matrix representation A. a^. The proof is by in-
and the theorem
9.5.
is
/ is the identity
/(A)
0.
Now assume the theorem holds for polynomials of degree less than n. algebra isomorphism, ^if(T))
the matrix
is
+
a^t
mapping and
the identity
/' is
that /(A)
=
0(tt„r"
=
a„0(r) 0(r»-i)
=
o„AA«-i + (a„_jA"-i +
+
a„_irn-i
+
+
•
•
+
•
+
0(a„_ir"-i •
+
aiT
•
+
•
since ^ is
an
a^') •
+
aiT
aiA
+
a<,/)
•
•
Then
+
a^')
=
/(A)
proved.
Prove Theorem 9.1: Let / and g be polynomials over K. Let A be a square matrix over K. Then: (i) (/ + fir)(A) = /(A) + flr(A); (ii) {fg){A) = /(A) g{A); and (iii) (fe/)(A) = kf(A) where fc G K. Suppose /
=
Suppose
•
•
+ Oi* + Oq
•
= a„A» +
f(A)
(i)
+
a„t"
m—n
and
•
•
+
•
let
+
/
+
ttiA
=
ftj
=
and g
if
•
•
+ bit +
•
=
^(A)
Oq/
and
i>
m. Then •••
= (a„+6„)t"+
sr
+
b^t^
+K +
+
ft^A"
+
6i)t
Then by
bf,.
•
•
+
definition,
biA
+
bol
/(A)
+
flr(A)
(ao+6o)
Hence
+ g)(A) = =
(/
{a„
+ 6„)A" +•••+(«! + 6i)A + +
a„A»
6„A«
+
•
•
+
•
ajA
+
By
definition,
=
fg
c„ +„ t" +>"+•••+ Ci*
+
Ojcfro
2
=
Hence
aj6fc_i.
1=0
/n /(A)f,(A)
(iii)
By
= kf
definition, (fe/)(A)
N/™
=
A;a„t"
+
fca„A»
+
•
•
•
•
•
•
+ +
fcajt
fettiA
+
6o)^
tto^
+
=
where
c^t''
c^
fcoo.
fcoo/
=
eio^fc
+
"'i^fc-i
+
and
CfcA^
n+
m
2 2 2»M*+^ = fc=0 t=0 j=0
=
+
60/
fc
nm
\
OiAMf 2 6jAi) /\j=o /
2 \i=o
(
=
= 2
(/ff)(A)
+
m
2=
=
n+m
fc
+
Co
+
6iA n+
(ii)
(a<,
CfcA"
=
(/ff)(A)
and so
=
A;(a„A»
+
•
•
+ ajA + o,/) =
•
k /(A)
EIGENVALUES AND EIGENVECTORS 9.6.
Let
A,
be an eigenvalue of an operator T:V^V. Let Vx denote the set of all eigenT belonging to the eigenvalue X (called the eigenspace of A.). Show that a subspace of V.
vectors of
Vx
is
Suppose v,w
&
Vx,
T(av
Thus av
+ bw
is
that
is,
+ 6w) =
T(v)
a
=
r(i;)
\v
+
an eigenvector belonging
and
6 T(w) to X,
r(w)
i.e.
=
aiw) av
\w.
+
Then for any scalars a,b
b{\w)
+ bw G
V^.
& K,
- Mav + bw) Hence Vx
is
a subspace of V.
EIGENVALUES AND EIGENVECTORS
206
9.7.
Let (ii)
(i)
A =
1
4\
2
g
I
(i)
.
Find
all
P such
Find an invertible matrix Form
the characteristic matrix
=
A(t)
The roots of
We
A(t) are 5
and the corresponding eigenvectors. diagonal.
is
tj
A
A(t) of
«-
\2
is its
3J
determinant:
-4
1
«-
-2
=
«2
_
_
4t
(1) to
First substitute
5.
_ /0\ ~ \o)
4x-4y =
f
\^2x +
""^
t
=
into
5
The eigenvectors belonging
obtain the matrix
homogeneous system determined by the above matrix,
-4\/x\ 2)\y)
4
{t-5){t+l)
-4-
4
solution of the
.-2
=
5
3
obtain the eigenvectors belonging to the eigenvalue
form the
U)
t-3
-2
\
and —1, and so these numbers are the eigenvalues of A.
the characteristic matrix 5
9
-AoiA:
*/
=
\tI-A\
P-^AP
that
\0 The characteristic polynomial
A
eigenvalues of
[CHAP.
—
X
=
2y
to
i.e.,
=
2/
form the kernel of the operator tl — A for = 5.) The above system has only one independent solution; for example, x = 1, y = 1. Thus " = (1, 1) is an eigenvector which generates the eigenspace of 5, i.e. every eigenvector belong(In other words, the eigenvectors belonging to 5 t
ing to 5
We
is
a multiple of
v.
obtain the eigenvectors belonging to the eigenvalue —1.
to obtain the
~2x -2x -
-4 -4
-2 -2
The system has only one independent is
(ii)
= —1
t
into {1)
= =
4.y 4i/
X
or
solution; for example, x
2,
3/
=
+
=
2j/
-1.
Thus
w=
(2,
-1)
an eigenvector which generates the eigenspace of —1.
Let
P
be the matrix whose columns are the above eigenvectors:
B — P~^AP
{Remark. Here
P
vectors {v, w}.
Hence
For each matrix,
P
1
2
1
-1
Then
the diagonal matrix whose diagonal entries are the respective eigenvalues:
is
B = P-^AP =
9.8.
Substitute
homogeneous system
is
'1/3
2/3\/l
1/3
-1/3 JV 2
/5
U
-1
the transition matrix from the usual basis of R2 to the basis of eigenB is the matrix representation of the operator A in this new basis.)
find all eigenvalues 1
(i)
4\/l 2\ _ sjil -1/ "
A =
3
-3 -5
and a basis of each eigenspace:
/-3
3\ 3
5 =
(ii)
,
1
-l'
5-1
-7
I
6-6 Which matrix can be (1)
diagonalized,
A
and compute «
A(t)
The roots of
A(t) are
=
\tI-A\
—2 and
4;
=
6
-2
and why?
the characteristic matrix tl — polynomial A(t) of A:
Form istic
-6
4
-3 -6
its
-3 -3
3
1 t
+ 6
determinant to obtain the character-
5 f
-
=
(t
+ 2)2(t-4)
4
hence these numbers are the eigenvalues of A.
CHAP.
EIGENVALUES AND EIGENVECTORS
9]
We
find a basis of the eigenspace of the eigenvalue —2.
acteristic
matrix
tl
—A
to obtain the
207
Substitute
t
= —2
into the char-
homogeneous system
+ + +
f-Sa; or
i
—3a;
[—6a;
— —
32/ 3j/ 62/
3z
= =
6z
==
32
—
x
or
+
2/
=
z
The system has two independent solutions, e.g. a; = l, 2/ = l, z = and a; = 1, j/ = 0, z = —1. Thus u = (1, 1, 0) and v = (1, 0, —1) are independent eigenvectors which generate the eigenspace of —2. That is, u and v form a basis of the eigenspace of —2. This means that every eigenvector belonging to —2 is a linear combination of u and v.
We
find a basis of the eigenspace of the eigenvalue 4.
acteristic
matrix
tl
—A
to obtain the
Substitute
t
=
4
into the char-
homogeneous system
+ + +
3x Sx
6x
By
--
9y
-- 3z
= = =
3z
6y
+
X
y
2y
— -
= =
z z
variable; hence any particular nonzero solution, e.g. x = 1, y = 1, generates its solution space. Thus w = (1, 1, 2) is an eigenvector which generates, and so forms a basis, of the eigenspace of 4.
The system has only one free z
=
2,
Since A has three linearly independent eigenvectors, A is diagonalizable. be the matrix whose columns are the three independent eigenvectors:
In fact, let
P
/-2 Then P-^AP = \
As
expected, the diagonal elements of
P~^AP
are the eigenvalues of
A
corresponding to the
columns of P. t
=
A(t)
(ii)
The eigenvalues
We
find
to obtain the
of
B
\tI-B\
are therefore
-1
1
7
t-5
1
6
-6
+
—2 and
3
=
+
t
(t
+
2)2(t-4)
2
4.
a basis of the eigenspace of the eigenvalue —2. homogeneous system
+
z
Ix--ly + 6a; -6y
z
X
--
The system has only one independent solution, forms a basis of the eigenspace of —2.
y
e.g.
a;
=
7a;
-
6a;-
The system has only one independent solution, forms a basis of the eigenspace of 4.
2/ 2/
62/
+ + +
e.g.
2 z
6Z a;
y
1,
4.
2/
=
1,
z
=
Substitute
= = = =
= -2
t
= = —
We find a basis of the eigenspace of the eigenvalue obtain the homogeneous system 7a;
Substitute
0.
t
7a;
+
=0
Thus
u—
=
—
4
2/
into tl
+
2/
=
1, z
=
1.
Thus
-B
=
z
y
z a;
0,
into tl
(1, 1, 0)
—B
to
= = v
(0.1,1)
Observe that B is not similar to a diagonal matrix since B has only two independent Furthermore, since A can be diagonalized but B cannot, A and B are not similar matrices, even though they have the same characteristic polynomial. eigenvectors.
EIGENVALUES AND EIGENVECTORS
208
9.9.
A =
Let
Find
A
eigenvectors of
B
and
9
eigenvalues and the corresponding
all
viewed as matrices over
[CHAP.
the real field R,
(i)
(ii)
the complex
field C.
Mt) =
(i)
Hence only 2
3
1
t
=
2
=
(2
—A
into tl
=
4
4i
t-1
-1
Put
an eigenvalue.
is
t-
=
\tI-A\
- 2)2
(t
and obtain the homogeneous system
—X + y = —X + y =
—
X
y
The system has only one independent solution, e.g. a; = 1, y = 1. Thus i) = (1, 1) is an eigenvector which generates the eigenspace of 2, i.e. every eigenvector belonging to 2 is a multiple of
V.
We
also have ^B(t)
Since
+
t2
i
=
B
has no solution in R,
t-
=
\tl-B\
1
1
-2
t
=
+
t2
+
1
1
has no eigenvalue as a matrix over R.
Since A^(t) = (t — 2)2 has only the real root 2, the results are the same as in (i). That is, 2 is an eigenvalue of A, and v = (1, 1) is an eigenvector which generates the eigenspace of 2, i.e. every eigenvector of 2 is a (complex) multiple of v.
(ii)
The
characteristic matrix of
B
is
=
Ajj(t)
\tl
— B\ =
t^
+ i.
Hence
i
and —i are the eigen-
values of B.
We
find the eigenvectors associated with
=
t
Substitute
i.
=
t
in tl
i
—B
to obtain the
homogeneous system
'i-1 -2
1
\/x\ _ /0\ [oj
{i-l)x
/
i+lAy) ~
""^
The system has only one independent solution, e.g. an eigenvector which generates the eigenspace of i.
Now '-i
—
1
substitute 1
t
=
\/x\
—i
into tl
—B
x
to obtain the
/0\
=
y
=
1,
y
+
{-i
The system has only one Independent solution, e.g. x an eigenvector which generates the eigenspace of —i.
-l)x + y =
(i
°
—
1 -
w=
Thus
(l,l
— i)
is
homogeneous system
(—i—l)x + y =
{
-2x
9.10.
+
\-2x + {i+l)y =
Q
(-i
- l)y =
1,
y
=
1
+
i.
-l)x + y
Thus w'
Find all eigenvalues and a basis of each eigenspace of the operator by T{x, y, z) = {2x + y,y- z, 2y + 4«).
T
:
R^
=
-*
=
(1,1-1-
i)
is
R^ defined
First find a matrix representation of T, say relative to the usual basis of R^:
/2
The characteristic polynomial
A =
m
T
then
A(*) of
is
t-2 A(t)
=
=
\tI-A\
-1
t-1 -2
Thus
2
We
and 3 are the eigenvalues of
0^
1
=
=
1
t-
(t-2)2(t-3)
4
T.
find a basis of the eigenspace of the eigenvalue 2.
Substitute
t
=
2
into tl
—A
the homogeneous system /
™
\
/a\
r
«.
—
n
=0
y y
+
z
=
to obtain
CHAP.
EIGENVALUES AND EIGENVECTORS
9]
The system has only one Independent solution, a basis of the eigenspace of 2.
We the
=
=
0.
Substitute
t
y
1,
find a basis of the eigenspace of the eigenvalue 3.
Observe that T
Show
that
We
9.12.
=
0, z
Thus u
—
3
=
(1, 0, 0)
into tl
—A
Thus
v
forms
to obtain
homogeneous system
The system has only one independent solution, forms a basis of the eigenspace of 3.
9.11.
x
e.g.
209
T{v)
=
Let
A
an eigenvalue of T
is
have that
Ov
=
0,
i.e.
T
not diagonalizable, since
is
if
an eigenvalue of T
is
that
T
is
if
and only
T
By Problem
9.11
—
1,
z
=
—2.
=
(1, 1,
—2)
if
is
singular.
there exists a nonzero vector v such that
singular.
B be w-square matrices. Show that AB
and
y
1,
has only two linearly independent eigenvectors.
and only if
—
x
e.g.
BA
and
have the same eigenvalues.
and the fact that the product of nonsingular matrices is nonsingular, the fol(i) is an eigenvalue of AB, (ii) AB is singular, (iii) A or B is singular, (v) is an eigenvalue of BA.
lowing statements are equivalent: singular,
(iv)
BA
is
Now suppose X is a nonzero eigenvalue of AB. Then ABv = Xv. Set w = Bv. Since \ # and v ¥= 0,
Aw = ABv = But
w
an eigenvector of
is
BA
\v
Hence X
belonging to the eigenvalue X since
Suppose of
B\v = \Bv = \w
any nonzero eigenvalue of
Similarly,
BA
is also
an eigenvalue
AB. Thus
9.13.
an eigenvalue of BA.
is
w #
and so
¥=
BAw - BABv = of
there exists a nonzero vector v such that
AB
A.
is
and
BA
have the same eigenvalues.
Show that A~*
an eigenvalue of an invertible operator T.
is
an eigenvalue
T-K Since
By
T
is invertible, it is also
definition of
nonsingular; hence by Problem 9.11,
X
# 0.
an eigenvalue, there exists a nonzero vector i; for which T(v) — Xv. Applyv = T-i(\v) = xr-i(i;). Hence r-i(v) = X-iv; that is, X"!
ing r-i to both sides, we obtain is an eigenvalue of r~i.
9.14.
Prove Theorem 9.3: Let Vi, .,Vn be nonzero eigenvectors of an operator T:V ->V Then vi, belonging to distinct eigenvalues Ai, A„. .,Vn are linearly independent. .
.
.
The proof Assume « > 1.
is
by induction on Suppose
OiV,
where the
Oj
Applying T
are scalars.
aiT(Vi)
But by hypothesis
r(i;j)
=
XjVj;
+
-I-
.
n
—
02^2
+
Ii
n.
. ,
.
to the
then Vj
1,
•
•
+
•
a„v„
above relation,
+
+
a„T(v„)
.
linearly independent since
is
=
#
0.
{1)
we
=
Vi
obtain by linearity T{0)
^
hence ajXiVi
+
02X2^2
+
•
•
•
+
ttn^n^n
=
(2)
EIGENVALUES AND EIGENVECTORS
210
On
the other hand, multiplying
Now
subtracting
(5)
from
ai(Xi
By
by
(1)
X„,
(2),
- Xj^i +
a2{\2~K)'"2
+
induction, each of the above coefficients is
Hence
aj
=
•
•
•
=
a^-i
[CHAP. 9
=
•
+
•
•
0.
Substituting this into
0.
a„_i(\„_i
Since the
we
(1)
Xj
get
- Xjiin-i =
are distinct, a„i;„
=
0,
Xj
— X„ 9^
and hence a„
for
=
i
# w. Thus
0.
the Vi are linearly independent.
CHARACTERISTIC POLYNOMIAL, CAYLEY-HAMILTON THEOREM 9.15.
Consider a triangular matrix 'an
ai2
a22
...
Find
polynomial A{t) and
its characteristic
Since
A
is
triangular and tl
diagonal, tl
is
is
/*-«!! tl
- A
eigenvalues.
its
—A
also triangular with diagonal elements
-<»12 t
^
CUnj
•
=
A(t)
|t/
— A|
is
"-in
t-anni
the product of the diagonal elements
=
A(t)
Hence the eigenvalues of
1 9.16.
Let
A =
I
2
3\
2
3
A
.
are an,
Is
A
tt22,
(t .
.
— ««:
-a In
•
0.22
\
Then
•
t
t — ««:
- aii)(t - a22) •••(*- «nn) ««„,
. ,
its
i.e.
diagonal elements
similar to a diagonal matrix?
If so, find
one such matrix.
3/ Since A are distinct,
is
A
A
are the diagonal elements 1, 2 and 3. Since they triangular, the eigenvalues of similar to a diagonal matrix whose diagonal elements are 1, 2 and 3; for example,
is
/i
o^
2
\0
9.17.
For each matrix
find a polsoiomial
3;
having the matrix as a root /l
«^
= (i -3> <«)« = (' :4)' By
<'")
^ =
»
4-3
\\I
the Cayley-Hamilton theorem every matrix is a root of we find the characteristic polynomial A(t) in each case.
Therefore (i)
(ii)
A(t)
A(«)
=
==
\tI-A\
\tI-B\
= =
t-2 -1
-B f
t-2 -7
A(t)
=
\tI-C\
=
+t-
+
=
-2
(2
+
2t
+
13
4
-4 t
11
3
3 t
t-1 (iii)
+
3 3
-1
t+1
(t
- l)(t2 -2t-5)
its
characteristic polynomial.
CHAP.
9.18.
EIGENVALUES AND EIGENVECTORS
9]
Prove the Cayley-Hamilton Theorem
Every matrix
9.5:
211
a zero of
is
its
characteristic
polynomial. Let
A
be an arbitrary w-square matrix and let A(i) be
=
A(t)
Now
let B(t)
= <«+
\tI-A\
a„_it"-i
polynomial; say,
its characteristic
+••+«!* +
do
tl — A.
denote the classical adjoint of the matrix The elements of B{t) are cofactors — A and hence are polynomials in t of degree not exceeding n — 1. Thus
of the matrix tl
B(t)
=
B„_it"-i
where the Bj are re-square matrices over of the classical adjoint (Theorem 8.8),
K which
+
•••
(i/-i4)(B„_it«-i+
-
(<"
Bo t.
By
the fundamental property
+ a„_it»-i+
l-ai«
+ aoK
corresponding powers of
coefficients of
Bn-l =
a„-il
- ABi = -ABo =
Multiplying the above matrix equations by A", A"-^,
t,
1
Bn-2~AB„_i —
Bo
+
\tI-A\I
+Bi« + Bo) =
Removing parentheses and equating the
B^t
are independent of
{tI-A)B{t) or
+
a,/
agl
.
.
A, I respectively,
.,
A"Bn-i = A" An-iB„_2 - A«J?„_i 3^ a„-iA»~i
A»-2B„_3-A«-iB„_2 = a„_2A"-2 ABo - A^Bi - a^A
—ABo =
»o^
Adding the above matrix equations,
= A» + In other words, A(A)
9.19.
Show By
=
that a matrix
That
0.
A
and
its
the transpose operation,
have the same determinant,
\tI
is,
A
is
a„_iA«-i a zero of
+
•
•
•
+
aiA
+
oo/
its characteristic
polynomial.
transpose A' have the same characteristic polynomial.
(t/-A)'
— A\ -
=
tl*
|(t7-A)*|
— A* :=
tl
- A*.
\tI-A*\.
Since a matrix and its transpose A and A* have the same char-
Hence
acteristic polynomial.
9.20.
Suppose
M
=
/Ai \
^
B\
^1
acteristic polynomial of
Ai.
where Ai and Az are square matrices.
M
is
Show
that the char-
the product of the characteristic poljmomials of Ai and
Generalize.
/tl-A,
tl-M = |t/ — A| \tl — B\, I
By
~B
\
tl-A/'
^®"*=^ ^y Problem 8.70,
\tI-M\
=
as required.
induction, the characteristic poljmomial of the triangular block matrix
tl
-
Ai tl
-B - A^
EIGENVALUES AND EIGENVECTORS
212
lA,
M
=
B
C
A2
D
where the Aj are square matrices,
the product of the characteristic polynomials of the Aj.
is
MINIMUM POLYNOMIAL 9.21.
2
The characteristic polynomial of
and
1
1
-2
4
-1
-1
t-1
-1
t-2
2
i-4
t-2.
t-
1
=
(t-3)(t-2)3
The minimum polynomial m{t) must divide A(i). Also, each irreducible factor — 3, must be a factor of m{t). Thus m(t) is exactly one of the following:
of A(t),
i.e.
t
—2
<
=
/(t)
We
2
is
t-2
=
A(f)
A
1
A =
Find the minimum polynomial m{t) of
t-2
A^l
...
\
[CHAP. 9
(t-3)(t-2),
flr(t)
=
(i-3)(«-2)2,
/i(«)
=
(t-3)(t-2)3
have -1
=
= (A-37)(A-2/)
/(A)
-2 -2 1
ff(A)
=
(A
o\ 1' '0
1
-1
1
1/ \: o\ lo
1
g{t)
=
Remark.
(t
- 37)(A - 27)2 =
— 3)(t — 2)2
the
is
We know that
h{A)
-2
minimum polynomial
—
A(A)
=;
¥-
-1
1
-2
2
-1
1
1
-1 1
0-2 Thus
1
=
,0
of A.
by the Cayley-Hamilton theorem. However, the degree is the minimum poljmomial of A.
of g{t) is less than the degn:ee of h(t); hence g(f), and not h(t),
9.22.
Find the minimal polynomial m{t) of each matrix (where a¥^Oy.
^
(i)
(ii)
The
characteristic polynomial of
A(t)
=
*/
A
is
\o A(t)
{t
-
— X)2. We
find
A — \/ #
0;
hence
— X)^. (Note m(t) is exactly one of t — \, = A(t) = (t - X)S. = (t - X)*. We find (C - X/)3 # 0; hence
is A(i) = (t characteristic polynomial of find (B X/)2 ¥- 0; thus OT(t) X)3.)
-
{t
A, m{t)
—
(t-\)K
B
The or
(iii)
\o
We
-
The
characteristic polynomial of
A(t)
=
(t-X)'«.
C
is
A(t)
{t
— X)2
m{t)
=
CHAP.
9.23.
EIGENVALUES AND EIGENVECTORS
9]
M
Let
(A
=
0\
g
I
Q
M
polynomial m(f) of
and
fe(t)
of
A
Since m(V) mifi) = divides m{t).
and
Now
is
Thus m(t)
is
the
is
multiple of the
minimum
"m,(A.)
M, m(Af) =
of
m(B)^
minimum polynomial
a multiple of
and
g(t)
and
g{t)
But m{t) is the minimum polynomial of multiple of g{t) and /i(t).
h(t);
M; hence
g(f)
=
and hence m(A)
=
of A, g{t) divides m{t).
then /(M)
m(t) divides
=
/(t).
minimum polynomial
then have, by induction, that the
M
polynomials
Similarly,
h{t)
h(t).
/f(A)
/O
\
0^
=
/(B),
V
We
minimum
that the
Generalize.
be another multiple of
let f{t)
Show
are square matrices.
common
the least
minimum polynomial
the
Since g{t)
0.
B
and
B respectively.
and is
A
where
1
213
Thus m{t)
is
the least
0.
common
of
=
\o where the Aj are square matrices,
is
the least
common
minimum polynomials
of
The minimum polynomials
of
multiple of the
the A,.
9.24.
Find the minimum polynomial m(i) of '2
M
Let
2
A
8
c The
2
A,
C and D
are
(t
— 2)^,
and
t2
it is
also the
—5 t-
=
\tI-B\
and so
t
-1
M
Show
that the
+
is
= (t-2)(t-5)
10
'\ Thus m(t)
C C and
7t
t2
B
of B.
B
= ,0
polynomials of A, B,
9.25.
t-3
(5).
characteristic polynomial of
-2
4:
minimum polynomial 'A
Observe that
respectively.
D =
is
the least
common
multiple of the
minimum
2?/
D. Accordingly, m(t)
minimum polynomial
= tm - 2)2(t — 5)
of a matrix (operator)
A
exists
and
is
unique.
By the Cayley-Hamilton theorem, A is a zero of some nonzero polynomial (see also Problem 9.31). Let n be the lowest degree for which a polynomial f(t) exists such that /(A) = 0. Dividing f(t) by its leading coefficient, we obtain a monic polynomial m(t) of degree n which has A as a zero. Suppose m'(t) is another monic polynomial of degree n for which m'{A) = 0. Then the difference m{t) — m'(t) is a nonzero polynomial of degree less than n which has A as a zero. This contradicts the original assumption on n; hence m(t) is a unique minimum polynomial.
EIGENVALUES AND EIGENVECTORS
214
9.26.
Prove Theorem
[CHAP.
The minimum polynomial m(t) of a matrix
9.10:
divides every polynomial which has
A
as a zero.
(operator)
9
A
In particular, m{t) divides the char-
acteristic polynomial of A. f{A) = 0. By the division algorithm there exist polyor deg r(t) < deg m(t). Suband r(t) = for which f{t) = m{t) q(t) + r(t) and 'm{A) = 0, we obtain r(A) = 0. If stituting t = A in this equation, and using that f(A) = r{t) ¥= 0, then r(t) is a polynomial of degree less than m(t) which has A as a zero; this contradicts and so f{t) = m{t) q(t), i.e. m(t) divides /(*)• the definition of the minimum polynomial. Thus r(t) =
Suppose
nomials
9.27.
q{t)
f(t) is
and
a polynomial for which
r{t)
Let m{t) be the minimum polynomial of an %-square matrix A. acteristic polynomial of A divides (m(i))''. Suppose
m(t)
= f+
Cjf-i
-|-
•
•
-I-
c^-it
+
that the char-
Consider the following matrices:
c^.
= / = A + cj Bo = A^ + CjA +
Show
Bo Bi
Then
B^_i
=
A"--!
B^
=
I
+
Cg/
CiA'-2
4-
•
•
+
c^_i7
- ABg = cj B^-AB^ = C2I Bi
-AB^^i = C^ - {Ar+CiAr-i+
Also,
= c^ — = Set
B{t)
+Cr-iA+CrI)
«i(A)
c^I
= f-i^o + f-^Bi +
•
•
+
tBr-2
+
Br-1
Then {tI-A)-B(t)
= (t'-Bo+t'-'^Bi+ •• +tBr-i) - (t'-^ABo + tr-^ABi+ ••• +ABr-i) = t^Bo+ tr-i{Bi-ABo)+ t'-2(B2-ABi)+ -f t(B,._i - AB^-a) - AB^-i ••• = f/ Cif-l/ + C^f-^I + + Cr-itl + C^ = m(t)I The determinant of both sides gives \tl — A\ \B{t)\ = \m(t) I\ = (TO(t))". Since \B(t)\ is a polynomial, 1*7 — A divides (m(t))"; that is, the characteristic polynomial of A divides (»n(t))".
•
-1-
I
9.28.
Prove Theorem 9.11: The characteristic polynomial A{t) and the minimum polynomial m{t) of a matrix A have the same irreducible factors. Suppose divides A(t). (m(t))^.
But
an irreducible polsmomial. If f{t) divides m{t) then, since m(t) divides A(t), f(t) the other hand, if f(t) divides A(t) then, by the preceding problem, /(«) divides Thus m{t) and A(t) have the same is irreducible; hence f(t) also divides m{t).
f{t) is
On f(t)
irreducible factors.
9.29.
be a linear operator on a vector space V of finite dimension. Show that T is and only if the constant term of the minimal (characteristic) polynomial of T is not zero. + a^t + a^f. Suppose the minimal (characteristic) polynomial of T is f(t) = f + a„_it'— 1 Each of the following statements is equivalent to the succeeding one by preceding results: (1) T is is not a root of m(t); (v) the is not an eigenvalue of T; (iv) invertible; (ii) T is nonsingular; (iii) constant term Uf, is not zero. Thus the theorem is proved.
Let
r
invertible if
-I-
•
•
•
.
CHAP.
9.30.
EIGENVALUES AND EIGENVECTORS
9]
dimF =
Suppose
r
—
T:V ^V
Let
».
T
equal to a polynomial in
215
he an invertible operator.
Show
that T-*
is
of degree not exceeding n.
= f + a^-if— i +
Let m(t) be the minimal polynomial of T. Then m{t) Since T is invertible, Kq # 0. We have
•
•
•
+
a^t
+
a^,
where
n.
y + a^_ir'-i +
=
m(r)
+
OiT
=
do/
Hence
+ aJ)T =
r-i
and
I
1
=
(Tr-i
+ ar-i r'-2 +
•
•
•
+ aj)
tto
MISCELLANEOUS PROBLEMS 9.31.
Let T be a linear operator on a vector space V of dimension n. Without using the Cayley-Hamilton theorem, show that T is a zero of a nonzero polynomial.
N
- rfi. Consider the following N-\-l operators on V: I, T,T^, .... T^. Recall that the Let vector space A(V) of operators on V has dimension — rfi. Thus the above iV + 1 operators are linearly dependent. Hence there exist scalars a^, Oj, ., aj, for which a^^T^ + •• + a^T + a^ = 0. Accordingly, T is a zero of the polynomial f{t) = a^^t^ + ••• + a^t + Oq.
N
.
9.32.
•
.
T:V ^V. The geometric
Prove Theorem
9.13: Let A be an eigenvalue of an operator multiplicity of X does not exceed its algebraic multiplicity.
Suppose the geometric multiplicity of X is r. Then X contains r linearly independent eigenvectors Extend the set {dJ to a basis of V: {v^, ...,v^, w^, .,Wg}. We have
Vi, ...,Vr.
.
.
= =
T{vi) 1'(^2)
I'(^2)
= = =
7'(W.)
=
1\Vr)
T(w,)
XVi \V2
W, ...
a2iVi
+ +
O'sl'Vl
+
...
aiiVi
The matrix of T in the above basis
a2rVr
+ 6nWi + + b2tWi +
O'sr'Or
+
+ +
...
+
airVr
&SIW1
•
•
•
•
+
•
+ +
bi,w.
+
bss^s
b2s'Ws
is
/^
«ii
«21
••
«'12
»22
•
l
ttgr
*"
621
...
6,1
&12
622
•
•
6r2
hs
hs
•
0-sl\
{
X 1
•
•
«s2
•
•
O'sr
\
r. X
1
M
= »
1
1
".
\. A =
where
(0.;^)'
By Problem
and
B=
9.33.
Show
r,
6ss
Vo
A^
5/
/
(»«)••
M and
which is (t — \Y, must divide the charThus the algebraic multiplicity of X for the operator T is
hence T.
as required.
that
The
•
(^'l
9.20 the characteristic polynomial of X/,,
acteristic polynomial of
at least
1
=
A =
is
not diagonalizable.
characteristic polynomial of
A
is
A(t)
=
find a basis of the eigenspace of the eigenvalue 1.
the homogeneous system
(t
— 1)^;
hence 1
Substitute
t
=
1
is the
only eigenvalue of A. We matrix tl — A to obtain
into the
EIGENVALUES AND EIGENVECTORS
216
o)(^) = (I) The system has only one independent of the eigenspace of
Since
9.34.
Let
A
F be
«'•
9
=
{
=
x
solution, e.g.
[CHAP.
=
y
1,
Hence u
0.
—
forms a basis
(1, 0)
1.
has at most one independent eigenvector,
an extension of a
A
Let
K.
field
A
cannot be diagonalized.
Note that
be an w-square matrix over K.
A may also be viewed as a matrix A over F. Clearly and A have the same characteristic polynomial. Show
\tl
~ A\ =
that
A
\tl
- A\,
and
A
that
also
is,
A
have the
same minimum polynomial. Let m{t) and m'(t) be the minimum polynomials of A and A respectively. Now m'{t) divides every polynomial over F vifhich has A as a zero. Since m(t) has A as a zero and since m{t) may be viewed as a polynomial over F, m'{t) divides m(t). We show now that m(t) divides m'(t). Since m'(t)
F
a polynomial over
is
=
m'(t)
which
fi(t)hi
where /i(«) are polynomials over K, and We have
=
m'{A)
Let alP denote the y-entry of
64
.
+
/i(A)bi
an extension of K, we may write
+
/aft) 62
.
6„ belong to
,
.
+ •• +
/2(A)62
+
F
and are linearly independent over K.
+
•••
fnit)b„
+
aif 62
are linearly independent over
=
/i(A)
K
+
+
•
=
=
a^f &„
...,
0,
(1)
that, for each pair {i,j),
and since the a\P S K, every
/2(A)
0,
•
=
/„(A)6„
The above matrix equation implies
/fc(A).
a'6i
Since the
61,
is
/„(A)
tty'''
=
0.
Then
=
K
which have A as a zero and since m(t) is the minimum polyare polynomials over as a matrix over K, m(t) divides each of the /;(«)• Accordingly, by (1), m(t) must also That is, divide m'(t). But monic polynomials which divide each other are necessarily equal. m(t) = tn'{t), as required. Since the
/i(t)
A
nomial of
9.35.
Let
{vi,
T{V2)
=
T"-
0.
.
.
.
,
Let
v„} be a basis of 7. T{vs)
chiVi,
It suffices to
-
aaivi
+
asiVi,
.
T F -* 7 Tivn) = :
.
.,
i
—
1,
.
.
For then
.,n.
THvj)
it
We
prove
{*)
=
•
•
•
+
an,n-iv„-i.
T{vi)
Show
=
0,
that
=
-
2,
T»-i(Ti(Vj))
=
T"
by induction on .,n} from .
(*)
follows that
and, since {v^, ...,«„} is a basis,
follows (for j
+
show that Ti{Vj)
for
be an operator for which a„iVi
=
r"-5(0)
The case
j.
=
0,
for ^
=
1
n
0.
j
=
1
is
true by hypothesis.
Ti{Vj)
=
Ti-HT(Vj))
=
aj^Ti-Hvi)
=
ajiO
= TS-HajiVi+ +aj^j.iVj^i) +
+ •• +
•••
+aj.,_irJ-i(Vj_i)
=
aj,j_iO
Remark. Observe that the matrix representation of T in the above basis diagonal elements
The inductive step
.
0:
/O
a^i
ttsi
...
a-ni
032
..
a„2
.
...
\0
...
a„,„_i
is
triangular with
J
CHAP.
EIGENVALUES AND EIGENVECTORS
9]
217
Supplementary Problems POLYNOMIALS OF MATRICES AND LINEAR OPERATORS 9.36. Let f(t) = 2*2 - 5t + 6 and g(t) = t^ - 21^ + t +
9.37.
9.38.
r:E2^R2
Let
Let
V
be defined by
T(x,y)
=
{x
+
y,2.x).
be the vector space of polynomials v(x) = ax^ Let f{t) = fi + 2t ~ 5. Find f(D){v{x)).
Find f(A), g{A), f(B)
i.
Let
+
=
/(i)
+
hx
-
t2
¥\n& f(T)(x,y).
3.
D:V ^V
Let
c.
+
2t
and g(B) where
be the differential
operator.
9.39.
Let
A
Find A2, A3, A". .0
'8 9.40.
Let
B =
0^
12
Find a real matrix
12
8
I
A
such that
B=
As.
8/
9.41.
Consider a diagonal matrix
M and
'
«!
M
=
a triangular matrix N:
...
a. "-2
"
and
•
...
Show
any polynomial
that, for
f(t),
f(M)
=
\
and
I
/Ai
M
=
f{M)
that, for
...
/(A2)
B
C
A,
D
N =
A„,
...
'/(Ai)
=
/Ai and
where the Aj are square matrices. Show
=
/(AT)
\
...
\
X
a block triangular matrix N:
...
A2
//(ai)
f{aj
M and
Consider a block diagonal matrix
c
d
f(M) and f(N) are of the form
...
9.42.
...
...
\
f^"'^^
^^
b
02
AT
a
...
'/(tti)
a^
\
any polynomial
/(*),
f(M) and f{N) are of the form
X
...
/(A2)
...
//(Ai)
\
^^^
...
^(^)
^
I
F
\
I
...
9.43.
Show
/(A„)/
...
\
that for any square matrix (or operator) A, (P^iAP)" = P-iA"F where show that f{P-^AP) = p-if{A)P for any polynomial f(t).
P
/(Aj/
is invertible.
More
generally,
9.44.
Let f{t) be any polynomial. then /(A) is symmetric.
Show
that:
(i)
/(Af)
=
(/(A))*;
(ii)
if
A
is
symmetric,
i.e.
A*
EIGENVALUES AND EIGENVECTORS 9.45.
For each matrix,
find all eigenvalues
and linearly independent eigenvectors:
/22\
« ^ =(1
3)'
/42\
(")
^
=(3
3)'
Find invertible matrices Pj, Pg and P3 such that P~iAPi,
/5-1
(-)
^=(13
P-^BP^ and P~^CPs
are diagonal.
=
A,
EIGENVALUES AND EIGENVECTORS
218
9.46.
For each matrix,
A
(i)
When
/3
1
l\
2
4
2
\1
1
3/
/
B =
(ii)
,
9
and a basis for each eigenspace:
find all eigenvalues
:=
[CHAP.
/l
2\
1
2
1
2
-1
14/
\-l
0^
1
C =
(iii)
,
1
ly
\0
P2 and P3 such that P~^APi, F^'BPj ^"d Pj^CPs are
possible, find invertible matrices Pj,
diagonal.
9.47.
A =
Consider
I
)
^
4y
Vl
and
B =
I
as matrices over the real field R.
)
\1Z
-3/
Find
all
eigen-
Find
all
eigen-
values and linearly independent eigenvectors.
9.48.
9.49.
Consider A and B in the preceding problem as matrices over the complex values and linearly independent eigenvectors.
field C.
For each of the following operators T -.K^ -* R^, find all eigenvalues and a basis for each eigen(i) T{x,y) = (3x + 3y,x + 5y); (ii) T{x,y) = (y,x); (iii) T(x,y) = (y,-x).
space:
9.50.
For each of the following operators T{x, y,z) — (x + y + (i)
eigenspace: (iii) t\x, y,
9.51.
-> R3,
:
2y
+ z,
2y
find
+ 3«);
""(J
;)•
Suppose v
is
eigenvalues
all
T(,x,
(ii)
y,z)
=
and a basis for each (x + y,y + z, —2y — z);
= (x-y,2x + 3y + 2z, x-^y + 2z).
2)
For each of the following matrices over the complex independent eigenvectors:
<"(: 9.52.
T R^ z,
D-
C, find all
field
eigenvalues and linearly
«G:r). «(;:?;
an eigenvector of operators S and 6 are any scalars.
T.
Show
that v
is
also
an eigenvector of the operator
aS + bT where a and 9.53.
Suppose v is an eigenvector of an operator T belonging to the eigenvalue is also an eigenvector of T" belonging to \".
X.
Show that
for
n >
0,
V
an eigenvalue of an operator T. Show that
9.54.
Suppose X
9.55.
Show
that similar matrices have the
Show
that matrices
9.56.
is
A
/(X) is
an eigenvalue of
f(T).
same eigenvalues.
and A* have the same eigenvalues.
Give an example where
A
and A' have
different eigenvectors.
9.57.
9.58.
be linear operators such that ST — TS. Let X be an eigenvalue of Show that is invariant under S, i.e. S(W) C W.
W
eigenspace.
y
Let of
9.59.
S and T
Let its
V
be a vector space of finite dimension over the complex invariant under a linear operator T V -* V. Show that :
let
W
be
C. Let W # {0} be a subspace W contains a nonzero eigenvector of T. field
.yV^G K" be linearly independent eigenvectors of Let A be an ii-square matrix over K. Let v^, belonging to the eigenvalues Xj, X„ respectively. Let P be the matrix whose columns are the vectors Vi,...,v„. Show that P~^AP is the diagonal matrix whose diagonal elements are the eigenvalues Xj, X„. .
A
.
.
.
.
.
.
.
,
,
CHARACTERISTIC AND MINIMUM POLYNOMIALS 9.60.
T and
For each matrix,
find a polynomial for
which the matrix
is
a root: '2
^^
^ = (4
D'
(")
^ = (3
3)'
^"^
'^='[1
3
-2^ 4
i_i
1.
.
CHAP.
9.61.
EIGENVALUES AND EIGENVECTORS
9]
219
Consider the w-square matrix
A =
Show 9.62.
that
/(«)
=
(t-
X)™
both the characteristic and
is
5
3
0\
4
2
3
5
B
and
B =
1/
2
2
Show
.
that
A
9.66.
Show
9.67.
be an n-square matrix for which
that a matrix
Suppose
f(t) is
A
and
A^ =
is
and
B
if
and only
k>
for some
Q
minimum
called the scalar
k&K
transpose A* have the same
its
A
have different characteristic
1/
\0
The mapping T:V -*V defined by T{v) = kv Show that T is the scalar mapping belonging to T is m(t) = t-k. Let
x/
VO
0\
polynomials (and so are not similar), but have the same matrices may have the same minimum polynomial.
9.65.
0X0
1 3
/2
0\
2
\0
c =
1
7/
1
1
A =
3
3
\0
Let
0X000 0X00
3
A =
0\
IX
1
2
9.64.
of A.
Find the characteristic and minimum polynomials of each matrix: /2
9.63.
minimum polynomial
mapping belonging if
minimum
=
to
k
&
K.
the minimal polynomial of
Show that A" =
n.
an irreducible monic polynomial for which f(T) that f(f) is the minimal polynomial of T.
Thus nonsimilar
polynomial.
0.
polynomial.
where
T
Is
a linear operator
T:V ^V. Show
9.68.
Consider a block matrix acteristic matrix of Af
M
=
V
I
\^
Show
that
tl
—
"/
M
= (
\
_
fj
^
_ n
)
is
the char-
V
9.69.
Let r be a linear operator on a vector space of finite dimension. Let Tf' be a subspace of V invariant under T, i.e. T(W) cW. Let Tyf-.W-^W be the restriction of T to W. (i) Show that the characteristic polynomial of Tt„ divides the characteristic polynomial of T. (ii) Show that the minimum polynomial of Tyf divides the minimum polynomial of T.
9.70.
Let
A = ,
A(«)
ail
«i2
<*i3
'*21
"^22
"'•23
a^i
a^2
<^33
(ail
Show
that the characteristic polynomial of
"ll
+ 022 + a33)f2 +
"^21
9.71.
''12
"•11
"'13
"22
A
is
t
'*22
A
"31
"33
"32
Hi
^13
"21
"22
"23
"31
032
"S3
"23 "33
Let be an m-square matrix. The determinant of the matrix of order n — obtained by deleting the rows and columns passing through diagonal elements of A is called a principal minor of degree n~m. Show that the coefficient of t™ in the characteristic polynomial A(t) = \tI — A\ is the sum of all principal minors of A of degree n — multiplied by (— l)"-™. (Observe that the preceding problem is a special case of this result.)
m
m
m
EIGENVALUES AND EIGENVECTORS
220
9.72.
Consider an arbitrary monic polynomial f(t) = t" + a„_if"^i n-square matrix A is called the companion matrix of f{t):
1 1
Show 9.73.
that
A
Find a matrix
minimum polynomial
the
f{t) is
.
.
.
.
—do -«1
.
.
-«£
.
.
1
[CHAP.
+
•••
+
a^t
+
9
The following
a^.
-ftn-l
of A.
whose minimum polynomial
is
t^
(i)
-
St^
+
6t
+
8,
(ii)
t^
-
51^
-
2t
+
7t
+
4.
DIAGONALIZATION 9.74.
A =
Let
9.76.
be a matrix over the real
)
,
dj
\c
A
and d so that
a, b, c
9.75.
(
is
diagonalizable,
i.e.
field R.
that a matrix (operator)
A
diagonalizable
is
sufficient conditions
on
has two linearly independent eigenvectors.
Repeat the preceding problem for the case that
Show
Find necessary and
is
if
a matrix over the complex
and only
field C.
minimal polynomial
if its
is
a product
of distinct linear factors.
9.77.
Let
K
such that (i) AB = BA and (ii) A and B are both be Ji-square matrices over Show that A and B can be simultaneously diagonalized, i.e. there exists a basis of in which both A and B are represented by diagonal matrices. (See Problem 9.57.)
A
B
and
diagonalizable.
K^ 9.78.
E iV ^V
Let
be a projection operator,
can be represented by the diagonal matrix
Answers -26
=
-3 _27
f(A)
9.37.
f(T)(x, y)
=
(4a;
9.38.
f(D){v(x))
=
-5aa;2
"'•
--':
I)'
9.40.
Hint.
5
(
)
.
+
(4a
(ii)
9.45.
(i)
(ii)
(iii)
Using
(i),
\
a
61 c
we have
2 (
_^
)
/3
39\
=
= =
4,
-u
6,
1;
Xg
X2
1\ J
(2a
where r
is
diagonalizable and, in fact,
the rank of E.
/3
12
= (o
15
6\
,)'
^(^)
+ 26 - 5c).
B = A^
Set
.
(/(A))«
= 1, M = (2, -1); = 1, M = (2, -3); Xi \ = 4, u= (1,1).
=
j
is
and then obtain conditions on
a, 6
and
c.
2j
Xi
Let Pi
^
E
that
^-ii:
t)'
2 .0
9.44.
i
Show
-27)' /(^)=(o
-65
- 5b)x +
^'-(i
A =
E.
- y, -2x + 5y).
'2
Let
9{A)
A =
=
Supplementary Problems
to
-40
9.36.
E^
i.e.
)
and P2
/(A«)
=
=
/(A).
= =
(1, 1).
/
2
(1, 1).
1\
(
eigenvector, and so cannot be diagonalized.
1.
P3 does not exist since
C
has only one independent
CHAP.
9.46.
EIGENVALUES AND EIGENVECTORS
9]
= 2, M = (1, -1, 0), V = (1, 0, -1); Xj = 6, w = (1, 2, 1). = 3, M = (1, 1, 0), V = (1, 0, 1); X2 = 1, w = (2, -1, 1). = 1, tt = (1, 0, 0), V = (0, 0, 1).
(i)
Xi
(ii)
Xi
(iii)
X
1
/
=
Let Pj
/I
1\
1
—1 -1
\
9.47.
(i)
X
=
3,
M
9.48.
(i)
X
=
3,
M=
(i)
Xi
=
2,
=
1/
(ii) (iii)
9.51.
(i)
V
9.56.
=
—1
\0 and
has no eigenvalues
-1).
(ii)
Xj
=
6,
1)
M=
-1); X2
(3,
=
2i,
P3 does not
.
C has
at most two
u
=
3
(1,
=: (1, 1).
(in R).
- 2i); =
Xi
(ii)
Xg
=
-2i,
1,
m
=
i;
=
(1, 1);
(1,
X2
3
+ 2i>
=
-1, v
=
(1,
2,
u=
-1).
(iii)
1;
(in R).
!;
=
1,
-t).
(1,
M=
X2
(1,0);
(iv)
=
Xi
t,
=
i, 1)
M =
=
(1,1
(2, 1 - i);
+ i). Xg =
X v
(ii)
—
i,
= =
w=
3,
1,
m=
(2,
1
+
(1,
(1,0).
-2, -1).
(iii)
Xj
A = ( ^ ) Then X = 1 is the only eigenvalue and v = (1, 0) \^ /I 0\ X = 1 is still the only of X = 1. On the other hand, for A* = j ( ^ = X 1. eigenspace of the generates Let
=
(3,i);
=
-2,
generates the eigenspace
.
,
9.58.
Xj
i).
''
9.57.
There
(in B).
= 1, M = (1, 0, 0); Xj = 4, = (1, 1, 2). = X 1, M = (1, 0, 0). There are no other eigenvalues = (2, -2, -1); X3 = = 1, M = (1, 0, -1); X2 = 2, Xi Xi
exist since
so cannot be diagonalized.
B
(1,
I
1/
1
(ii)
Xi
(i)
\
2\
1
1
(1,-1);
are no eigenvalues
9.50.
—
and P2
2
linearly independent eigenvectors,
9.49.
221
eigenvalue, but
w=
(0, 1)
Let v G W, and so T{v) = Xv. Then T(Sv) = S(Tv) - S(\v) = \(Sv), that is, Sv is an eigenvector and thus S(W) C W. of T belonging to the eigenvalue X. In other words, Sv e
W
Let T:W-*W be the restriction of T to W. The characteristic polynomial of T is a polynomial an over the complex field C which, by the fundamental theorem of algebra, has a root X. Then X is T. eigenvector of an which also is in eigenvector nonzero T has a eigenvalue of T, and so
W
9.59.
Suppose T(v)
9.60.
(i)
/(<)
9.62.
(i)
A(t)
=
Xv.
Then
= t^-St + 43,
(t-X)5;
=
(ii)
{kT){v)
g(t)
(t-2)3(t-7)2; m(f) m(«)
-
kT{v)
=
k(\v)
^t^-8t + 23,
=
(iii)
(t-2)2(t-7).
Use the
Hint.
result of
Use the
=
Ht)
t^
=
- 6*2 + 5f - 12. (<-3)«; m(<)
= «-X.
Problem
9.72.
(i)
A =
-8\ -6
h \0
9.77.
(k\)v.
h(t)
(ii)
/O 9.73.
=
result of Problem 9.57.
1
5/
,
(ii)
A =
=
(t-3)».
(iii)
A(«)
=
chapter 10
Canonical Forms INTRODUCTION r be a linear operator on a vector space of finite dimension. As seen in the preceding chapter, T may not have a diagonal matrix representation. However, it is still possible Let
of ways. This is the main topic theorem, and the decomposition In particular, we obtain the primary of this chapter. triangular, Jordan and rational canonical forms. to "simplify" the
We
matrix representation of
T
in a
number
comment that the triangular and Jordan canonical forms
exist for
T has all its roots in the base field K. C but may not be true if K is the real field R.
the characteristic polynomial A{t) of true
if
K is the complex field
T
if
and only
This
is
if
always
introduce the idea of a quotient space. This is a very powerful tool and will be used in the proof of the existence of the triangular and rational canonical forms.
We also
TRIANGULAR FORM Suppose T can be rep-
be a linear operator on an n-dimensional vector space V. resented by the triangular matrix
Let
r
(an
ai2
...
ttin
(122
.
0,2n
.
.
ann
Then the
\
I
characteristic polynomial of T, A{t)
=
|*/-A|
a product of linear factors. namely, is
Theorem
Alternate
10.1:
Form
=
{t
- an){t - a^i)
The converse
is
.
.
also true
.
[t
- ann)
and
is
an important theorem;
be a linear operator whose characteristic poljmomial factors Then there exists a basis of V in which T is into linear polynomials. represented by a triangular matrix.
Let
of
T:V-^V
Theorem
10.1:
Let A be a square matrix whose characteristic polynomial factors into linear polynomials. Then A is similar to a triangular matrix, i.e. there exists an invertible matrix P such that P'^AP is triangular.
say that an operator T can be brought into triangular form if it can be represented by a triangular matrix. Note that in this case the eigenvalues of T are precisely those entries appearing on the main diagonal. We give an application of this remark.
We
222
CHAP.
CANONICAL FORMS
10]
Example
10.1
Let
:
223
A
be a square matrix over the complex field C. Suppose X is an eigenvalue of A2. that a/x or -Vx is an eigenvalue of A. We know by the above theorem that similar to a triangular matrix
Show
A
is
/Ml
=
B Hence A^
is
1^2
similar to the matrix
/.?
52
=
Since similar matrices have the same eigenvalues, X = ya? for some = VX or ^j = — -y/x; that is, Vx or - Vx is an eigenvalue of A.
i.
Hence
/jj
INVARIANCE Let T:V-*V T-invariant restricted
defined
if
be linear.
W
T maps
A
subspace IF of
into itself,
7
is
vGW
if
i.e.
said to be invariant under T or T{v) G W. In this case T
implies
toW defines a linear operator on W; that is, T induces a linear operator f:W-*W = T{w) for every w GW.
by T{w)
Example
10.2:
Let
T K^ ^ R3 :
by an angle
be the linear operator which rotates each vector about the z axis
e:
T(x, y,
z)
=
{x cose
—
y
Observe that each vector w = {a, b, 0) in the xy plane remains in under the mapping T, i.e. is r-invariant. Observe also that the z axis U is invariant under T. Furthermore, the restriction of T to rotates each vector about the origin O, and the restriction of T to TJ is the identity mapping on U.
W
W
sin
e,
x sine -
.fj
W
+
y cos
e,
z)
T(v)
^^
'
W
Example
10.3:
Nonzero eigenvectors of a linear operator T V ^ V may be characterized as generators of T-invariant 1-dimensional subspaces. For suppose T{^v) = \v, v 0. Then = {kv, k e K}, the 1-dimensional subspace generated by v, is invariant :
#
W
under
T
because T{kv)
=
k T(v)
=
k(\v)
= kWEiW
^
Conversely, suppose dim 17 = 1 and m generates U, and U is invariant under T. Then T{u) e V and so T(u) is a multiple of u, i.e. T(u) = /lu. Hence u is an eigenvector of T.
The next theorem gives us an important
Theorem
10.2:
Let
T:V^V
class of invariant subspaces.
be linear, and
let f{t)
be any polynomial.
Then the kernel
of f{T) is invariant under T.
The notion
Theorem
10.3:
of invariance
Suppose
W
is
related to matrix representations as follows.
an invariant subspace of T:V-^V. Then T has a block A B' matrix representation where A is a matrix representation of q r^ is
]
[
the restriction of
T
to
W.
.
CANONICAL FORMS
224
INVARIANT DIRECT-SUM DECOMPOSITIONS A vector space V is termed the direct sum of its if
every vector v
GV
subspaces Wi,
.
.,Wr, written
.
can be written uniquely in the form 1)
The following theorem
Theorem
[CHAP. 10
= wi + W2+
•
•
+
•
G Wi
with Wi
iVr
applies.
Suppose Wi, ...,Wr are subspaces of V, and suppose
10.4:
{Wn,
.
.
.
,
Wm^},
.
{WrU
.,
.
.
.
.
V
are bases of Wi,...,Wr respectively. Then .,wi„i, Wi if and only if the union {wn, .
.
.
.
tVrn^}
,
the direct
is
..wn,
.
.
.,w™J
sum is
of the
a basis
of V.
Now
suppose T-.V-^V .,Wr: subspaces Wi, .
linear
is
V
sum
the direct
is
of (nonzero) T-invariant
.
®Wr
V = Wi® Let
and
T{Wi)
and
i^l,...,r
cWi,
T to Wi. Then T is said to be decomposable into the operators Also, the subTr. direct sum of the Ti, written T = Ti ©
Ti denote the restriction of
Ti or
T
is
said to be the
spaces Wi, ...,Wr
•
are said to redvxe
•
•
Tor to form a T-invariant direct-sum decomposition of F.
W
reduce an operator T:V-^V; Consider the special case where two subspaces U and W2, ws) are bases of [/ and = and (wi, u^} = suppose {ui, and S dim 2 and say, dim C/ respectively, then and C7 T to of restrictions denote the T2 and respectively. If Ti
W
W
W
T2{wi) Ti
(ui)' '^
Tl^(U2)
= =
anUi
+
ai2U2
0.21^1
+
a22U2
„
,
,
rr,
,
X
T2{W2) T2(W3)
-^
= bnWi + = &21W1 + = bsiWl + ,
,
I.
,
+
h\2W2
bi3W3
,1,
r,
&22W2
+
&23W3
U
+
J> .„ O33W3
hz2W2
I
Hence
A =
an ""
a2i "''
ttl2
0^22
f
and
^
B =
'&n
&21
^31
612
^22
b
613
&23
^"33
I
I
1
{mi, M2, wi, W2, wz) are matrix representations of Ti and Ta respectively. By the above theorem T in this basis is matrix of = the = r2(Wj), and r(Wi) T,{Ui) Since r(tti) is a basis of V.
the block diagonal matrix
A generalization Theorem
10.5:
„
I
1
argument gives us the following theorem.
of the above
and V is the direct sum of T-invariant submatrix representation of the restriction of a spaces Wu ••, Wr. If Ai is by the block diagonal matrix represented T to Wi, then T can be Suppose
T:V^V
is
linear
[Ai
M
...
A2
...
A,
...
M with
The block diagonal matrix direct
sum
of the matrices Ai,
.
.
.
,
diagonal entries Ai, = Ai Ar and denoted by .
M
.
.,
©
A. •
•
is •
sometimes called the
© Ar.
CHAP.
CANONICAL FORMS
10]
225
PRIMARY DECOMPOSITION The following theorem shows that any operator is decomposable into operators whose minimal polynomials are powers of irreducible pols^omials. This is the first step in obtaining a canonical form for T,
T:V^V
Primary Decomposition Theorem
T:V^V
Let
10.6:
be a linear operator with minimal
polynomial
=
m{t)
where the direct
T
.
Moreover,
fi{T)"'.
V is the the kernel of the minimal polynomial of the restriction of
are distinct monic irreducible polynomials. of T-invariant subspaces Wi, .,Wr where Wi
fi{f)
sum
/i(f)">/2(t)"^... /.(*)"'
/i(;t)«i
is
.
Then
is
to Wi.
Since the polynomials /i(i)"* are relatively prime, the above fundamental result follows (Problem 10.11) from the next two theorems.
Theorem
10.7:
Suppose T:V^V is linear, and suppose f{t) = git)h(t) are polynomials such that f{T) = and g{t) and h(t) are relatively prime. Then V is the direct sum of the T-invariant subspaces U and W, where U = Ker g{T) = Ker h{T). and
W
Theorem
10.8:
In Theorem 10.7, if f{t) is the minimal polynomial of T [and g{t) and h{t) are monic], then g{t) and h{t) are the minimal polynomials of the restrictions of T to U and respectively.
W
We
will also use the
primary decomposition theorem
to prove the following useful
characterization of diagonalizable operators.
Theorem
10.9:
Alternate
Form
^V
A
linear operator T -.V has a diagonal matrix representation if and only if its minimal polynomial m{t) is a product of distinct linear polynomials.
Example
10.4:
10.9: A matrix A is similar to a diagonal matrix if and only minimal polynomial is a product of distinct linear polynomials.
Theorem
of
if its
Suppose
A
is
A#
7
is
a square matrix for which
similar to a diagonal matrix
complex
if
A
is
A^
=
I.
a matrix over
Determine whether or not (i)
the real field R,
(ii)
the
field C.
Since A^ - I, A is a zero of the polynomial f(t) = t^-1 = {t- l){t^ +t + The minimal polynomial m(t) of A cannot be t — 1, since A ¥' I. Hence m{t)
=
t2
+
t
+
1
or
m(t)
=
t^
-
l).
X
Since neither polynomial is a product of linear polynomials over R, A is not diagonalizable over R. On the other hand, each of the polynomials is a product of distinct linear polynomials over C. Hence A is diagonalizable over C.
NILPOTENT OPERATORS A linear operator T F -» V
is termed nilpotent if T" = for some positive integer n; k the index of nilpotency of T if T'' — but T''-^ ¥= 0. Analogously, a square matrix A is termed nilpotent if A" = for some positive integer n, and of index fc if A'' = but yj^k-i ^ Clearly the minimum polynomial of a nilpotent operator (matrix) of index k is m{t) — f"; hence is its only eigenvalue. :
we
call
The fundamental
Theorem
10.10:
result
on nilpotent operators follows.
Let T:V-^V be a nilpotent operator of index k. Then T has a block diagonal matrix representation whose diagonal entries are of the form
[CHAP. 10
CANONICAL FORMS
226
1 1
.
.
.
.
.
.
.
.
N 1
except those just above the main diagonal where are of orders of order k and all other ^ k. The number of of each possible order is uniquely determined by of all orders is equal to the nullity T. Moreover, the total number of (i.e. all
entries of A^ are
they are
There
1).
is
at least one
N
N
N
N
of T.
N
of order i is In the proof of the above theorem, we shall show that the number of — mi+i — Mi- 1, where mj is the nullity of T\ We remark that the above matrix is itself nilpotent and that its index of nilpotency is of order 1 is just the 1 x 1 zero equal to its order (Problem 10.13). Note that the matrix
2mi
N
N
matrix
(0).
JORDAN CANONICAL FORM An operator T can be put into Jordan canonical form if its characteristic and minimal polynomials factor into linear polynomials. This is always true if K is the complex field C. In any case, we can always extend the base field Z to a field in which the characteristic and minimum polynomials do factor into linear factors; thus in a broad sense every operator has a Jordan canonical form. Analogously, every matrix is similar to a matrix in Jordan canonical form. Theorem
Let T:V ->¥ be a linear operator whose characteristic and minimum polynomials are respectively m{t) = (i - Ai)"' ...{t- Xr)™and A{t) = (t- Ai)"' ...(*- XrY'
10.11:
Then T has a block diagonal matrix Ai are distinct scalars. representation / whose diagonal entries are of the form
where the
/A;
Ai «/ ij
0\
...
1
...
1
— Ai
Ai/
For each
A.
the corresponding blocks Ja have the following properties: at least one Ja of order mi;
(i)
There
(ii)
The sum of the orders of the Ja
(iii)
The number
(iv)
The number of Ja of each
is
is
all
other Ja are of order
^ mi.
m.
of Ja equals the geometric multiplicity of
Ai.
possible order is uniquely determined
Ai
A
1 Ai
1
T.
in the above theorem is called the Jordan canonical form of the block diagonal Ja is called a Jordan block belonging to the eigenvalue Ai.
The matrix J appearing operator T. Observe that
by
.
.
.
.
.
.
^\
1
...
Ai
Ai
..
1
...
..
+ ..
.
Ai
1
A
J
... .
.
.
Ai
'
Ai
.. ..
1
.
CHAP.
That
CANONICAL FORMS
10]
227
is, Jtj
=
Xil
+
N
N
where is the nilpotent block appearing in Theorem 10.10. In fact, we prove the above theorem (Problem 10.18) by showing that T can be decomposed into operators, each the sum of a scalar and a nilpotent operator. Example 105:
Suppose the characteristic and minimum polynomials of an operator T are respectively A(«)
=
(f-2)4(t-3)3
and
Then the Jordan canonical form of T
m{t)
=
(«-2)2(t-3)2
one of the following matrices:
is
or
The first matrix occurs if T has two independent eigenvectors belonging to its eigenvalue 2; and the second matrix occurs if T has three independent eigenvectors belonging to 2.
CYCLIC SUBSPACES Let r be a linear operator
on a vector space V of finite dimension over K. Suppose V and v ^ 0. The set of all vectors of the form f{T){v), where f{t) ranges over all polynomials over K, is a T-invariant subspace of V called the T-cyclic subspace of V generated by v;we denote it by Z{v, T) and denote the restriction of T to Z{v, T) by r„. We could equivalently define Z{v,T) as the intersection of all T-invariant subspaces of V containing v.
GV
Now
consider the sequence V, T{v),
T\v), T\v),
.
.
of powers of T acting on v. Let k be the lowest integer such that T''{v) bination of those vectors which precede it in the sequence; say, T^iv)
Then
=
m„(i)
-ttfc-i
=
t"
T'^-^v)
+
-
ak-it^-^
-
...
+
aiT{v)
+
ait
+
the unique monic polynomial of lowest degree for which mv(T) T-annihilator of v and Z{v, T).
is
The following theorem
Theorem
10.12:
Let Z(v,
is
a linear com-
aov ao (v)
=
0.
We
call m„(i) the
applies.
T),
T^ and m„(i) be defined as above.
(i)
The set
(ii)
The minimal polynomial of T„
(iii)
The matrix representation
{v, T{v), ..., r'=-i (v)} is
Then:
a basis of Z{v, T); hence dim Z{v, T) is
m„(i).
of Tv in the above basis is
=
fe.
CANONICAL FORMS
228
.
1 1
is called
— tto
.
.
.
-ai
.
.
-tti
.
.
.
The above matrix C
[CHAP. 10
— aic-2 — aic-i
1
.
the companion matrix of the polynomial m„(t).
RATIONAL CANONICAL FORM In this section we present the rational canonical form for a linear operator T:V^V. emphasize that this form exists even when the minimal polynomial cannot be factored into linear polynomials. (Recall that this is not the case for the Jordan canonical form.)
We
Lemma
Let
10.13:
T:V-*V
f{t) is
be a linear operator whose minimal polynomial
Then V
a monic irreducible polynomial.
V =
©
Z{vi, T)
•
•
•
e
is
the direct
is /(*)"
where
sum
Zivr, T)
of T-cyclic subspaces Z{Vi, T) with corresponding T-annihilators
-, fit)"',
/(*)"', /(«)"^
Any
other decomposition of
V
n =
Ml
^ %2 -
•
•
•
- Wr same number
into jT-cyclic subspaces has the
of components and the same set of T-annihilators.
above lemma does not say that the vectors vi or the T-cyclic subdetermined by T; but it does say that the set of T-annihilators uniquely spaces Zivi, T) are T. Thus T has a unique matrix representation are uniquely determined by
We emphasize that the
Cr
\
where the
d
polynomials
are companion matrices.
In fact, the Ci are the companion matrices to the
/(*)"*.
Using the primary decomposition theorem and the above lemma, we obtain the following fundamental result.
Theorem
10.14:
Let
T:V^V
be a linear operator with minimal polynomial m{t)
=
fi{tr^f2{tr- ... fsitr-
Then T has a
are distinct monic irreducible polynomials. unique block diagonal matrix representation
where the
/{(«)
'Cn
\ Clrj
Cs
where the C« are companion matrices. panion matrices of the polynomials
mi =
nil
— ni2
=
ni: \'
In particular, the C« are the com-
/i(t)"« rris
where
=
TCsl
—
^52
—
•
•
•
— Msr.
CHAP.
CANONICAL FORMS
10]
The above matrix representation of T nomials
called its rational canonical form.
The
poly-
are called the elementary divisors of T.
/i(i)»"
Example
is
229
Let V be a vector space of dimension 6 over R, and let T be a linear operator whose minimal polynomial is m{t) = (t^-t + 3)(« - 2)2. Then the rational canonical form of T is one of the following direct sums of companion matrices:
10.6:
(i)
C(t2-( +
3)
(ii)
C{f2-t +
3)
(iii)
C(t2-t +
where
3)
e © ©
C(f(t)) is the
C(«2-t
+ 3)
C((t-2)2) C((t-2)2)
©
© ©
C((<-2)2)
C((t-
2)2)
©
C(f-2)
companion matrix of
/(t);
C(t-2)
that
is,
/O
-3
I
.L__
12 (i)
(ii)
QUOTIENT SPACES Let F be a vector space in V,
we
write v
K
over a field and let T7 be a subspace of V. for the set of sums v + w with w GW:
+W
W of W
+
V
=
We
If
v
show (Problem
any vector
10.22) that these cosets
W be the subspace of R2 defined W = b): a=b} That W the line given by the equation x — y = We can view V + W as translation of the Let
10.7:
is
wGW)
+ w:
{V
These sets are called the cosets in V. partition V into mutually disjoint subsets. Example
(iii)
v
by
+
w
{(a,
is
is,
0.
line,
a,
obtained by adding the vector v to each point in W. As noted in the diagram on the right, v is also a line and is parallel to W. Thus the cosets of in R2 are precisely all the lines parallel to W.
+W
W
In the next theorem we use the cosets of a subspace vector space; it is called the quotient space
W of a vector space
ofVhyW and is denoted by
new
Theorem
Let
10.15:
W he a subspace of a vector space over a field K.
V
to define
Then the
WinV form a vector space over K with the following operations tion
and scalar multiplication:
(i)
{u
(ii)
kiu
+ W) +
{v
+ W) =
{u
+ v) +
+ v) +
W
=
of addi-
+ W) = ku + W, where kGK. it is first
W = u' + W
{u
cosets of
W
We note that, in the proof of the above theorem, operations are well defined; that is, whenever u + (i)
a
V/W.
(u'
+ V') +
W
and
(ii)
ku+W
-
necessary to show that the = v' + W, then and v +
ku'
W
+ W,
for any
k&K
CANONICAL FORMS
230
we have
In the case of an invariant subspace,
Theorem
W
Suppose
10.16:
[CHAP. 10
the following useful result.
a subspace invariantunder a linear operator^
is
T V -» :
V.
Then T induces a linear operator f on V/W defined by T{v -\-W) = T{v) + W. Moreover, if T is a zero of any polynomial, then so is T. Thus the minimum polynomial of T divides the minimum polynomial of T.
Solved Problems
INVARIANT SUBSPACES 10.1. Suppose T:V -^V is linear. Show (i)
(ii)
{0},
We
(i)
have
(iii)
(ii)
For every v Let
is
T.
€
Then
=
is
the intersection Suppose V
€:W=
W=
HiWi is v e Wi
then
riiWi and so
i.e.
Thus
a subspace of V.
is
G
v
eV,
it
is
v
certainly true if
G Im
T.
Hence the image of
Ker/(r),
f(T)(T(v))
=
W
of f{T)
is
f(T)(v)
i.e.
Wj
is
T-invariant,
T(v)
be any linear operator and invariant under T.
=
We
0.
Since f{t)t=tf(t),
0.
Show
that
also T-invariant. for every i. Since is T-invariant.
T:V-^V
Let
10.2:
Then the kernel
Suppose V of /(r),
T
since the kernel of
a collection of T-invariant subspaces of a vector space V.
eW;
Prove Theorem nomial.
invariant under T.
is
S Ker T
invariant under T.
Suppose {Wi}
Thus tIv)
invariant under T:
is
invariant under T.
is
V
V; hence
^(m)
that each of the following image of T.
(iv)
hence {0}
{0};
Since T{v) G Im T for every T is invariant under T.
(iv)
10.3.
G
V, T(v)
G.
u e Ker
kernel of T,
=
T(Q)
(iii)
Ker T
10.2.
V,
=
f(T)T(v)
need to show that
^(i;)
G Wj
let f{t)
for every
i.
be any poly-
also belongs to the kernel
we have f(T)T=Tf(T). Thus
Tf(T){v)
=
T(0)
=
as required.
10.4.
Find
all
invariant subspaces of
A -
viewed as an operator on
(
R^.
J
First of all, subspaces, then
that R^ and {0} are invariant under A. Now if A has any other invariant must be 1-dimensional. However, the characteristic polynomial of A is
we have it
A(t)
=
=
\tI-A\
t-2 -1
5
+
t
=
t2
+
1
2
A has no eigenvalues (in R) and so A has no eigenvectors. But the 1-dimensional invariant subspaces correspond to the eigenvectors; thus R2 and {0} are the only subspaces invariant under A.
Hence
10.5.
Prove Theorem
10.3:
Suppose
W
is
an invariant subspace of T:V-^V.
We We
T
of
T
to
W.
choose a basis {wi, ....wj of
have
W
^
and extend
A
is
a matrix representa-
J
(
tion of the restriction
where
„
has a block diagonal matrix representation
Then T
^ it
to a basis {w^,
.
.
.,Wr,Vi
v^}
of V.
CHAP.
CANONICAL FORMS
10]
231
A. A.
T{W2)
=
T{Wr)
—
T(W2)
=
a2iWi
+
r(Wr)
a^iWi
T{V<;^
= = =
621«'l
+ + +
^^(ys)
=
6slWl
+
^(i;!)
But the matrix of T
•
+
•
•
•
+ + +
•
•
•
•
•
•
•
•
Oar^r
ftrr^r
hr'^r
+ +
ftsr^r
+
bi^w^
+
Therefore
the matrix of coefficients for the obvious subsystem.
T
10.6.
T denote the restriction of an operator T T{w) — T{w) for every w GW. Prove:
(ii) (i)
•
•
•
•
•
•
•
•
+
c^^v^ Cas^j
Css^s
coefficients in the )
above system
is
the transpose of
A
is
the matrix of
i.e.
= fiT)(w).
/(f)(w)
f(t),
an invariant subspace W,
to
The minimum polynomial of T divides the minimum polynomial of T. If /(*) = or if f{t) is a constant, i.e. of degree 1, then the result clearly holds. Assume deg/ = n > 1 and that the result holds for polynomials of degree less than n. Suppose that + ajt + oo /{*) = a„t" + a„_i f»-i + •
•
Then
for every
T
of
= = = =
f(T){w)
Let m(t) denote the
(ii)
+
(
Let
For any polynomial
CjiVi
+ +
•
W.
relative to the basis {Wj} of
(i)
+ +
where A C^ By the same argument,
has the form
it
C21^1
Cij-Wj
matrix of
in this basis is the transpose of the
(See page 150.)
of equations.
b^jWi
•
w S W;
(a„r»
+ o„_ir"-i +
+ ao/)(w) + + aoI)(w) (o„_ir«-i + + oo/)(w)
(a„r»-i)(r(w))
•
•
•
(a„_i rn-i
+ +
(a„h-i)(T(w))
•
.
.
•
.
•
fiTHw)
minimum polynomial of T. Then by (i), m(T)(w) = m{T)(w) = 0(to) = is, T is a zero of the polynomial m(t). Hence the minimum polynomial
that
divides m{t).
INVARIANT DIRECT-SUM DECOMPOSITIONS 10.7.
Prove Theorem i
=
l,
.
.,r,
.
and only ,
IS
if
.
.
.,
is
Wi„^}
.
.
the union
B =
„ ,,
.
.,Wr are subspaces of V and suppose, for Wi, a basis of Wu Then V is the direct sum of the Wi if
Suppose
10.4:
{wii,
{Wu,
.
.
.
Win,,
,
.
.
.
Wrl,
,
.
.,
.
Wm.)
a basis of V.
B is a basis of V. Then, for any v &V, = duWii + + ai„jWi„j + + a^iW^i + Wj = a-ai^n + + ai„.Wi„. G PTj. We next show
Suppose V
where
•
•
•
•
•
.
.
. ,
—
w'l
+
Win.} is a basis of Wi,
V
=
•
•
V
Since {wji,
•
•
+
611W11
•
•
+ + w'r = w[ b^Wn + + 6i„jWi„^ +
W2
+ a^w^^ =
•
that such a
where W;
•
•
•
•
•
•
+
•
+
Since B is a basis of V, Oy = 6y, for each i and each unique. Accordingly, V is the direct sum of the PFj.
6i„.W{„.
ftrl^rl
+
Hence
j.
Wi
+ W2+
• + w^
is
unique.
Suppose
sum
S Wi and so
+ Kn^^m^ w^ = w,' and
•
so the
sum
for v
is
+ w, Conversely, suppose V is the direct sum of the W^. Then for any v GV, v = Wj + G PFj. Since {Wy.} is a basis of Wi, each w^ is a linear combination of the Wy. and so v •
•
•
where Wj
is a linear combination of the elements of B. independent. Suppose
"llWli
+
•
•
•
+
«!„ Win,
+
Thus •
+
B
spans V.
an^ri
+
We now
show that
B
is
linearly
CANONICAL FORMS
232
Note that
aa«'ii
such a sum
+
for
•
•
is
+ ain.Wm. G
•
We
W'j.
+
•
+
•
T:V^V
m{t)
(ii)
A{t)
are the
(i)
Ai{t) A2{t), where A{t), and T2 respectively.
By Problem
10.6,
each of
=
/(T) V
r
is
a zero of
/(r) f(t).
and
Ai(t)
G
Since
Wi-
.
.
,
r
Thus
B
=
w +
with respect to a T-invariant
Now
=
and /(r2)(W)
f(T)
w =
suppose
m{t), mi{t)
/(T2)
then
w =
and so m{t)
f{t),
f{t) is
vGV;
Let
0.
+
M
/(Ti)
Hence m(t) divides
linearly independent
is
and
in2{t)
are the characteristic polynomials of
A2(t)
and m^it) divides m(t).
TOi(t)
and m2(t); then f{Ti){U) C/ and w G W. Now
is,
0.
.
common multiple of mi{t) and m2{t) where minimum polynomials of T, Ti and T2 respectively;
-
That
the a's are
and suppose T = Ti © ^2 V = U ®W. Show that:
Wi(()
Me
all
1,
the least
is
T, Ti
=
for
linear
is
direct-sum decomposition (i)
i
=
ai„.Wi„.
The independence of the bases {wy.} imply that and hence is a basis of V.
Suppose
+ 0+---+0 where
=
also have that
unique,
aaWti
10.8.
[CHAP. 10
is
a multiple of both v - u + w with
=
+
the least
common
multiple of
Wi(t) and m2{t). (ii)
By Theorem
10.5,
T has
=
tl
=
\tI-M\
=
(
j
Then, by Problem
representations of T^ and T2 respectively.
A(t)
M
a matrix representation
-A tl
=
-B
where
A
and
B
are matrix
9.66,
\tI-A\\tI-B\
=
Ai(t)A2(t)
as required.
10.9.
Prove Theorem
T:V-*V
Suppose
10.7:
and suppose
is linear,
=
and g{t) and polynomials such that /(T) is the direct sum of the T-invariant subspaces = Kerh{T).
U
= g{t) h{t)
f{t)
are
Then V Kerflr(r) and
are relatively prime.
h{t)
W
and
=
where
C7
since g(t)
and
W
W
Now
are T-invariant by Theorem 10.2. Note first that U and prime, there exist polynomials r(t) and s(t) such that
Hence for the operator Let
veV;
But the
first
then by
term
T,
+
s{t) h(t)
r(T) g(T)
+
s{T) h(T)
=
v
(*),
in this
sum
belongs to
h(T) r{T) g(T) V
-
1;.
V
is
I
r(r)ff(r)i> (*) to
w
=
r(T)g(T)u
alone and using h{T)
w =
r(r)flr{r)w
+
w =
r(T)f(T)v the
sum
of
r(r)fl'(r)w 0,
(*)
since
must show that a sum v Applying the operator r{T)g(T) to
obtain
Also, applying
Hence
=
8(T) h(T) v
W — KerhCT) =
are relatively
1
V = U ®W, we
uniquely determined by
we
+
r(T) g(T) v
r(T) g{T) h(T) v
Similarly, the second term belongs to U.
To prove that
=
r(t) sr(t)
h(t)
we
=
r(T)(iv
=
U and W. — u + w with u&V, w &W, v = m + w and using g(T)u = =
,„
0,
__
r(r)s'(r)w
obtain
+ 8(T)h(T)w =
r(2')flr(T)w
Both of the above formulas give us w = t(T) g(T) v and so w is uniquely determined by M is uniquely determined by v. Hence V — V @W, as required.
ilarly
is
v.
Sim-
CHAP.
10.10.
CANONICAL FORMS
10]
233
Prove Theorem 10.8: In Theorem 10.7 (Problem 10.9), if f{t) is the minimal polynomial of T (and g{t) and h{t) are monic), then g{t) is the minimal polynomial of the restriction Ti of T to U and h(t) is the minimal polynomial of the restriction Tz of
rto W. Let mi(t) and mgCf) be the minimal polynomials of T^ and T2 respectively. = because U = Ker g(T) and = Kerh(T). Thus
W
and h(T2)
and
mi(t) divides g(t)
By Problem
Note that 9(Tj)
=
m2{t) divides h{t)
(1)
common multiple of mi(t) and nizit). But mi{t) and m2(t) are and h{t) are relatively prime. Accordingly, f(t) = mj(t) m,2(t). We also have that f{t) — g(t) h(t). These two equations together with (1) and the fact that all the polynomials are monic, imply that g(t) — •mi(t) and h{t) = m^^t), as required.
10.11.
10.9, f{t) is the least
prime since
relatively
g{t)
Prove the Primary Decomposition Theorem with minimal polynomial -mit)
=
10.6:
Let
T 7 -» F :
be a linear operator
/l(i)"i/2(i^.../r(<)"'
Then V is the direct sum fi{t) are distinct monic irreducible polynomials. of T-invariant subspaces Wi, ...,Wr where Wi is the kernel of fi{TY\ Moreover, /i(i)"' is the minimal polynomial of the restriction of T to Wu
where the
The proof proved for r —
is
and Fi where
W^
Theorem
by induction on
By Theorem
1.
10.8, the
The case r
r.
10.7
we can
= 1 is trivial. Suppose that the theorem has been V as the direct sum of T-invariant subspaces W^
write
is the kernel of /i(r)"i and where V^ is the kernel of fziT)"^ By /r(r)"'. minimal polynomial of the restrictions of T to TFj and Vi are respectively /i(f)"i .
.
.
and/2(«)"2 ... /r («)"'.
Denote the restriction of
T
to
V^ by
By
T^.
the inductive hypothesis, V^ is the direct
sum
of
subspaces W2, .,'W^ such that 'W^ is the kernel of /{(Ti)". and such that /((<)»! is the minimal polynomial for the restriction of T^ to PT,-. But the kernel of fi{T)"i, for i = 2, .,r is necessarily contained in V^ since /;(*)"' divides /2(t)"2 /r(*)"''- Thus the kernel of /i(r)»i is the same as the kernel of fi{T^^i, which is W^. Also, the restriction of T to W^ is the same as the restriction of T^ to Wi (for i = 2, .,r); hence /;(*)"» is also the minimal polynomial for the restriction of T to WiThus V = is the desired decomposition of T. .
.
.
.
.
.
Wi®W2®
10.12.
.
Prove Theorem 10.9: tion if and only if
®Wr
A linear operator T.V^V its
minimal polynomial m{t)
has a diagonal matrix representais a product of distinct linear
polynomials. Suppose m{t)
is
a product of distinct linear polynomials; say, m{t)
=
(t-Xi){t-X2)
... (t-X,.)
Xj are distinct scalars. By the primary decomposition theorem, V is the direct sum of subspaces Wi,...,Wr where Wj = Ker(7'-Xi/). Thus ii v e Wi, then (T-\iI){v) = or T(v) — Xj-y. In other words, every vector in TFj is an eigenvector belonging to the eigenvalue Xj. By Theorem 10.4, the union of bases for Wi, This basis consists of eigenvectors ., W^ is a basis of V.
where the
.
and so T
Xj,
.
is
.
diagonalizable.
Conversely, suppose T is diagonalizable, i.e. V has a basis consisting of eigenvectors of T. Then the operator ., Xj be the distinct eigenvalues of T.
Let
.
= {T-\J)(T-X2l)...(T-Kl) Thus f{T) = and hence the minimum
f(T)
maps each
basis vector into the polynomial
Accordingly,
m,(t) is
0.
m
=
(t-Xi)(i-X2)...(t-X,/)
a product of distinct linear polynomials.
polynomial m(() of
T
divides
CANONICAL FORMS
234
[CHAP. 10
NILPOTENT OPERATORS, JORDAN CANONICAL FORM 10.13. Let T:V^V be linear. Suppose, for vGV, T''{v) = S =
(i)
The
(ii)
The subspace
(iii)
(iv)
set
{v, T{v), ..., T'^-^iv)}
is
¥- 0.
Prove:
linearly independent.
W generated by T-invariant. The restriction T of T to W nilpotent of index k. -S is
is
Relative to the basis {T''-^{v),
.
.,T{v),v} of
.
1
1
Hence the above (i)
but f'-^v)
matrix
/c-square
..
.
..
.
..
.
..
.
W,
the matrix of
T
is
of the
form
1
nilpotent of index k.
is
Suppose av
+
+
di T{v)
+ a^.^n-Hv)
+
T^v)
02
(*)
and using r'=(i;) = 0, we obtain aT'<^-i(v) = 0; since Ti'-'^(v) ^ 0, a - 0. Now applying T^-z to (*) and using P'iv) = and a = 0, we find a^ r'=-i(i;) = 0; hence Next applying T''-^ to (*) and using T<^(v) = «! = 0. and a = ai = 0, we obtain a2T^~^{v) = 0; hence Og = 0. Continuing this process, we find that all the a's are 0; hence Applying
S (ii)
T'^-i to (*)
independent.
is
veW.
Let
Then V
Using THv)
=
=
(iii)
By
W
is
+
+
biT(v)
=
bT{v)
+
biT2(v)
T''{v)
=
0.
Hence, for
i
applying
of index at most
exactly
(iv)
+ •• +
=
b^_iT'^-Hv)
^
b^.^^T'^-H'")
W
k—1,
Tk{Ti(v)) is,
+ •• +
T-invariant.
hypothesis
That
biT^v)
we have that
0,
T{v)
Thus
bv
T'^ to fc.
On
each generator of the other hand,
=
+ «(i;)
r''
W, we
Tf^-^v)
= =
and so T
obtain 0; hence
T'^
=
hence
T''-^v)
¥= 0;
T
is
fc.
For the basis
{T'<'-^v), Ti'-^v),
.
.
T(T^-^(v))
=
r(rfc-3(^))
=
.,T{v),v} of
W,
=
rk(i;)
r'=-2(-u)
T(T{y))
T'^(v)
T(v)
T(v)
Hence the matrix of T
in this basis is 1 1
.
.
.
.
.
.
.
.
1
is
nilpotent
nilpotent of index
CHAP.
10.14.
CANONICAL FORMS
10]
Let T-.V-^V be linear. (ii)
T{W) C
U=
Let
235
W = KerT+\
KerT' and
Suppose ueU = Kern Then THu) = and so MGKerr* + = W. But this is true for every m G f/;
(i)
wG
Similarly, if
= Ker
W'
r*+i,
T'+Mw) =
then
=
T'+Mm)
i
(ii)
Show
that
(i)
UcW,
U.
hence
=
T(0)
r'+Mw) = r*(r(w)) =
Thus
0.
=
T(,Ti(u))
0.
Thus
UcW. r«(0)
=
and so r(W') c U.
10.15.
Let r F ^ F be linear. preceding problem,
X = Ker r*-^ Y = Ker 7*-^
Let
:
XcY cZ.
{Ml, .... Mr},
Y
are bases of X,
{Ml,
.
Z
and
.
.
Mr, Vi,
,
is
Y
contained in
By
and
is
the
.
{Ml,
Show
.,
.
.
{Mi,
Vs},
,
Mr,
.
.
,Ur, Vi,
.
.
.
.
Vs,
,
Wi,
.
.
.
,
Wt}
that
r(wi),
.
.
r(M;t)}
.,
c Y and hence S CY. Now suppose S
T(Z)
is linearly
dependent.
exists a relation
+
aiUi
•
•
+
•
a^Mr
+
6i
+
+
T(wi)
where at least one coefficient is not zero. Furthermore, since must be nonzero. Transposing, we find 6fc bi T{wi)
+
•
•
•
+
Hence
6t
T(wt)
r->(6iWi
+ •• +
6tWt)
=
•
- aiUi -
=
Ti-^(biT(wi)
Thus
By
linearly independent.
the preceding problem,
Then there
.
.
respectively.
s =
Z = Ker T*.
and
Suppose
+
•
+
and so
•
independent, at least one of the
X =
e
a^u^
Ker P'^
=
btT(wt))
+
=
T{wt)
{u^} is
-
5iWi
b^
•
•
+
6,Wt
G r = KerT*-!
{mj, Vj} generates Y, we obtain a relation among the Mj, i»j and Wj; where one of the coefficients, one of the 6^, is not zero. This contradicts the fact that {Mj, Vj, w^} is independent. Hence S must also be independent.
Since
i.e.
10.16.
Prove Theorem 10.10: Let T.V^V be a nilpotent operator of index k. Then T has a block diagonal matrix representation whose diagonal entries are of the form 1
1
.
.
.
.
N .
.
.
.
1
There is at least one N of order k and all other N are N of each possible order is uniquely determined by T. N of all orders is the nullity of T. Suppose for
i
—
Problem
dimy =
1,.. .,k.
Let
n.
Since
T
is
Wi = Ker of index k,
of orders
^ k.
The number
of
Moreover, the total number of
W2 = Ker ra W^ = Ker T". Set m^ = dim W^j, W^ = V and Wj^-i # V and so m^_i
T,
10.17,
WiCW^C Thus, by induction,
we can
choose a basis {mj,
.
•••
.
CW^ = V
.,m„} of
V
such that
{u^,
.
.
.,«„
> is
a basis of
PFj.
We now choose a new basis for V with respect to which T has the desired form. It will be convenient to label the members of this new basis by pairs of indices. We begin by setting v{l,k)
=
u^^_^ +
i,
w(2,
fc)
=M„^_j
+
2,
...,
•y(mfc-«tfc_i,
fc)
=M„j^
CANONICAL FORMS
236
[CHAP. 10
and setting
v{l,k-l) = Tv{l,k), v{2,k-l) = Tv(2,k),
By
vim^-m^.-i, k-1)
^ Tv{m^-m^^i,k)
the preceding problem,
=
Si is
...,
u^^_^, v{l,k-l),
...,
{Ml
We
a linearly independent subset of W^-i(if necessary) which we denote by
vCmfc-mfc^i, fe-1)}
...,
extend Sj to a basis of Wfc-i by adjoining
new
ele-
ments
y(mfe-m;,_i
Next we
+
l,
v(m^-mk-i +
fc-1),
set v(l,
k-2) =
2,
k~l),
v(m^_i-- m^^^tk-V)
...,
- 1), v(2, k-2) = Tv{2, k-1), ..., - mfc_2. k-2) = Tv(m^_i - m^.g, - 1)
Tv(l, k
v(in^_i
fc
Again by the preceding problem,
=
Si
{Ml,
...,
u^^_^, v{l,k-2),
a linearly independent subset of W|c-2 which elements
is
vim^-.i-
+
711^-2
1,
u(mfc_i-w^^2. ^^-2)}
...,
we can extend
k-2), y(mfc_i-mfc_2+2, fc-2),
V
Continuing in this manner we get a new basis for
to a basis of TFfc_2
...,
by adjoining
vim^^^-ink-s, ^-Z)
which for convenient reference we arrange
as follows:
— TOfc_i,
v{l, k),
...,
'y(mfc
v{l,k-l),
...,
v{m^-mk_i,k-l),
...,
•u(mfc_i
- Wfc-a,
fe
i;(l, 2),
...,
v(mfc-mfe_i,
2),
...,
•u(mfc_i
- mfc_2,
2),
...,
^(ma-mi,
2)
v(l, 1),
...,
i;(mfc-mfc_i,
1),
...,
•u(mfe_i
-mfe_2,
1),
...,
i;(m2-mi,
1),
A;)
- 1)
...,
v(mi,
1)
The bottom row forms a basis of Wi, the bottom two rows form a basis of W2, etc. But what is important for us is that T maps each vector into the vector immediately below it in the table or into if the vector is in the bottom row. That is,
Now
it is
(v{i, j
=
Tv(i,i)
— 1)
for for
T
clear (see Problem 10.13(iv)) that
}
.
^
)
> =
1 .
1
have the desired form
will
if
the v{i,]) are ordered
lexicographically: beginning with v(l, 1) and moving up the first column to ^(l, k), then v{2, 1) and moving up the second column as far as possible, etc.
Moreover, there will be exactly m^,
(mfc_i
— m;;_2) —
(m^
— mfc_i) =
—
2m2 2mi
— m^ —
to
diagonal entries of order k
mfc_i
2mfc_i
jumping
Wfc_2
— mi — m^ — 7^2
diagonal entries of order
fc
—
1
diagonal entries of order 2 diagonal entries of order
1
m^ are uniquely as can be read off directly from the table. In particular, since the numbers mj, determined by T, the number of diagonal entries of each order is uniquely determined by T. Finally, .
.
. ,
the identity
mi = (mfc-mfc_i) + shows that the nullity mj of T
1
10.17.
Let
A =
hence
A
which
is
1
(2mfc_i is
-m^ -TOfc-a) +
the total
nilpotent of index
similar to A.
+ (2m2-mi-m3) + (2mi-m2)
of diagonal entries of T.
/O
1\
0011l\ 00000/
Then
^00000/ is
number
•••
A^
=
1
1
Find the nilpotent matrix
and
A3
=
0;
0/
\0 2.
1
/ooooo looooo M
in canonical
form
CHAP.
CANONICAL FORMS
10]
Since 2.
A
M
nilpotent of index 2, contains a diagonal block of order 2 and none greater than A - 2; hence nullity of = 5 - 2 = 3. Thus contains 3 diagonal blocks. must contain 2 diagonal blocks of order 2 and 1 of order 1; that is,
is
M
M
A
Note that rank
Accordingly
237
1
j
_0_^j^_0^
M=
OOlOllo |_0_^
'_0 I
10.18.
Prove Theorem By
r =
the
©
Ti
10.11,
page 226, on the Jordan canonical form for an operator T.
primary decomposition theorem, T is decomposable into operators T-^, ., r^, where (t — Xj)"»i is the minimal polynomial of Tj. Thus in particular, .
•
•
- Xi/)"«, =
(Ti
= Ti- Xj7.
Set Ni
Then for
i
=
l
Ti
That
is,
since
(t
.
T^,
i.e.
©
•
sum
Tj is the
— Xj)*"!
=
(i
r,
= Ni+
where
\I,
Nr't
=
of the scalar operator Xj/ and a nilpotent operator
the minimal polynomial of
is
(T^-\J)r«-r
...,
0,
iV{,
which
of index mj
is
Tf.
Now
by Theorem 10.10 on nilpotent operators, we can choose a basis so that iVj is in canonical In this basis, Ti = N^ + \I is represented by a block diagonal matrix Mj whose diagonal entries are the matrices J^. The direct sum J of the matrices Mj is in Jordan canonical form and, by Theorem 10.5, is a matrix representation of T, form.
Lastly we must show that the blocks Jy satisfy the required properties. Property (i) follows from the fact that A^j is of index mj. Property (ii) is true since T and J have the same characteristic polynomial. Property (iii) is true since the nullity of Ni= Tj — \I is equal to the geometric multiplicity of the eigenvalue Xj. Property (iv) follows from the fact that the Tj and hence the N^ are uniquely determined by T.
10.19.
Determine
possible Jordan canonical
all
characteristic polynomial is A{t) Since 5
t
—2
must appear
has exponent 3 in twice.
Thus the
A(t),
— 2
forms for a linear operator
T:V ->V
must appear three times on the main diagonal. Similarly
possible Jordan canonical
forms are
11
2
2
1
2
I
1
2
!_2 5
1
I
5
1
5
1
5
5 (ii)
(i)
2
2
1
2
1
2
1
(iii)
I
2,'
L?.i
— 5
I
(iv)
whose
— 2)^(t — 5)^.
(t
(V)
5
1
-(
I
(vi)
5
CANONICAL FORMS
238
10.20.
[CHAP. 10
Determine all possible Jordan canonical forms / for a matrix of order 5 whose minimal polynomial is in(t) — {t — 2y. J must have one Jordan block of order 2 and the others must be of order 2 or are only two possibilities: 2
1
2
I
1
1.
Thus there
I
I
.i^_.
=
J
2
I
1
^
I
I
i
'
2
!__ + __ I
Note that
all
the diagonal entries must be 2 since 2
2
the only eigenvalue.
is
QUOTIENT SPACE AND TRIANGULAR FORM 10.21.
Let (i)
W he a subspace of
uGv + W, Suppose
a vector space V. Show that the following are equivalent: (iii) v + W.
Gu
u-v GW,
(ii)
uG v + W.
Then there exists w^eW such that u = v + Wq. Hence u — v = WoSW. Then u — v = Wq where Wq S 1^. Hence u = v + WgSv + W.
Conversely, suppose u — Thus (i) and (ii) are equivalent.
vGW.
We
u
have: also equivalent.
10.22.
also
The
Prove:
— vGW
any two cosets u +
(ii)
each v
gV
partition
W and
v
V
+W
v
+W
if
and only
iflf
v
& u+ W.
Thus
(ii)
into mutually disjoint sets.
and
(iii)
That
is:
are
are either identical or disjoint; and
belongs to a coset; in fact, v
+W—
Furthermore, u for any
V
cosets of PF in
(i)
— (m — v)=v — uGW
iff
Gv + W.
u — vGW, and
if
so {v
+ w) +
W
=
+
v
W
w GW.
= v + E v + W which proves Now suppose the cosets u+W and v + W are not disjoint; say, the vector « belongs to both u+W and v + W. Then u — xGW and x — vGW. The proof of is complete if we show that u + W = v + W. Let M + Wq be any element in the coset u+W. Since u — x, x — v and Wq belong Let
1)
e
G W, we have
Since
V.
v
(ii).
(i)
to
W,
+ Wq) —
(u
Thus u + W(,Gv is
contained in
+
W
u+ W
V
=
(u
— x) +
u+W
and hence the coset = v + W. and so u+
W
is
{x
— v) +
Wo S
contained in the coset v
u+W - v + W
The last statement follows from the fact that by the preceding problem this is equivalent to u — v G W.
10.23.
W
Let be the solution space of the homoDegeneous equation 2x + By + 4:Z = 0.
W in R^.
scribe the cosets of
a plane through the origin O = (0, 0, 0), are the planes parallel to W. Equivalently, the cosets of are the solution sets of the family of equations
TF
is
and the cosets of
2x
+
W
Sy
+
W
4z
=
kGR
k,
In particular the coset v + W, where v is the solution set of the linear equation
2x or
2(x
+
Sy
-a) +
+
Az
3(y
=
2a
- 6) +
+
4(2
36
+
4c
- c) =
=
(a, b, c),
W
if
+ W.
and only
if
Similarly v
uGv + W,
+
W
and
CHAP.
10.24.
CANONICAL FORMS
10]
239
W
Suppose is a subspace of a vector space V. Show that the operations in Theorem = u' + 10.15, page 229, are well defined; namely, show that if u + and v +
W
v'
+ W,
W
W
then {u
(i)
W
+ v) +
=
{u'
W
+ V') +
and
W
+
ku
(ii)
=
+ W,
ku'
for any
k&K
u + W ^ u' + W and v + W = v' + W, both u — u' and v — v' belong to W. But then + v) - (u' + v') - {u- u') + {v- v') e W. Hence (u + v) + W = (m' + v') + W. Also, since u — u' S W implies k(u — u') G W, then ku — ku' = k(u — u') G W; hence ku+W = ku' + W. Since
(i)
(u
(ii)
10.25.
Let
F be a vector space and W a subspace of V. Show that the natural map by
defined
rj{v)
=
For any u,v ^
+ W,
v
V
+ v) =
and
10.26.
u
+
+
V
=
v{kv) is
r)
:
F -» V/W,
and any k G K, we have
v{u
Accordingly,
ij
is linear.
W
-
u
+
W
=
kv
W+V+W
+
k(v
+ W) =
= k
v{u)
+
v{v)
ri(v)
linear.
W
W W
and he a subspace of a vector space V. Suppose {wi, Wr} is a basis of the set of cosets {vi, Vs}, where Vj = Vj + W, is a basis of the quotient space. + Show that B = {vi, .,Vs, Wi, Wr} is a basis of V. Thus dim V = dim Let
.
.
.
.
.
.
,
.
,
.
.
.
,
.
dim (7/TF). Suppose M
e
Since {5^}
y.
is
u = u Hence u
—
aiVy +
Accordingly,
We now
B
•
•
•
+ a^v^ + w u —
W
+
a.2'U2
+
•
w G W. Since {w;} + + a^Vg + bjWi +
a^Vi
•
show that
B
is linearly
+
•
+
•
•
ttj^s
a basis of
is
•
+
•
W,
b^w^
Suppose
independent.
+
•
CgVs
+
Cj'Di
is
+
di'i'i
where
Then is
—
generates V.
e^Vi
Since {Vj} Since {wj
V/W,
a basis of
•
•
+
djWi
+
c^Vs
+
•
•
•
-
+
dfWr
=
W
=
(1)
independent, the c's are all 0. Substituting into (1), we find djWi + + d^w^ = 0. independent, the d's are all 0. Thus B is linearly independent and therefore a basis •
•
of y.
10.27.
Prove Theorem 10.16:
W
a subspace invariant under a linear operator f on V/W defined by f{v + PF) = T{v) + W. Moreover, if T is a zero of any polynomial, then so is T. Thus the minimum polynomial of T divides the minimum polynomial of T.
T:V^V. We
u+W
Suppose
Then T induces a
first
=
show that f
v + Accordingly,
W
then
is
is
linear operator
well defined,
i.e.
and, since
T{u+W) =
T{u)
+
u+W = v + W
if
u-vGW
W
W
is
=
T(v)
then
T-invariant,
+
W
=
t(u+W) = f(v + W). If = T(u) - T{v) G W.
T(u -v)
T(v
+ W)
as required.
We
next show that
t
is linear.
+W) +
(v
+
t{{u
W))
= =
We
have
+ v + W) = T(u + v) + W = T(u) + Tiv) + T(u) + W+ T(v) + W = f{u + W) + T(v + W) f(u
and f{k{u
Thus f
is
+
W))
linear.
=
f(ku
+ W) =
T{ku)
+
W=
kT(u)
+
W
=
k{T(u)
W
+ W) = kf(u+ W)
CANONICAL FORMS
240
Now, for any
coset
m
+
f2(u+W) = THu) +
—
Hence T^
=
Similarly T"
T^.
VIW,
IF in
W
T(T(u))
=
=
=
=
10.28.
a„t"
=
and so 7(r) root of
Thus the theorem
Accordingly,
/(r).
is
T
if
+
=
•
•
t{T{u)
•
+
ao
=
2 afi,
+
W
=
"2 diiTKiA
^aifi(u+W) a root of
is
+ W) ^ t(f{u+W)) = t^u+W)
Thus for any polynomial
n.
'^.a^Tiiu)
=
^ajFCw+W)
=
f(t).
W
+
f(T)(u)
W
+
T" for any
/(«)
HT)(u+W)
[CHAP. 10
f{t)
+
'^)
= {^ a.ifi)(u + W) = f{f)(u+W) = W = /(f), i.e. f is also then 7(f) =
a
proved.
Prove Theorem 10.1: Let T .V -^V be a linear operator whose characteristic polynomial factors into linear polynomials. Then V has a basis in which T is represented by a triangular matrix. The proof tion of
r
is
is
by induction on the dimension of V.
a 1 by 1 matrix which
is
If
dim
V=
then every matrix representa-
1,
triangular.
n>
Now suppose dim V — 1 and that the theorem holds for spaces of dimension less than n. Since the characteristic polynomial of T factors into linear polynomials, T has at least one eigenbe the 1-dimensional subvalue and so at least one nonzero eigenvector v, say T(v) — a^^v. Let
W
space spanned by also that
W
V = VIW.
Set
v.
invariant under T.
is
W
= to - 1. Note Then (Problem 10.26) dim V = dim V — dim f on V whose T linear operator induces a By Theorem 10.16,
minimum polynomial divides the minimum polynomial of T. Since the characteristic polynomial of r is a product of linear polynomials, so is its minimum polynomial; hence so are the minimum and characteristic polynomials of f. Thus V and f satisfy the hypothesis of the theorem. Hence, by induction, there exists a basis
{v^,
.
let
{v,V2,
V2,
is
a basis of
V
But
W
is
spanned by
v;
T(V2)
Similarly, for T{Vi)
-
t
=
ai2V2
assVs
which belong
=
Since
to the
f(v2)
and so
— a22'''2 —
a2i'"
a^Vi
e
a'22V2
such that
is
-
1)2,
we have
2X^2) "" «22'"2
a multiple of
and so
v,
T^kv^)
«„ respectively.
cosets a22'V2,
^
.
.
.
,
Then
^
say
—
"21^
+
«22'"2
n,
3
-
a'22^2
hence T{v2)
—
+
(Problem 10.26).
f («2) —
V
0„} of
as2.V2
V
be elements of
-tVn
...,vj
. ,
=
f(vs)
Now
.
Ojs-ys
-
Thus
•
•
•
-
W
and so
= =
a,iv
T{V2)
a2iV
+
0.22^2
T(Vn)
=
a„iV
+
a„2«2
T(v)
and hence the matrix of T in this basis
is
+
T{Vi)
•
•
+
=
a^v
+
0(2^2
-!-•••+
a^iV^
ann-yn
triangular.
CYCLIC SUBSPACES, RATIONAL CANONICAL FORM 10.29.
Prove Theorem Z{v,T), and
10.12:
m„(«)
^
t^
Let Z{v, T) be a T-cyclic subspace, T^ the restriction of + Oo the T-annihilator of v. Then: + 0.^-1*"-' +
set {v, T{v), ..., r'=-i(v)} is a basis of Z{v,T); hence
(i)
The
(ii)
The minimal polynomial
(iii)
The matrix
of Tv
is TO„(f).
of T« in the above basis is
dimZ(t;,r)
=
k.
T
to
CHAP.
CANONICAL FORMS
10]
.
1
By
(i)
241
— fto
.
.
.
.
.
.
.
-ai
1
— afc-2 — Ctlc-l
definition of m^Ct), T''{v) is the first vector in the sequence v, T{v), T^v),
.
.
which
.
is
a
vectors which precede it in the sequence; hence the set B — {v, T{v), ., r''-i(i;)} is linearly independent. now only have to show that Z(v, T) = L(B), the linear span of B. By the above, T^v) e L{B). prove by induction that T^{v) &L(B) for every n. Suppose and T^-^(v) E. L(B), i.e. !r"-i(v) is a linear combination of V, ..,T^-i{v). Then r"(v) = r(r«-i(v)) is a linear combination of T{v), -tT^v).
combination
linear
.
of
those
We
.
We
n>k
.
.
THv) G L{B); hence T^{v) £ L(B) for every n. Consequently f(T)(v) polynomial /(<). Thus Z{v, T) = L(B) and so B is a basis as claimed. But
Suppose
(ii)
we
m(t)
=
i*
^
^
(v, ),
Thus m„(r)
+
6j_i<«~i
+
=
m(T^){v)
•
•
•
+
is
&o
=
m(T){v)
T^(v)
and so m^(T^)
hence w„(t) (iii)
=
=
Then
0.
+
L(B)
polynomial of r„.
h^^iT^--^(v)
+
•
+
•
.
for any
h^v
.,
.
.
Then, since
T^-i{v), and therefore k ^ s. m(t) divides m„(t) and so s — k. Accordingly
T^{v) is a linear combination of v,T{v),
=
the minimal
G
fc
However, = s and
tn{t).
T„(v)
=
T„{T{v))
=
T„{T''-Hv))
=
T(v)
THv)
=
T^v)
-
-oov
a^T{v)
-
a^^^n'Hv)
a^n(v)
By definition, the matrix of T„ in this basis is the transpose of the matrix of coefficients of the above system of equations; hence it is C, as required.
10.30.
Let r y -» y be linear. Let TF be a T-invariant subspace of V and T the induced operator on VIW. Prove: (i) The T-annihilator of v G V divides the minimal polynomial of T. (ii) The T-annihilator of v G VIW divides the minimal polynomial of T. :
GV
The r-annihilator of v by Problem 10.6,
(i)
therefore,
the minimal polynomial of the restriction of
is
it
The f-annihilator of p S VIW divides the minimal polynomial of T by Theorem 10.16.
(ii)
T
to Z(v, T)
and
divides the minimal polynomial of T. f,
which divides the minimal
f(t) is
a monic irreducible poly-
polynomial of
Remark. In case the minimal polynomial of T is /(t)" where G V and the T-annihilator of where m — n. nomial, then the T-annihilator of v
10.31.
i)
G VIW
are of the form /(t)™
Let T F -* V be a linear operator whose minimal polynomial is /(t)" where f{t) is a monic irreducible polynomial. Then V is the direct sum of T-cyclic subspaces Zi = Z{vi, T), i = l, ., r, with corresponding T-annihilators
Prove
Lemma
10.13:
:
.
/(f)"i,
Any
other decomposition of
number the
/{f)%
Y
.
.
. ,
.
/(i)"',
n =
into the direct
of components and the
same
sum
^ »2 -
•
•
•
- «r
of T-cyclic subspaces has the
same
set of T-annihilators.
The proof is by induction on the dimension of V. lemma holds. Now suppose dim V > 1 and that
dimension less than that of V.
ni
If
the
dim
V=
lemma
1,
then
V
is itself T-cyclic
and
holds for those vector spaces of
CANONICAL FORMS
242
there exists v^GV such that f{T)^~i(vi) ¥= 0; Zi = Z{vi,T) and recall that Zi is T-invariant. Let be the linear operator on V induced by T. By Theorem 10.16, the minimal poly-
T
Since the minimal polynomial of hence the r-annihilator of Vi is /(«)".
V=
is /(<)",
Let
V/Zi and let f nomial of f divides /(<)"; hence the hypothesis holds for the direct
sum
We
e
Z(%, r)
where the corresponding f-annihilators are
f{t)"2,
w
•
•
.
.
=
f(T)n.{w)
Since /(<)"
is
©
•
But /(«)" is the r-annihilator of some polynomial h(t). We set
1^1
w — Uj =
'i(^) (^1)
^
^1.
hence
;
On
=
Similarly, there exist vectors Vj is /(t)"i,
Let d denote the degree of
= w-
,
,„r^
h{T)
— n^
•
=
.
v^,
=
Thus the T-annihilator
-
f(T)^(w)
of V2 is a
=
g{T){vi)
such that ViGvl and that the T-annihilator of
.,Vf&V
Z{V2,T),
Z,
...,
=
Z(i;„r) since /(t)"! is both the r-annihilator
Then
that
{%
and
i
f(v^,
=
.
.
.s^
{Vi, ..., rd"i-l(Vi), V2, .... Tdr^^-HV2),
V=
10.4,
remains to show that the exponents
Wj,
Z(vi, T) .
.
.
,
®
.
,
f d^ii- 1 (iTj)}
©
=
THv)
(see
..., Td-r-l
...,V„
Z(v^
f)
©
•••
©
,,
.
fi(v)
••
V=
But
2,...,r.
a basis for V. Therefore by Problem 10.26 and the relation
the degree of
for
f(t)«2 h(t)
as claimed.
T*"i- 1 (Vi)}
Thus by Theorem
=
(1),
.
a basis for V.
g{t)
set
we know
Z(vZ, t); hence
and so
>
(vi)
are bases for Z(Vi, T) and Z(v^, f) respectively, for
Z(v^, T),
Problem
10.27),
(i;^)}
as required.
w^ are uniquely determined by T.
Since d denotes
fit),
dimy =
dimZj
and
h n^)
d{nx^
=
is
&V
f(T)Hv) /(r)«(Wj)
€
f(Ty{Zi).
Let
t
= f{Ty(Wi)+
%1
f(Ty{V)
=
dim(/(r)'(V))
S,
...,
Jlt
>
S,
Wt + i
f(T)HZi)
©
=
- s) +
d[{ni
Wi
=
-
w-^+
s. •
•
•
+ Wr where
w^
£ Zj.
+f(T)s(w,)
be the integer, dependent on
>
l,...,r
a cyclic subspace generated by if
can be written uniquely in the form v Now any vector v Hence any vector in f(T)^{V) can be written uniquely in the form
=
i
drii,
Also, if s is any positive integer then (Problem 10.59) f(T)^(Z>i f{T)s(Vi) and it has dimension d(Mj-s) if «i > s and dimension
and so
•
(1)
f(t)«-''2 g(t) ,
the other hand, by
.
We
{Vi, T(v,), ...,
Then
•
(1),
so that /(*)"« has degree dwj.
/(t)
and the f-annihilator of
where
is
f(T)«-n2g{T)(vi)
/(i)" divides
is /(t)"2
v^,
the f-annihilator of ^.
Z2
It
—
n2
9{T){Vi)
f(T)'H(w-h(T)(vi))
Consequently the T-annihilator of v^
is
—
ti
^2 also belongs to the coset 82-
multiple of the if-annihilator of V2f{T)«^{v2)
=
f{T)^{w)
U2
is
V
Z(v„ f)
/(*)">,
.,
we have by
the minimal polynomial of T,
=
Vi
Consequently, by induction,
f.
is a vector V2 in the coset V2 whose T-annihilator is /(<)"2, the T-annihilator be any vector in Dg- Then f{T)"i (w) e Z^. Hence there exists a polynomial g(t) for
which
Since
and
claim that there
Let
of V2.
V
of f-cyclic subspaces; say,
V =
of
[CHAP. 10
•••
© •
-
s,
for which
s
/(DM^t)
+
{n^
- s)]
(*)
The numbers on the left of (*) are uniquely determined by T. Set s = re — 1 and (*) determines the number of TOj equal to re. Next set s = re — 2 and (*) determines the number of re, (if any) equal to and determine the number of % equal to 1. Thus n-1. We repeat the process until we set s = lemma is proved. and the and T V, by the Wi are uniquely determined
CHAP.
10.32.
CANONICAL FORMS
10]
243
Let F be a vector space of dimension 7 over R, and with minimal polynomial m{t) = {t^ + 2){t + 3f.
let
T.V^V
Find
be a linear operator
the
all
possible
rational
canonical forms for T.
The sum
of the degrees of the companion matrices
must add up to 7. Also, one companion and one must be (t + 3)3. Thus the rational canonical form of T is exactly one of the following direct sums of companion matrices:
matrix must be
t^
(i)
C(t2
(ii)
C(«2
+ 2) © + 2) ©
(iii)
C(t2
+ 2)
That ^0
+
2
+ 2) © C((t + 3)3) + 3)3) © C((t + 3)2) C((« + 3)3) © C(t + 3) © C(t2
C((t
©
C(t
+
S)
is,
-2
-2
/° -2
\
A
\ -27 -27 -9
1 1
1 1
-9 1 (i)
-3
/
V
-6/
PROJECTIONS 10.33. Suppose V = Wi® ® Wr. The projection of V into its ping E:V ^ V defined by E{v) = Wk where v — wi+ that (i) E is linear, (ii) E^ = E. (i)
Since the defined.
V
-{-
-3/ (iii)
(ii)
•
•
•
sum v = Wi+ Suppose, for m
u =
(wi
+ w() +
•
•
+ w^, WiG W ~ w^i-
S •
V, u
+
(Wr
•
+ w'r)
-27 -27 -9
is
•
uniquely determined by
+ w^,
and
subspace Wk is the map+ Wr, Wi e Wi. Show
w[ S W^.
A:t;
=
fcwj
+ u and kv. Hence = E{v) + E(u) and
v,
the
mapping
E
is
well
Then
+
•
•
•
+
kwf, Wj
kw^,
+ w,' G
PFj
are the unique sums corresponding to v
E(v and therefore (ii)
We is
have that
the unique
Thus E^
10.34.
=
+ u) = Wk +
wl^
E(kv)
=
kw^
=
kE(v)
£7 is linear.
0+---+0 +
+ 0+---+0 sum corresponding to w^ G Wk'-, hence E(w^) = w^. Then EHv) = E(E(v)) = E(Wk) = w^ = E(v) w^ =
TOfc
for any v
G
V,
E, as required.
E:V-*V is linear and E^ - E. Show that: (i) E(u) = u for any uGlmE, the restriction of to its image is the identity mapping; (ii) V is the direct sum of the image and kernel of E: V — ImE KerE; (iii) is the projection of V into Im E, its image. Thus, by the preceding problem, a linear mapping T V -> V is a projection if and only if T^ = T; this characterization of a projection is frequently Suppose
E
i.e.
@
E
:
used as (i)
If
its definition.
M G Im E, then
there exists
E(u)
e V for which E(v) = u; hence = E{E{y)) = EHv) = E{v) = u
v
as required. (ii)
Let V
V
SV. We
— E(v) G
= E{v) + v — E(v). Now E(v - ^(1;)) = E(v) - E^(v) = E{v) - E(v) = Accordingly, V = Im + Ker E.
can write v in the form v
Ker E.
jE?
E(v)
e.ImE
and, since
CANONICAL FORMS
244
[CHAP. 10
n Ker E. By (i), E{w) = w because w G Im £/. On the other Now suppose w and so ImE n Ker E = {0}. These two because w G Ker E. Thus w = hand, E{w) = conditions imply that V is the direct sum of the image and kernel of E.
GImE
Let v
(iii)
by
(i),
That
10.35.
and suppose v = u + w where uGlmE and w e Ker E. because w G Kerfi". Hence and E{w) — = E{u + w) = E(u) + E{w) = u + = m £?(-!;)
eV E
is,
U®W
V=
Suppose
r-invariant
Observe that
E{v)
if
for every
G
t>
that U and TF are both the projection of V into U.
is
and that
Y,
Show
is linear.
=
E(v)
(i)
v
v€:U,
iff
=
K(i;)
(ii)
v&W.
iff
ET =
Suppose
Hence
U
T-invariant.
is
=
E{T(w))
Hence
W
= Now
-
=
(TE)(u)
w GW.
let
= w, = {ET){u) =
Since E{u)
E(T(u))
= 0, = r(0) =
G U
Since E{w)
=
(TE){w)
T{E{w))
T(w)
and so
G
W
T-invariant.
is also
G r and w G
T^.
'^^'"^
=
(ET){v)
(ET){v)
=
U
and
Then
W are both G
?(«)
i7
Let « hence
T-invariant.
and
r(w)
G W;
G Y and S(r{w))
suppose
TM
=
•y
= w+w
and
+ TO) = (Er)(M) + (Br)(w) = E(T(u)) + E{T{w)) = {TE)(v) = (r£7)(M + w) = r(S(M + «;)) = T{u) for every
(TE)(v)
-y
G
ET = TE
F; therefore
where
E(T(w))
=
0.
T(u)
(Br)(M
and is,
U.
T{E(u))
(ET){w)
Conversely, suppose
That
G
TE. Let u r(M)
M
T:V-^V
TE - ET where E
G U
=u
into its image.
and suppose
and only
if
V
the projection of
is
Note that E(u)
as required.
Supplementary Problems INVARIANT SUBSPACES
W
invariant under
10.36.
Suppose
10.37.
Show
10.38.
is invariant under Suppose S + r and ST.
10.39.
Let
that every subspace of
V
T:V -^V. Show is
W
W 10.40.
is
r y :
is
Let
->
y
be linear and
let
that
W
invariant under I and
S:V^V
and
T
:
0,
V -^
10.42.
the identity and zero operators.
F.
Show
that
W
is
also invariant
W be the eigenspace belonging to an eigenvalue X of
under
Show
T.
that
r-invariant.
y
be a vector space of odd dimension (greater than V has an invariant subspace other than
linear operator on
10.41.
invariant under f{T) for any polynomial
is
Determine the invariant subspaces of
V=
Show
that there exist T-invariant subspaces
Suppose dim
n.
A =
_A
f
1)
V
Show
over the real field E. or {0}.
viewed as a linear operator on
(i)
that any
R2,
(ii)
C^.
T:V^V
has a triangular matrix representation if and only if cW^ = V for which dim TFfc = fc, k = l,...,n. WiCWzC
INVARIANT DIRECT-SUMS 10 43
The subspaces Wi,...,Wr are said
L(Wd =
Wi®
each Wi = 0. Show that LiWi) denotes the linear span of the Wi-) 10.44.
10.45.
that V = W^ + i,...,Wr) =
Wi®
• ®Wr
{0},
=
Show Show
that
L(Wi)
fe
l,
Wi + and only
to be independent if
if
@Wr
and only
if
if
(i)
V=
•
•
-|-
•
if
L{Wi)
w^
the
and
=
Wj G Wi, implies that are independent. (Here
0,
Wi (ii)
W^nL{Wi,
.
.
.,Wk-i.
...,r.
= W,®---®Wr
if
and only
if
dimLd^i)
=
dim Wi
+
•
•
•
-t-
dim W^.
CHAP.
10.46.
CANONICAL FORMS
10]
Suppose the characteristic polynomial of T V -» V are distinct monic irreducible polynomials. Let position of
V
into r-invariant subspaces. 7"
V =
Show that
=
fAt)«' where the be the primary decomthe characteristic polynomial of the
A(t)
is
:
fi(t)
restriction of
245
/i(f)"i /2(f)»2
• ®Wr
WiQ
/((t)™! is
.
.
.
to PFj.
NILPOTENT OPERATORS 10.47.
Suppose
S and T
are nilpotent operators which commute,
ST =
i.e.
Show
TS.
that
S+T
ST
and
are also nilpotent. 10.48.
Suppose
A
that 10.49.
Let is
10.50.
A
is
a supertriangular matrix,
i.e. all
entries on
V
be the vector space of polynomials of degree nilpotent of index n + 1.
Show
and below the main diagonal are
Show
-
Show
n.
that the differential operator on
V
that the following nilpotent matrices of order n are similar: 1 1
.
.
.
.
.
.
.
.
1
and
10.51.
0.
is nilpotent.
1
.
.
.
.
.
.
1 ...
\0
1
0/
Show that two nilpotent matrices of order 3 are similar if and only if they have the same index of nilpotency. Show by example that the statement is not true for nilpotent matrices of order 4.
JORDAN CANONICAL FORM 10.52.
10.53.
Find all possible Jordan canonical forms for those matrices whose characteristic polynomial and minimal polynomial m{t) are as follows:
A(t)
= = =
(t-7)5, (t-2)7,
A(t)
^
(t-3)*(t-5)\
(i)
A(t)
(ii)
A{t)
(iii)
(iv)
Show
(t-2)4(f-3)2, m(t) m(t)
10.54.
Show
10.55.
Suppose
(t-2)2(t-3)2
(t-7)2 (t-2)3 m(t) =: («-3)2(i-5)2
that every complex matrix
Problem
=
m{t)
= =
A(t)
similar to its transpose.
is
{Hint.
Use Jordan canonical form and
10.50.)
that
A
all
is
complex matrices
A
of order
n
for which
A" =
/
are similar.
a complex matrix with only real eigenvalues. Show that
A
is
similar to a matrix with
only real entries.
CYCLIC SUBSPACES 10.56.
Suppose
T:V -*V
containing
is linear.
Prove that Z{v, T)
is
the intersection of all T-invariant subspaces
v.
10.57.
Let fit) and g{t) be the T-annihilators of u and v respectively. atively prime, then f(t)g(t) is the T-annihilator of u + v.
10.58.
Prove that
10.59.
Z(m, T) r-annihilator of u.
=
Z(v, T)
if
and only
if
g(T)(u)
=
v
Show
where
that if
g(f)
is
f(t)
and
relatively
g{,t)
are rel-
prime to the
W
= Z{v, T), and suppose the T-annihilator of v is /(*)" where f(t) is a monic irreducible polyLet nomial of degree d. Show that f{T)^{W) is a cyclic subspace generated by f(Ty(v) and it has dimension d{n — s) if n > s and dimension if n — s.
RATIONAL CANONICAL FORM 10.60.
Find
10.61.
all
possible rational canonical forms for:
(ii)
6X6 matrices with minimum 6X6 matrices with minimum
(iii)
8
(i)
X 8 matrices with minimum
polynomial m(t) polynomial mit) polynomial m(t)
= = =
+ 3){t + 1)2 + 1)3 {t^ + 2)^(t + Z)^
(t^ (t
Let be a 4 X 4 matrix with minimum polynomial m(t) = (t^ + \){fi — 3). Find the rational canonical form for A if A is a matrix over (i) the rational field Q, (ii) the real field B, (iii) the complex field C.
A
CANONICAL FORMS
246
10.62.
Find the rational canonical form for the Jordan block
10.63.
Prove that the characteristic polynomial of an operator
[CHAP. 10
T :V
-^
V
a product of
is
elementary
its
divisors. 10.64.
Prove that two
10.65.
Let
3X3
minimum and
matrices with the same
characteristic polynomials are similar.
denote the companion matrix to an arbitrary polynomial
C(f(t))
Show that
f(t).
f(t) is
the char-
acteristic polynomial of C{f(t)).
PROJECTIONS 10.66.
10.67.
=
^ i;
0, i
E -.V ^V
Suppose
)
=
J
Bj
V
E
the rank of
is
•
that
E
•
EiE^
(i)
=
0,
projections;
(i)
that
E^ = E. Prove
i.e.
Prove:
into Wi-
Ef = £/;, i.e. the Ei are V = Im Ej © © Im B^-
such that:
+E^. Prove
H
a projection,
is
where r
^
(
(iii)
V
Let Ei denote the projection of
be linear operators on
.,Er
..
EiEj
form 10.69.
®Wr• +E^.
(ii)
Let El, (ii)
10.68.
V = Wi® I = E^+
Suppose
i^j;
•
has a matrix representation of the
/^ is the r-square identity matrix.
and
Prove that any two projections of the same rank are similar.
Use the result of Problem
{Hint.
10.68.) 10.70.
E -.V -^V is a projection. Prove: I-E isa projection and V = ImE ®
Suppose (i)
Im (I-E);
I
(ii)
+E
1
+1 #
0).
IF} in
V/W
is
invertible
is
(if
QUOTIENT SPACES 10.71.
linearly independent. 10.72.
Suppose the set of cosets {vi + W,V2 + that the set of vectors {v^, V2, ..., vj in V
Let IF be a subspace of V.
Show
and that
nW
L(Ui)
=
{mi, Wg that the set of cosets {mi
Suppose the set of vectors
Let IF be a subspace of V.
Show
{0}.
+
W, ...,«„+
also linearly independent.
is
m„} in
V
is
IF,
m„
+
.
.
.
,
linearly independent,
IF}
V/W
in
is
also
linearly independent. 10.73.
®W
ii„ + IF} is Show that {ui + W, tt„} is a basis of U. and that {mj, Suppose V = U a basis of the quotient space V/W. (Observe that no condition is placed on the dimensionality of
V 10.74.
let
=
v
Let
by
V
U
+
S
of the
form
Ugt*
X".
Prove that the coset v
+
•
•
•
+
+
a^Xn
•
•
•
=
V
such that
F
?7/IF is a subspace of
(i)
U
since
V/W,
R
V/{UnW)
and
Show
.
.
,
IF of IF in K"
i._„j,j^ 6 - ajfti +
^V
the solution set of the
+
.a. •
•
•
Note that any coset
implies
dim {V/W)
the cosets of
«,
w e V; hence U/W
-
dim(i7/IF)
UnW
{{v+U)n{v'+W):
in
F
v,v'
eV}
U
v
:
:
i(u)
=
u.
(See diagram.)
U
V
+
IF of IF in ?7 may a subset of V/W.
is
can be obtained by inter-
:
i.e.
„ h a„6„
= dim{V/U}.
in V:
be linear with kernel IF and image U. Show that T:V under the mapping V/W is isomorphic to quotient space the Furthermore, show that e V/W -> U defined by e{v -I- IF) = T{v). T = io 0OTJ where i; F -> V/W is the natural mapping of V into C is the inclusion mapping, V/W, i.e. r,{v) = -y -t- IF, and t Let
is
be the subspace of polynomials divisible that the quotient space V/W is of dimension 4.
by each of the cosets of IF
=
K
let IF
WcUcV. (ii)
+
where
b
O'i^
0.
,
u&U
Show that
F
an^n
,
h a„_^t^.
-\
and IF be subspaces of V. secting each of the cosets of t/ in
Let
020:2
,
+ a^t^
and IF be subspaces of
Prove that
=
+
,
a2X2
+
ajXi
also be viewed as a coset of IF in
10.78.
.
,
be the vector space of polynomials over
t*, i.e.
Let
6„)
(5i, 63
aiXi
10.77.
.
Let IF be the solution space of the linear equation
and
10.76.
.
or IF.)
linear equation
10.75.
.
yjy^
CHAP.
CANONICAL FORMS
10]
Answers 10.41.
(i)
10.52.
(i)
R2 and {0} 2
(ii)
C\
247
Supplementary Problems
to
{0}, PTj
=
L((2, 1
1
- 2i)), W^ = 2
1
L{{2, 1
+
2i))
•
I
2 4_.
2
1
L2_.
I
2
^2^
I
i_lL
r3
1 i
3
1
3
3
I
(ii)
7
7
1
i
7
I
;
1'
7
L_i4-2
1
I
7
I
I
111. 7 "[7-
I
7
'
(iii)
1
7
2
I
1
2
2
1
2
I
2
1
2
2
1
-I
1
H 2
2
I
1
1
I
2
I
2
I
r
2
1
I
2
I
1
I
2
I
2
I
I
111-
TiL.
2 /
I
3
(iv)
1
I
3 3
1
1
2
I
1
2
2
I
2
I
I
r3
I
Wi^
1
I
1
5
1 I
L. \
5
I
1
5
I
I
3
1
3
3
I
1
3
I
5
/
I
I
rL-
[5
n
-
3 _ L"-' r5""i"i
f-
I
'
5
I
|5 I
1
5
^-in_ I
5
/
CANONICAL FORMS
248
10.60.
-3
'0
(i)
[CHAP. 10
-3
-3 1
1
1
1
-1
-1 -2
1
-2 -1
-1 1
(ii)
0-1
'0
10-3
1
'0
1-3
-2/
0-1 0-3 1-3
1 1
1
1
-1 -3 -3
-1 -2
-1 -1,
-1, (iii)
2
2
-4
-4 1
1
1
-9 -6
-9 1
-6/
2 1
-4
1
1
1
-9 -6 -3.
10.61.
(i)
(ii)
/O
-1
(iii)
V^
y/^
-W 10.62.
-x*\
^0
10 10 \0
1
4\3
-6\2 4X
-Vsl
chapter
11
and the Dual Space
Linear Functionals INTRODUCTION
K
In this chapter we study linear mappings from a vector space V into its field of scalars. (Unless otherwise stated or implied, we view as a vector space over itself.) Naturally all the theorems and results for arbitrary linear mappings on V hold for this special case.
X
However, we treat these mappings separately because of their fundamental importance and 7 to Z gives rise to new notions and results which do not
because the special relationship of apply in the general case.
LINEAR FUNCTIONALS AND THE DUAL SPACE Let F be a vector space over a field K. A mapping tional (or linear form)
GV
for every u,v
if,
—
4,{au -^hv)
F
In other words, a linear functional on Example
Let
11.1:
is
Example
11.2:
jt-j
:
K"
linear
Let
V
-»
K
and so
11.3:
Let
V
a
+
4,{u)
b
termed a linear func-
is
{v)
V
a linear mapping from
is
be the ith projection mapping, a linear functional on X".
into K.
7rj(ai, aj,
i.e.
=
...,a„)
a^
Then
ttj
it is
be the vector space of polynomials in
=
operator defined by ^{p(t)) a linear functional on y.
Example
K
:V -*
and every a,b G K,
p{t) dt.
j
t
over R.
Let
Recall that
^
^ iV -^ R is
be the integral
and hence
linear;
it is
o
be the vector space of n-square matrices over K.
Let
T iV ^
K
be the trace
mapping
- «„ +
T{A)
a22
+
•
•
•
+
where
«««.
A =
(ay)
A
That is, T assigns to a matrix the sum of its diagonal elements. linear (Problem 11.27) and so it is a linear functional on V.
By Theorem
6.6,
a vector space over
the set of linear functionals on a vector space V over a addition and scalar multiplication defined by
+ (T){v) =
(j>{v)
+
where ^ and a are linear functionals on V and is denoted by V*. 11.4:
field
K
map
is
is also
K with {(t>
Example
This
and
(t{v)
V
and k G K.
{!i)iv)
=
This space
k
is
called the diuil space of
Let V = K", the vector space of ti-tuples which we write as column vectors. Then the dual space V* can be identified with the space of row vectors. In particular, = {a^, any linear functional a„) in y* has the representation .
.
.
.
.
.,Xn)
249
,
=
(tti.aa,
.
.
.,a„)
LINEAR FUNCTIONALS AND THE DUAL SPACE
250
[CHAP.
11
or simply
=
«„)
0(a;i
+
a^Xi
02*2
+
•
•
+
•
was termed a
Historically, the above formal expression
In^n
linear form.
DUAL BASIS Suppose y is a vector space of dimension n over K. By Theorem 6.7, the dimension of In fact, each basis of V is of dimension 1 over itself.) the dual space V* is also n (since determines a basis of V* as follows:
K
Theorem
11.1:
Suppose
{Vi,
.
.
linear functionals defined
Let
over K.
^j,
^
=
^«
6.2,
.
.,^„
G V*
be the
jo if^^i
^„} is a basis of V*.
{^i
The above basis {<|>^) is termed the basis dtial to (Vi) or the dval mula which uses the Kronecker delta Si, is a short way of writing
By Theorem
.
by
^'(^^^
Then
V
.,v„} is a basis of
4>^(vJ
=
1,
,{v^)
=
0,
^(v^)
=
0,
.
.
.,
.I>,ivj
^^{v,)
=
0,
,,(t;,)
=
1,
,{v,)
=
0,
.
.
.,
,ivj
The above
basis.
for-
-
these linear mappings ^, are unique and well defined. {v^
Consider the following basis of R^:
Example Hii:
=
(2,1), Uj
-
Find the dual basis
(3,1)}.
{^i> ^2}'
We
seek linear functionals ^i('i;i)
= = 0iK)
Thus
0,(vi)
11.2:
Let
{vi,
Then
.
.
0i(2, 1)
0j(3,l)
02(2,1)
=
02(3, 1)
Hence the dual basis
Theorem
1,
02K) = 02(^2)
The next theorems give
=
is
^i(a;,
y)
=
01(^2)
=
0,
= = = =
{4>i(x, y)
2a 3a 2c 3c
Let
.
,
v™} be a basis of
for any vector
{ffj,
P
is
{vi, .
.
.
,
6
d
d
+
V and let
by and 02(*. y)
02(^1)
= =
=
0,
11
9*2(^2)
=
=
ex
+ dy
1
^^
a
=
^^
e
= l,d = -i
-1, 6
=
3
= Oj = ij 4>2(.x,
y)
and their {^,
=
x
- 2y}.
duals.
...,4>Jhe the dual basis of V*.
uGV,
4,^{u)v^
+
4,^{u)v^
€
+
•
•
•
+
Su)v^
V*,
...,Vn} and {wi, ...,Wn} be bases of V and let {<^,, .. o-„} be the bases of 7* dual to {Vi} and {Wi} respectively.
the transition matrix from
matrix from
{4>^}
to
such that
Oj
= -x + Sy,
and, for any linear functional a
11.3:
+ +
6
relationships between bases
u =
Theorem
+ +
ax
(o-.j.
[Vi) to {Wi}.
Then
.,<#>„}
and
Suppose
(P-i)* is the transition
CHAP.
LINEAR FUNCTIONALS AND THE DUAL SPACE
11]
251
SECOND DUAL SPACE We repeat: every vector
space V has a dual space F* which consists of all the linear has a dual space V**, called the second dual of V, which the linear functionals on V*.
Thus V*
functionals on V. consists of all
itself
We now show that each GV* we define
for any
v
GV
determines a specific element v
^
=
vi)
It
remains to be shown that this
any
linear functionals
^,
+
ba)
v(a(j>
That
V
is,
is
Theorem
linear If
11.4:
o-
G
=
V*,
map v.V* we have
+ b(T)(v) —
(acf>
and so v GV**.
V has finite
-^
GV**.
First of
all,
iv)
K
For any scalars a,b
is linear.
a (v)
+
-
b a(v)
The following theorem
av{)
+
GK
and
bv{a)
applies.
dimension, then the mapping v
^
v
is
an isomorphism of
V
onto V**.
The above mapping v i^ t; is called the natural mapping of V into V**. We emphasize mapping is never onto F** if 7 is not finite-dimensional. However, it is always
that this
linear and, moreover,
it is
always one-to-one.
Now suppose V does have finite dimension. By the above theorem the natural mapping determines an isomorphism between V and V**. Unless otherwise stated we shall identify V with V** by this mapping. Accordingly we shall view V as the space of linear functionals on V* and shall write V = V**. We remark that if {^J is the basis of V* dual to a basis {Vi} of V, then {vi} is the basis of V = V** which is dual to (^J.
ANNIHILATORS Let
^gV* We is
W he a subset (not necessarily a subspace) of a vector space V. A linear functional called an annihilator of W = {0}. 4>{w) = for every w GW, is
if
show that the
set of all such
a subspace of V*.
a,b
gK
Clearly
(a
W
annihilator
Theorem Here
^,
{W)
if
called the annihilator of
G
W,
Then, for any scalars
W^.
is
W
is
b (t{w)
—
aO
+
bO
=
a subspace of V*.
a subspace of F,
we have
the following relationship between
W and
W.
11.5:
F
Suppose (i)
IF"" is
+ b
and so
In the case that IF its
i.e.
W and
w GW,
and for any
Thus a^ + baG
mappings, denoted by G W^. Now suppose
dim
W" = {vGV:
has
^(v)
viewed as a subspace of
The concept
of
finite
W + dim IF" = F
dimension and IF
= dim F and
(ii)
TF»»
is
a subspace of
for every ^ G W>} or, equivalently, under the identification of F and F**.
an annihilator enables us
F.
Then
= W. IF""
=
to give another interpretation of a
(TF")"
where
homogeneous
system of linear equations, anXi
+
ai2X2
+
•
•
•
+
ainXn
— (*)
amlXi
+ am2X2 +
•
•
•
+
UmnXn
—
LINEAR FUNCTIONALS AND THE DUAL SPACE
252
[CHAP.
11
A = (an) is viewed as an element Here each row {an, oa, ., (kn) of the coefficient matrix ., Xn) is viewed as an element of the dual space. of K" and each solution vector ^ = {xi, X2, In this context, the solution space S of (*) is the annihilator of the rows of A and hence of the row space of A. Consequently, using Theorem 11.5, we again obtain the following fundamental result on the dimension of the solution space of a homogeneous system of .
.
.
.
linear equations:
dimS = dimK" — dim (row
space of A)
= n - rank (A)
TRANSPOSE OF A LINEAR MAPPING Let U.
V
be an arbitrary linear mapping from a vector space V into a vector space for any linear functional ^ G U*, the composition ^ o T is a linear mapping from
T :V -^ U
Now into K:
That
(j)oT
is,
GV*. Thus
the correspondence h»
>
is a mapping from U* into V*; we denote words, T*:TJ* -^ V* is defined by
it
r'(0)
Thus
{T\4,)){v)
Theorem
=
11.6:
^{T{v)) for every v
T*{a
That
K
+ ba-) = =
=
call it
the transpose of T.
In other
4,oT
T' defined above is linear.
and any linear functionals
+ 6tr)or = a T\4,) + h T*{a) (a^
a{ci>oT)
+
<^,
a
G
f/*,
b(aoT)
T' is linear as claimed.
is,
We
G
by T' and
&V.
The transpose mapping
Proof. For any scalars a,b
oT
emphasize that
from U*
if
T
is
into V*:
a linear mapping from ^
The name "transpose" for the mapping
Theorem
11.7:
V
into U, then T* is a linear
mapping
j,,
T* no doubt derives
from the following theorem.
Let T-.V-^V be linear, and let A be the matrix representation of T relative to bases {Vi} of V and {Ui} of V. Then the transpose matrix A* is the matrix representation of T*:U*-* V* relative to the bases dual to {Mi}
and
{Vi}.
CHAP.
LINEAR FUNCTIONALS AND THE DUAL SPACE
11]
253
Solved Problems
DUAL SPACES AND BASES 11.1.
Let
:
and
11.2.
R2
(t{x,
+
R and a - Sx- y.
-*
y)
=
(i)
(^
(ii)
(40)(aj,2/)
(iii)
{2
»)(«,
J/)
=
4
R^
:
i>(x,y) 0(a;,2/)
=
R
-*
Find
+ =
be the linear functionals defined by ^{x, y) ^ + a, (ii) 4^, (iii) 2^ - 5
+
x
+ 2y) =
i(x
-
=
ha(x,y)
=
Oi*
+ azv + aaz,
such that
=
0i(i'i)
03(^1)
-
y
+ 2j/) -
4>^(x, y, z)
=
=
4x
5(3x
-
+
y
=
j/)
+
-13a;
(1,-1,3), Vi
=
+ h^y + bgz,
03(3;,
h-^x
= = =
0i(v2)
03(^2)
9j/
(0,1,-1), Vs
01(^3) 1
= = =
we
Solving the system of equations,
next find
= = =
02(^2) 02(1^3)
Solving the system,
we
0i(0, 1,
-1)
0i(0, 3,
-2)
obtain
a,^
= = =
3)
=
«!
we
obtain
61
=
7,
-1.
02(1.
02
= = =
02(0,1,-1) 02(0,3,-2)
h^—
=
—2, 63
=
61
= = =
03('"2)
03(^3)
Solving the system,
we
obtain
Cj
=
03(0,3,-2)
= = =
=
=
03(1,
-1,
3)
03(0,1,-1)
—2,
C2
1,
Cg
0, 03
-
=
+
62
(0,3,-2)}.
+
3C3
C3
= = 3c2-2c3 =
Thus
1.
1
0i(a;, V, 2)
«.
—
7x
—
2y
—
+
z.
3«.
1
03(x, y, z)
= —2x +
y
— 1,
i.e.
V =
=
<^i(/(i))
(We remark that Let Vi
=
1
be the vector space of polynomials over R of degree bt: a,b GR}. Let ^j:F-»R and ^2 ^"*'' be defined by
basis {vi,
+ CjW + CgZ
e^x
1
02(a', y, z)
C2-
C2
= = =
=
V
+
=
363
Hence
-
Ci
z)
j/,
Thus
0.
= 62- 63 = 362-263 =
—3.
find 03: 03(^1)
+ 303 = - tts = - 2a3 =
a2
3a2
1,
3)
-
fflj
02-
02(^1)
Finally,
-1,
0i(l.
?*2(''3)
03(1^3)
^^ as follows:
01(^3)
Let
+ 2y
8y
02('y2)
0l(^'2)
{a
3x
=
{vi
1
4>\ko\)
We
+
2(a;
= =
02(i;i)
And
x
seek linear functionals ^i(«, y, «)
We
+
2y
ix
Consider the following basis of R^: Find the dual basis {^j, 4)^, ^g}.
We
11.3.
=
a(x,y)
2
=
(i)
t;2}
=
a
of
Um)
and
and ^^ are linear and so belong which is dual to {<^j, ^g}.
^
V
+ bt
= S^'mdt
and
i^a
0i(^i)
= c + dt. By =
1,
02('yi)
==
S^'fi*)^^
to the dual space V*.)
definition of the dual basis,
=
and
0i(V2)
=
0,
02(^2)
=
1
Thus 0j(vj)
=
j
(a+6t)df
=
a
+
^b
=
1
or 02(''i)
=
I
(a
+
6t)
dt
=
2a
+
26
=
a
=
2, 6
= -2
Find the
LINEAR FUNCTIONALS AND THE DUAL SPACE
254
=
01(^2)
+ dt)dt -
{c
I
+ ^d -
c
or
MV2) =
— 2t, —^ +
In other words, {2
11.4.
+ dt)dt -
{c
(
-
+
2d
V
which
.
dual to
is
Then
{^j,
We
Set
show that
kii
+
•
{^i,
+
•
.
.
Similarly, for
=
+ It
•
•
•
+
i
=
2,
.
.
.
a{Vi)
fe„0„.
=
(fci0i
=
fei
=
fci
+
=
(fci0i
=
A;i0i('!;i)
Similarly, for
.
.
.
,
^
& V*
i
i
=
to v^,
2,
.
.
.
0,
.
.
•
•
+ fe„0„)(lJi)
•
+
+
^2
•
1
+
•
•
•
•••
+
k2
+
•
&„
+
fe„0„(Vl)
=
•
fci
+ /i:„0„)(i;i)
•
+
+
•••
+ •• +
ki^iivi)
=
fe„0n(-i'i)
^i
=
{0i,
.
.
we
.,a.„
.
J
,
+
a202
+
•
•
+
•
=
=
a„0„
obtain
=
0(vj)
{aj0i
ai0i(l'i)
+
+
a2
»!
•
1
ttz
+
•
02(^1)
+
•
+ a„0n)(i'i)
•
•
•
+ •• + O'nSi'nC'yi) = tti + a„ •
©(-Ui)
= +
a-l 0l('yt)
0.
+ a„^„)(Vi)
(ai0i ^
Hence
•
•
+
•
+
»i S6i(^i)
{0i, ...,>n} is
•
•
.
.
.
and, for any linear functional a a
= M
«n
S^nC^'i)
=
"i
linearly independent and so it is a basis of V*.
,
V
and
+
<7(i;J<^„
let
{^,
..
.,
GV*,
cr(i;>j
—
+
•
v„} be a basis of Prove Theorem 11.2: Let {vi, uGV, vector basis of V*. Then, for any
Suppose
aiVi
+
.7(1;,)^,
+
a^Vi
+
««
+
•
•
+ •
•
•
•
+
«„-«„
=
«!
(2) (5)
Then 0i(m)
=
di 0i(i;i)
+
=
Suppose
independent.
is linearly
cr
«,
,
=
=
be an arbitrary element of V*, and suppose
.
=
«!
^,
and a agree on the basis vectors, for i = 1, ..,n. Since Accordingly, {0i 0„} spans V*.
= = =
is,
=
i
Let
w,
,
remains to be shown that
Applying both sides
+
01(1^1)
ai0i
That
02)-
Then
fc„0„.
Thus
1
i
spans V*. Let
.,
a(Vi)
&101
=
d
.,^„} is a basis of V*.
.
first
—
.
if
fl
^
8..
{961,
V over K.
.
=
= —1,
c
1
Prove Theorem 11.1: Suppose {vi, .,Vn} isa basis of be the linear functionals defined by S.(v.)
11.5.
2c
the basis of
t} is
[CHAP. 11
^2 01(^2)
+
•
•
SilC'^n)
•
1
+
O2
•
+
•
•
•
+
««
•
==
«1
CHAP.
LINEAR FUNCTIONALS AND THE DUAL SPACE
11]
Similarly, for
i
=
2,
.
.
.,n,
=
>i{u)
That
is,
=
ipiiu)
»!
=
Oj, 02(w)
Next we prove
•
+
0nW =
•.
•
•
+
,j,i(vi)
a2'
Ui
= = =
•
+
i{Vi)
+
a„
=
0i(i;„)
»;
Substituting these results into
«n-
Applying the linear functional a
(2).
a{u)
(3),
we
obtain
(1).
to both sides of (1),
+ ^aM^K) + ••• + >n(u) l + ('(^2)02 + m G V, a — a{vi)
•
•
Since the above holds for every
11.6.
255
•
•
•
•
•
•
as claimed.
•
Prove Theorem 11.3: Let {vu...,Vn} and {wi,...,Wn} be bases of V and let CT„} be the bases of V* dual to {vi} and {Wt} respectively. {^1, ^„} and (<7j, Suppose P is the transition matrix from {Vi} to {Wi}. Then (P~»)' is the transition matrix from {(j>J to {
.
.
.
,
.
.
,
Suppose
=
OuUi
W2 =
(121^1
w„ =
a„ii)i
Wi
P=
where
(«„)
+ +
+ +
ai2V2 a22'«'2
•
+ +
•
•
a2nVn
02
= =
ftn^i
62101
+ +
6i202
+
^2202
+
+ +
•
"
"
"
and
•
•
•
'
•
.
.
•
.
.
•
•
6i„0„ hn'f'n
6„n0„
,
definition of the dual basis,
= =
where Sy
is
Q=
and hence
(6(101 6jiaji
the Kronecker delta.
(P«)-i
=
+ 6i202 + + 6j2aj2 + iJ,C2
...
^2^1
R2C2
•
\RnCl
Rvp2
(P-i)«
We
•
•
•
+ 6j„^„)(ajii)i + aj2V2 + + aj„v„) = = + 6i„a.j„ Sjj Rfij •
•
•
Thus
K„C„\ R2C„
•
•"1=7
!"'
-
^rflnl
as claimed.
Suppose V has finite dimension. GV* such that <^(v) # 0. mapping
•
/KiCi
=
QPt
11.7.
+ a„2i'2 + a„ - b^i,pi + 6„202 + + a„„v„ + Q = (6y). We seek to prove that Q = (P-i)«. tth row of Q and let Cj denote the ith column of P«. Then Rj = (6ti. 6i2. and Cj = (dji, aj2, a^^Y 6i„)
Let Ri denote the
By
ai„i;„
extend {v} to a basis {i), ^2. such that 0(1;) :V ^
K
•
•
-
Show >
•
1
that
^n) of V-
and
GV, v¥'0,
v
if
By Theorem
^(i;^)
=
=
0, i
2,
6.1, .
.
. ,
then there exists
there exists a unique linear Hence ^ has the desired
n.
property.
11.8.
Prove Theorem isomorphism of
We a,b
first
11.4:
V
has finite dimension, then the mapping v ^ v (Here v V* -* K is defined by v{) = ^(v).) map v \-^ v is linear, i.e. for any vectors v,w eV and any
If
V
onto V**.
prove that the
& K, av + bw = av + bw. For any av
av
Since
map
+ bw
()
=
At.
v
1-^ 1;
is
linear.
(at)
+ bw (0) +
6M))(0)
is
an
:
= =
linear functional
+
av(,)
for every
bw)
+
=
bw(4,)
G
+
b
(av
+
a ^{v)
=
scalars
V*, (f>{w)
bw)(ip)
= 0?+
6w.
Thus the
LINEAR FUNCTIONALS AND THE DUAL SPACE
256
[CHAP.
11
G V* for which V GV, v ¥= 0. Then, by the preceding problem, there exists ¥^ v¥=Q, the map v H- r v implies v Q. Since # and thus Hence v # (
Now
suppose
# 0.
ANNIHILATORS 11.9.
that if GV* annihilates a subset S of V, then ^ annihilates the linear span L{S) of S. Hence S» = (L(S))». .yW^G S for which v = a^w^ + a^w^ + + a^w^. Suppose V e L{S). Then there exist Wj,
Show
=
{v)
(12
+
+
0(W2)
was an arbitrary element
Since v
11.10.
+
Ui 0(Wi)
of L(S),
=
('('^r)
+
"l"
^^20
+
•
•
+
'
=
afi
W
be the subspace of R* spanned by vi = (1, 2, -3, 4) and v^ = (0, 1, 4, -1). Find Let a basis of the annihilator of W. By the preceding problem, it suffices to find a basis of the set of linear functionals {x, y, z, w) = and {vi) =
= =
0(0,1,4,-1)
The system Set tional
c
tional
The
Show (i)
(ii)
= 0,
02(«^.
2/.
a, h, c,
to obtain the solution
= = -1 «. *") = w)
11a;
d
—
4j/
+
&x
~
—
y
+
2& 6
that:
(i)
=
= 6, 6 = -1, c = 0, d = -1
a
Then
(i)
dim
11.5:
S
W>, the annihilator of W.
S^S"";
of V,
=
V
Suppose
W + dim W
has
Consider the dual basis
v
v(0) = 0(v) = Accordingly,
e S*, S S»o.
(SO)*.
S C S"*. annihilates every ele-
But S1CS2; hence
Sg.
G
v
Hence
0.
finite
dimension and
(ii)
W^ = W.
We
want
W
n.
it
to the following basis of V:
to
show that dim
is
& subspace of V.
W'>
{wi,
.
= n-r. We .
.,w„
Vi,
.
.
.,
choose v„_r}-
,
, •
•
•
>
0r> "it
•
•
• >
<'ti-r/
dual basis, each of the above a's annihilates each Wj-, hence that {ctj} is a basis of W'". Now {
definition
We next
the
show that <7
Thus {ffi dim V — dim
ffn-r)
W
=
{a^}
a(Wi)0i
001
W.
G W<>. By Theorem 11.2, +
spans
4-
•
4•
+
•
•
•
Let
•
•
-I-
•
•
•
TFOO.
•
•
+ aiv^-rW-r
spans PF" and so
it is
a basis of W^.
Accordingly,
dim
W^"
= m—
r
=
as required.
Suppose dimV^^w and diraW = r. Then dim F* by (i), dim TVO" = n- (n-r) = r; therefore dim
WC
S1CS2, then S^cS?.
-
\01i
so it
G
= dim V and
W
„!
«
for every
if
(ii)
Therefore S^ cSj.
Si.
=r Suppose dim V = w and dim a basis {wi, ...,w^}oiW and extend By
and hence the linear func-
w.
V e S. Then for every linear functional Therefore, under the identification of F and V**, (!;)
form with free variables c and d. and hence the linear funcc = 1, d =
= -4,
11, 6
z.
for any subset
0GS2. Then G Si, i.e.
= =
4d
4c-d
Let
Let
+
3c
is in echelon
set of linear functionals {0i, 02} is a basis of
Prove Theorem
(ii)
+
a
d o
to obtain the solution
ment of
(i)
unknowns
of equations in
= 1, d =
0i(a;, y, z,
Set c
11.12.
«r
•
•
annihilates L(S) as claimed.
0(1,2,-3,4)
11.11.
•
.
.
Accordingly,
W = WO".
=m
and, by
W = dim W".
(i),
By
dimTF<' = TO-r. Thus the preceding problem,
CHAP.
11]
11.13.
Let
LINEAR PUNCTIONALS AND THE DUAL SPACE
U
That
0G {U+W)0. e
is,
On then
W be subspaces of V.
and
Let
Then
G
and
t/o
Prove:
annihilates
£
hence
TF";
= M+w
where
U+W,
annihilates
a
i.e.
n
C/o
Wf =
+
n W^.
U°
U
and so, in particular, ,p annihilates Thus {U + W)" C U" n W.
PFO.
W
a e n W. Then a annihilates U and also W. m £ [7 and w e W. Hence ctCv) = (r{u) + a{w) = + e. (U + TF)". Accordingly, U'>+W'>c(U+ W)'>.
the other hand, suppose
-y
(C7
U+W
257
If
=
and y.
ve^U+W, 0.
Thus
a
Both inclusion relations give us the desired equality.
Remark: Observe that no dimension argument
is
employed in the proof; hence the result holds
for spaces of finite or infinite dimension.
TRANSPOSE OF A LINEAR MAPPING 11.14.
be the linear functional on R^ defined by ^(a;, y) - x - 2y. For each of the following linear operators T on R^, find iT%4,)){x, y): (i) T{x, y) = {x, 0); (ii) T{x,y) = (y,x + y); (iii) T{x,y) = (2x~Zy,hx + 2y).
Let
By vector
11.15.
of the transpose mapping,
definition
V.
2/)
= =
4>{T{x,y))
y)
=
(i)
{Tt{,p)){x, y)
(ii)
(rt(0))(x,
(iii)
{Tt(
= = =
,(T{x,y))
0(x, 0)
is
G
(Tt{4,)){v)
i.e.
=
for
every
x
T*:U*^ V*
let
its transpose. Show that the kernel Ker T* = (Im T)". o y = o. If m G Im T, then m = T(v) for some
e Ker
that
T*;
r'(0)
is,
=
be
i.e.
V; hence
have that
On
=
=
0(7(1;))
for every
m
the other hand, suppose o
G
(THamv)
We
.pof,
+ y) = y - 2{x + y) = -2x - y ,f,(2x 3y, 5x + 2y) = i2x - 3y) - 2i5x + 2y) = ~8x-ly.
0(m)
We
=
the annihilator of the image of T,
Suppose •y
=
{y,x
Let T-.V-^U be linear and of T'
rf(0)
Hence
have that
(r'(a))('y)
=
G Im (Im
=
!;
G
hence that
G
=
0{v)
=
(Im
2^».
Thus Ker T* C (Im
(4>°T){v)
{aoT)(v)
for every
Q(v)
T;
T)";
=
Y;
is,
=
^(Im T)
T*(a)
=
r)*.
Then, for every v
{0}.
=
a(T{v))
hence
=
=
0(v)
0.
Therefore
a
&V,
S Ker T*
and so
(Im r)o c Ker TK
Both inclusion relations give us the required equality.
11.16.
Suppose rank(r)
V and U have = rank(rO-
Suppose dim
V=n
dim ((Im
the preceding problem, claimed,
rank(r«)
11.17.
T
Suppose
is
linear.
U = m. Also suppose rank (T) = r. Then, by Theorem = dim 17 - dim (Im T) = m - rank (T) = m - r Ker Tt = (Im T)'>. Hence nullity (T') = m — r. It then follows
dim
11.7:
-
C/*
nullity (T')
=
m ~
(m-r)
=
r
=
rank
Prove:
11.5,
that, as
(T)
T:V -^ U
Let
relative to bases {vi,
matrix A* is [Ui] and {Vj}.
T:V ^ U
be linear and let A be the matrix representation Then the transpose it„} of U. Vm} of V and {ui, the matrix representation of T:JJ* ^ V* relative to the bases dual to
Prove Theorem of
=
dimension and suppose
and dim
T)0)
By
finite
.
.
.,
^(^i) T{V2)
.
= =
a-iiU-i
a2iUi
+ a^^u-i + + a22M2 +
.
•
•
•
•
•
.
,
+ ain^n + a2nU„
.^.
LINEAR FUNCTIONALS AND THE DUAL SPACE
258
We
want
r'(of„)
{(tJ
Let
V
T(v)
11
prove that
to
r'(o'2)
where
[CHAP.
and
=
«1201
+
<*2202
—
a-in't'l
+
'''•2ti02
{mJ and
{^j} are the bases dual to
=
+
•
•
+
•
•
•
+
ttm2'^m
+
0'mn>m
{2)
{vj} respectively.
+ fejVa + + fc„v„. Then, by (1), = + km TivJ r(-i;i) + ^2 r(i'2) + + C-mn'^'n) = &i(aiiMi + + «2n**n) + + fcm(a'mlMl + + aj„M„) + fc2(a2lMl + + k„a'mn)'>^n = (fciaii + fc2a21 + + fcmaml)^! + + (fcitlln + fc2*2n + e.V and suppose
v
k^v^ •
fci
•
•
•
•
•
•
•
•
•
•
•
•
•
•
'
•
•
'
•
"
'
'
n
2
=
(fciOii
+ fc2«2i +
•
•
+ kmOmdUi
•
i=l
Hence for
=
;/
1,
.
.
,
.
n,
(Tt(aj)(v))
On
Since v
+ a2j02 +
e.V was
•
•
11.18.
=
fciOij
.
•
arbitrary, r'(CTj)
which
ffjCrCv))
+
=
0-j
2
(
+
fe2a.2j
(ajj^i
k^aij
and
(4)
aij0i
+
is
+
fc2(l2i
fcm«mi)«t
1-
) (^)
++ •
•
+
•
+
amj0m)(fei'^i
A:2'U2
+
•
" '
+ km^m)
k^a^j
{i)
imply that a2j*2
+
•
•
•
+
i
a.mj't'm'
=
1,
.
.
.,
m
proved.
mxn
A
+
k^amj
+ a2j4'2 + k2a2j +
matrix over a be an arbitrary equal. A are column rank of
Let
+
•
•
(feiaii
.
(3)
Thus the theorem
is (2).
=
= 1, .,n, + a^j^J(v) = =
the other hand, for j (aij0i
=
Prove that the row rank and the
K.
field
and K^ Let T:K«-> K'n be the linear map defined by T{v) = Av, where the elements of X" to the usual bases are written as column vectors. Then A is the matrix representation of T relative Hence of if" and K", and the image of T is the column space of A. rank
By Theorem
11.7,
A«
is
(T)
=
column rank of
A
the matrix representation of T* relative to the dual bases.
rank
=
(T')
column rank of A*
=
row rank
of
Hence
A
hence the row rank and the column rank of A are But by Problem 11.16, rank(r) Theorem 5.9, page 90, and was proved in a direct way as earlier stated was result (This equal.
=
in
Problem
rank(r');
5.21.)
Supplementary Problems DUAL SPACES AND DUAL BASES 11.19.
Let
:
a(x, y, z)
11.20.
Let
-» R R3 -> R be the and = Ax-2y + 3z. Find (i) +
R3
:
<',
linear functionals defined (ii)
3^,
be the linear functional on R2 defined by
in particular, find
11.21.
0(— 2,
(iii)
0(2,1)
20
=
15
^(x, y,z)
and 0(1,-2)
7).
Find the dual basis of each of the following bases of (ii) {(1, -2, 3), (1, -1, (i) {(1, 0, 0), (0, 1, 0), (0, 0, 1)},
by
= 2x-By + z
and
- 5
R^: 1), (2,
-4,
7)}.
=
-10.
Find ^(x,y) and,
CHAP.
11.22.
LINEAR FUNCTIONALS AND THE DUAL SPACE
11]
V
Let
be the vector space of polynomials over
V
functionals on
of 11.23.
= a+bt + ct^eV
f{t)
V
which
dual to {^i,
is
Suppose u,vGV some scalar k. Suppose 0,ffGy* some scalar k.
11.25.
Let
V
11.26.
/'(f)
= m),
^{m)
11.28.
11.29.
Let •ith
11.30.
.p^mm =
denotes the derivative of
Let
and ^3 be the linear
2
m)
Find the basis
f(t).
{u)
and that
=
implies
=
[v)
=
implies
for
=
all
G
for all
fzd), ^(t)}
{/i(«),
Show that v = ku
V*.
vGV. Show
be the vector space of polynomials over K. For a^K, define
—
be the vector space of polynomials of degree
2.
^K
Let
a,b,c€.K be
/(a),
0b (/(«))
W he
a, subspace of V. For any linear functional such that a(w) =
V
that
by
=
a
for
k^,
for
=
/('')•
0a(/(*))
.
y be
=
which
/(c).
Show
is its
dual.
be the trace mapping:
is
a linear functional
the restriction of a to TF.
.
.
V
of
^K
Let
distinct scalars.
/(6), 0c (/(*))
on W, show that there is
, e„} be the usual basis of K". Show that the dual basis {ci, projection mapping: viia^, .,a„) = a.;. .
=
{/i(t), /2(t), fait)}
Let V be the vector space of square matrices of order n. Let T iV T{A) = ail + «22 + + "rm> where A = (o.^). Show that T is linear. Let a on
0i,
03}.
0a, 0b and 0e be the linear functionals defined by 0a (/(«)) = that {0a, 0b, 0c} is linearly independent, and find the basis
11.27.
Let
2.
that:
V
Let
and
2,
and that
11.24.
Show
-
of degree
defined by
Him) = f mat, Here
B
259
is
Wi,
.
.
., ir„}
where
vi is
the
.
a vector space over B. Let 0j, 02 S V* and suppose Show that either 0i = or 02 = 0.
y -> B
:
defined by a(v)
=
01(1;) 02('y)
also belongs to V*.
ANNIHILATORS
W
11.31.
Let be the subspace of B* spanned by (1,2,-3,4), (1,3,-2,6) and (1,4,-1,8). the annihilator of W.
11.32.
Let
11.33.
Show
11.34.
Let
11.35.
Suppose
W be the subspace of B^ spanned by any subset S of V, L{S)
that, for
U
and
W be
V - U
1,0)
(1,
=
V
@
W. Prove that V* =
Find a basis of the annihilator of W.
(0, 1, 1).
where L{S)
S""
subspaces of a vector space
and
is
Find a basis of
the linear span of S.
of finite dimension.
{U n W)"
Prove:
=
U"
+ W.
W ® WO.
TRANSPOSE OF A LINEAR MAPPING 11.36.
Let r B3 :
(i)
be the linear functional on B^ defined by
^ B2,
find (rK0))(a;, y,
T{x,y,z)
=
(x
+ y,y + z);
T(x,y,z)
(ii)
TiV^W
11.37.
Suppose
S:U-^V
and
11.38.
Suppose
T:V -*U
is linear
and
11.39.
Suppose
r y
is linear
and u
rt(0)
11.40.
=
:
-» [7
and 0(m)
=
(x,y)
=
Zx
—
2y.
For each linear mapping
z):
V
=
are linear.
has
G
finite
U.
(x
+ y + z,2x-y).
Prove that (ToS)t
Prove that Im T'
dimension.
Prove that u
= StoTK
GlmT
=
(Ker
or there exists
•/>
T)".
GV*
such that
1.
Let y be of finite dimension. onto Hom {V*, V*). (Here T
Show that is
the
mapping T
any linear operator on
h> Tt
V.)
is
an isomorphism from
Hom (V,
V)
LINEAR PUNCTIONALS AND THE DUAL SPACE
260
MISCELLANEOUS PROBLEMS 11.41. Let y be a vector space over R. The line wv — {tu + (1 — t)v: a — t — 1). A subset S Let
<6
Prove
11.42.
e y* and
[CHAP.
segment uv joining points m, v S V is defined by of y is termed convex if u,v GS implies uvcS.
let
W+ = {vGV: 4,(v) > W and W~ are that W +
Let y be a vector space of nonzero linear functional
finite
>
W
0},
= {vGV:
(v)
=
0},
W" =
{vi=,V:
(v)
A
number of hyperplanes.
6x-5y + 4z,
(i)
11.20.
y)
11.21.
(i)
{^i{x,y,z)
(ii)
{>t,i(x,
11.25.
(ii)
Let
11.26.
|/i(«)
11.31.
{0i(a;,
11.32.
{(x,
11.36.
(i)
=
Ax
f{t)
=
=
Supplementary Problems -16x + iy -
(iii)
t.
Then
^a(/(*))
5x
- + 2, 2/
.
ISz
41
= X, z{x,y,z) = z} = -3x -hy- 2z, ^ii'^, y, z) = 2x + y,
(„_6)(„_e)
z, t)
y,z)
=
to
ex-9y + Sz,
ly, 0(-2, 7)
y, z)
-
2/,
+
(ii)
=
h(t)
02(«.
a
^
6
-
1/. «. *)
=
0b(/(O).
>3(x,
22/
=
x
+
+ y-2z,
(ii)
.
/3W
- t}
(r'(0))(a;,2/,z)
2y
and therefore ^„
(6_„)(5_^)
=
y,z)
= x-y + z)
(Tt())(x,y,z)=Zx
0}
hyperplane i/ of y is defined to be the kernel of a dimension. Show that every subspace of V is the intersection of a finite
on V.
Answers
<
convex.
,
11.19.
11
= -x +
Sj/
+ 3x.
+ z]
#
0b-
(^_„)(,_6)
I
chapter 12
Quadratic and Hermitian Forms
Bilinear,
BILINEAR FORMS Let F be a vector mapping
for
space of finite dimension over a
f.VxV^K
which
+ bu2, v) = avi + bvi) =
(i)
f{aui
af{ui, v)
(ii)
f{u,
af{u, Vi)
A
K.
bilinear
+ +
12.1:
Let
=
bilinear
4>(u)
form = ^ (g)
written /
Example
12.2:
Let
form
V
y
is
linear in the
=
A = / on
(a^)
(ffly)
K
and v
be any
X" by
=
is,
= u'v = (h^).
nXn
a^bi
Then /
is
+
a^h^
+
•
+
•
•
a„6„
a bilinear form on R".
matrix over K.
Then
A may
be viewed as a bilinear
defining '
f(X,Y)
:
a.)
Let / be the dot product on R"; that
where u 12.3:
a
Then
f{u, v)
Example
is
X -» be defined by Let / (Such a / is bilinear because (p and a are each linear. and
and a be arbitrary linear f unctionals on V.
f(u,v)
V
bf{u, Vi)
a,b&K
Example
form on
bf{u2, v)
and all im, Vi e V. We express condition (i) by saying / variable, and condition (ii) by saying / is linear in the second variable.
all
first
field
satisfies
= XtAY = =
2
U-
{xi,X2
^ij^iVj
a;
«ll«l2/l
J +
an
ai2
a^i
0.22
*nl
°'n2
«12»'l2/2
...
a.i„
•
"2n
I
\
I
I
Vi 2/2
Vnl
+
The above formal expression in variables Xi,yi is termed the bilinear polynomial corresponding to the matrix A. Formula (i) below shows that, in a certain sense, every bilinear form is of this type.
We
B{V) denote the set of bilinear forms on V. placed on B(V) by defining f + g and kf by: will let
{f
for any f,g
Theorem
+ 9){u,v) = {kf){u,v) =
& B{V) and any kE^K.
12.1:
f{u,v)
+
A
vector space structure
is
g{u,v)
kf{u,v)
In fact,
F
be a vector space of dimension n over K. Let {4,^, ...,>„} be a basis of the dual space V*. Then {fij:i,j = 1,. .,w} is a basis of B{V) Thus, in particular, where fa is defined by fi}{u,v) = ^{u) .{v).
Let
•
dimB{V) =
w'.
261
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
262
[CHAP. 12
BILINEAR FORMS AND MATRICES Let / be a bilinear form on V, and
let {ei,
.
.
.
and suppose
u =
aiCi
+
•
•
•
+
=
v
a„e„,
e^}
,
Suppose u,v
be a basis of V.
biCi
+
•
•
•
+
eV
bnen
Then f{u, v)
—
f{aiei
=
ai&i/(ei, ei)
+
•
•
•
+ anCn,
biCi
+
•
•
+
•
bne„) n
+
0162/(61,62)
+
+
•••
=
a„b„/(en, e^)
^=
i,3
Thus
/
is
completely determined by the n^ values
aib}f{ei,ej) 1
/(ei, e,).
A
= {an) where an — f{ei, e,) is called the matrix representation of f relThe matrix ative to the basis {ei} or, simply, the matrix of f in {ei}. It "represents" / in the sense that '&i^
=
f{u,v)
^
=
aibjfiei, ej)
(ai,
.
.
.
,
a„)A
=
^|
|
Me^Me
(1)
^bnj
for
all
basis
We basis
u,v
GV.
(As usual,
[u]e
denotes the coordinate (column) vector of
u
GV
in the
{ei}.)
is
does a matrix representing a bilinear form transform when a new The answer is given in the following theorem. (Recall Theorem 7.4 that matrix P from one basis {e^} to another {el} has the property that [u]e = P[u]e'
how
next ask, selected ?
the transition for every
uG
Theorem
12.2:
V.)
Let P be the transition matrix from one basis to another. matrix of / in the original basis, then
If
A
is
the
B = P'AP is
the matrix of / in the
new
basis.
The above theorem motivates the following Definition:
definition.
is said to be congruent to a matrix A nonsingular) matrix P such that B — P^AP.
A matrix B (or:
if
there exists an invertible
Thus by the above theorem matrices representing the same bilinear form are congruent. We remark that congruent matrices have the same rank because P and P' are nonsingular; hence the following definition Definition:
is
well defined.
The rank of a bilinear form / on V, written rank(/), is defined to be the rank of any matrix representation. We say that / is degenerate or nondegenerate according as to whether rank (/) < dim V or rank (/) = dim V.
ALTERNATING BILINEAR FORMS A bilinear form / on 7 is said to be alternating f(v,v)
(i)
for every v
GV.
If / is alternating,
= and so
f{u
=
then
+ v,u + v) = (ii)
if
f{u, u)
f{u,v)
=
+
f{u, v)
-f{v,u)
+
f{v, u)
+
f{v, v)
CHAP.
BILINEAR, QUADRATIC
12]
AND HERMITIAN FORMS
263
A
for every u,v GV. bilinear form which satisfies condition (ii) is said to be skew symmetric (or: anti-symmetric). If 1 + 1 v^ in K, then condition (ii) implies f{v, v) = — f{v, v) which implies condition (i). In other words, alternating and skew symmetric are equivalent
when
1
+1
9^ 0.
The main structure theorem of alternating
Theorem
forms follows.
bilinear
Let / be an alternating bilinear form on V. Then there exists a basis of V in which / is represented by a matrix of the form
12.3:
Oil -1_04
^
l1
I
LzLjPJ, 1
I
1-1
j
I
-T-«4 lOj ro-
Moreover, the number of it is
equal to ^ rank
(
-1
,,
]
uniquely determined by / (because
is
(/)).
In particular, the above theorem shows that an alternating bilinear form must have even rank.
SYMMETRIC BILINEAR FORMS, QUADRATIC FORMS A bilinear form / on V is said to be symmetric if =
f{u,v)
for every u,v €.V.
If
A
is
a matrix representation of
f{X,Y)
(We use the fact that symmetric,
X*AY
f{v,u)
is
= X'AY = {X'AYY =
and since
this is true for all vectors
versely
A
is
The main
Theorem
symmetric, then /
result for
12.4:
is
symmetric bilinear forms
is
write
Y'A'^X its
transpose.)
Thus
if
/ is
Y'AX
A = A*
or
A
is
sjTnmetric.
Con-
given in
K
(in which 1 + 1^0). Let / be a symmetric bilinear form on V over .,v„} in which / is represented by a diagonal Then V has a basis {vi,
matrix,
Form
=.
X, Y it follows that symmetric.
.
Alternate
we can
a scalar and therefore equals
Y*A*X = f(X,Y) = f{Y,X)
if
/,
i.e.
f{vi, Vj)
=
.
for
i ¥- j.
Let A be a symmetric matrix over X (in which 1 + 1 ^^^O). of Theorem 12.4: Then there exists an invertible (or: nonsingular) matrix P such that P*AP That is, A is congruent to a diagonal matrix. is diagonal.
BILINEAR, QUADRATIC
264
P
AND HERMITIAN FORMS
[CHAP. 12
a product of elementary matrices (Problem 3.36), one way is by a sequence of elementary row operations and the same sequence of elementary column operations. These same elementary row operations on / will yield PK This method is illustrated in the next example. Since an invertible matrix
of obtaining the diagonal
is
form P*AP
2-3
1
Example
12.4:
Let
A
-
matrix (A,
2
5
—4
-3
-4
8
a symmetric matrix.
|,
It is
convenient to form the block
I)
(A.I)
1
2
2
5
-3 -4
3
-4
8
1 1
1
apply the operations R^ -» — 2/Ji + R^ and R^ -^ SR^ + R3 to {A, I), and then the corresponding operations C^ -^ — 2Ci + C^ and Cg -* 3Ci + C3 to A to obtain
We
1
2
-3
1
1
2
-2
2
-1
3
1
We next apply the operation C3 - — 2C2 + C3 to obtain /l
^3
->
—2R2 + R3 and then
1
-2
1
-5
7
-2
and then
A
We
has been diagonalized.
mapping q:V-*K
1
1 0! -2 -5 7 -2
set
Pt A
and then
is called
1
1
P -
Definition:
the corresponding operation
1
2
1
Now A
and then
1
P
a quadratic form
if
q{v)
= f{v,v)
for some
symmetric bilinear form / on V.
We 1
+
call
q the quadratic form associated with the symmetric bilinear form from q according to the identity
/.
If
in if, then / is obtainable
1 7^
f(u,v)
The above formula
Now
if / is
is called
= Uq{u + v) -
the polar form of
q{u)
-
q{v))
/.
A=
represented by a symmetric matrix
(an),
then q
is
represented in the
form
(an
ai2
...
ain\ /xi\
a21
^22
...
azn \l X2
ttnl
a„2
...
...
=
y u
ttiiXiXj
-
is,
022X2
+
•
•
•
+
annxl
+22
atiXiXj
termed the quadratic polynomial correspondObserve that if the matrix A is diagonal, then q has the
in variables Xi is
ing to the symmetric matrix A. diagonal representation
that
+
*<^
The above formal expression
q{X)
aiiccf
annj \Xnj
- X*AX - anxl +
a22xl
+
•••
+
annxl
the quadratic polynomial representing q will contain no "cross product" terms. 12.4, every quadratic form has such a representation (when 1 + 1^0).
Theorem
By
CHAP.
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
12]
Example
12.5:
265
Consider the following quadratic form on R^:
One way which X
v)
=
_
2x2
+
\2,xy
5j/2
is by the method known as "completing the square" Problem 12.35. In this case, we make the substitution obtain the diagonal form
of diagonalizing q fully described in
is
= s + Zt,
y
=
t
to
=
q(x, y)
+ 3t)2 -
2(s
12(s
+ U)t +
-
5(2
2s^
-
13*2
REAL SYMMETRIC BILINEAR FORMS. LAW OF INERTIA In this section we treat symmetric bilinear forms and quadratic forms on vector spaces over the real field R. These forms appear in many branches of mathematics and physics. The special nature of R permits an independent theory. The main result follows.
Theorem
Let / be a symmetric bilinear form on V over R. Then there is a basis of V in which / is represented by a diagonal matrix; every other diagonal representation has the same number P of positive entries and the same number of negative entries. The difference S = P — is called the signature of /.
12.5:
N
A
N
symmetric bilinear form /
real
said to be nonnegative semidefinite
is
=
q{v)
for every vector v; and
is
By
=
/
(ii)
/ is positive definite
where
the signature of
Example
12.6:
>
f{v,v)
the above theorem,
and only if S and only if S = dim V
nonnegative semidefinite
(i)
is
^
f{v,v)
said to be positive definite if q{v)
for every vector v ¥=0.
if
if
if
/.
Let / be the dot product on R"; that f(u, v)
=
where u
(ttj)
and v
—
wV
=
Furthermore, /
is
is,
=
eii6i
Note that /
(6j).
f(u, v)
=
WV
=
+
a2&2
is
+
•
"
•
+
"n^n
symmetric since
vu
=
f{v, u)
positive definite because f(u, u)
when M
— rank (/)
= a^+ al+
+
al
>
v^ 0.
In the next chapter we will see how a real quadratic form q transforms when the transiP is "orthogonal". If no condition is placed on P, then q can be represented in diagonal form with only I's and — I's as nonzero coefficients. Specifically, tion matrix
Corollary
12.6:
Any
form q has a unique representation
real quadratic qytVif
.
.
.
,
iXyn)
—
The above result for real quadratic forms or Sylvester's Theorem.
is
in the
form
2 vCi
I
'
*
*
~T'
i^g
s
+1
sometimes referred to as the
Law
of Inertia
BILINEAR, QUADRATIC
266
HERMITIAN FORMS Let y be a vector space
AND HERMITIAN FORMS
[CHAP. 12
of finite dimension over the complex field C.
be such that
+ bu2,v) =
(i)
f{aui
(ii)
f{u,v)
^
+
af{ui,v)
Let /
fiv,u)
a, 6 G C and im, v &V. Then / is called a Hermitian form on V. denotes the complex conjugate of A; G C.) By (i) and (ii),
That
+ bv2) = =
is,
(As usual, k
+ bv2, u) — af{vi, u) + b f{v2, u) df{vi,u) + bf{v2,u) = df{u,Vi) + bf{u,V2)
f{avi
f{u, avi
(iii)
V xV -* C
bf{u2,v)
where
f{u, avi
:
+ bv2) -
+
a f{u, Vi)
b f{u, Vi)
express condition (i) by saying / is linear in the first variable. On the other hand, we express cond ition ( iii) by saying / is conjugate linear in the second variable. Note that, by (ii), f{v, v) = f{v, v) and so f{v, v) is real for every v GV.
As
before,
we
Example
12.7:
A=
Let
(dy)
be an
Ji
X w matrix over
We
C.
write
A
taking the complex conjugate of every entry of A, that A* for A« = AJ. The matrix A is said to be Hermitian
A
If
lem
is
Hermitian, then f{X, Y)
^X'^AY
defines a
for the matrix obtained
by
A = (a^). We also write A* = A, i.e. if ay = a]^.
is,
if
Hermitian form on C" (Prob-
12.16).
The mapping q:V->-B, defined by q{v) = f{v, v) is called the Hermitian quadratic form or complex quadratic form associated with the Hermitian form /. We can obtain / from q according to the following identity called the polar form of /:
=
f{u, v)
l{q{u
+ v)-
q{u
- v)) +
j(q{u
+ iv) -
q{u
- iv))
= (fo«) where feij = /(ei, e,) is suppose {ei, e„} is a basis of V. The matrix By (ii), f{ei,ej) = f(ej,ei); hence {ej}. called the matrix representation of / in the basis Thus any diagonal repreal. are is Hermitian and, in particular, the diagonal entries of the complex analog of theorem is resentation of / contains only real entries. The next Theorem 12.5 on real symmetric bilinear forms.
Now
.
.
.
H
,
H
H
Theorem
., en} of Let / be a Hermitian form on V. Then there exists a basis {ei, for = i.e. matrix, f{ei, ei) diagonal by a represented V in which / is number same the of has representation ¥= diagonal / every Moreover, i j. .
12.7:
P
of positive entries, and the
difference
S = P-N
Analogously, a Hermitian form /
is
eV, and
is
N
of negative entries.
The
/.
said to be nonnegative semidefinite if
=
q{v)
for every v
same number
is called the signature of
.
f{v,v)
^
said to be positive definite if
=
q{v)
f{v,v)
>
for every v ¥=0. Example
12.8:
Let / be the dot product on C"; that f(u, V)
where u
=
(«{)
and v
=
=
zi«i
+
ZiWi
Then / any v # 0,
(Wj).
positive definite since, for f(u, u)
is,
— U'V =
Z2,Z2+
•+
is
z„z„
+
Z2W2
+
•
+
z„w„
a Hermitian form on C".
=
l^iP
+
I22P
+
•
•
•
Moreover, /
+ K\^ >
is
CHAP.
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
12]
267
Solved Problems BILINEAR FORMS 12.1. Let u = {xi, X2, xs)
and v
=
f{u, v)
Express Let
/ in
A
=
(t/i, 1/2,
Sxiyi —
and
ya),
+
2xiyi
let
+
5x22/1
7*21/2
-
3X3
matrix whose i;-entry
-
Xzys
Then
the coefficient of XiVy
is
/3 -2
= XtAY =
0\/j/i^
7-8
5
(^i.xa.ajg)
:
4 -l/\2/3i
\0
Let A be an « X « matrix over K. Show that the following mapping / form on K": f{X,Y) = X*AY. For any a,bGK and any Xj, Fj e K", = (aZi + 6X2)*^ Y = (aXj + 6X^) .4 F /(aATi + 6Z2, F) = aXlAY + bXlAY = a/(Zi, F) + 6/(X2, F) Hence /
Hence /
the first variable.
is linear in
/(X, aFi
12.3.
4x33/2
matrix notation.
be the
f{u,v)
12.2.
+
8a;2i/3
+ ftFa)
=
XtA(aFi
+
a bilinear
Also,
= aX^AYi + bX^AY^ =
6F2)
the second variable, and so /
is linear in
is
a /(X, Fj)
+
6 /(X, F2)
a bilinear form on K".
is
Let / be the bilinear form on R^ defined by f{{xu (i)
(ii)
(iii)
Set
3xi2/2
+
X22/2
Set
= /(ttj, Uj): = f(ui,ui) = an ai2 = /(mi,M2) = 021 = /(M2.M1) = 022 = /(M2.W2) =
where ay
(ay)
A =
B=
(
ft
(6y)
/
*^
where 6y
612 621
622
Thus
We
B =
/3
/((1,0), (1,1)) /((1,1), (1,0)) /((l.l), (1,1))
=2-0 + 0=
2
= 2-3 + = 2-0 + = 2-3 +
2
= -1
0= 1
=
= = = = =
/(Vj,-!;,):
/(^i.i'i) /(^i.'y2) f(->'2,'>'i)
/(^2.^2)
= = = =
/((2,1), (2,1)) /((2.1), (1,-1))
/((l.-l), (2,1)) /((I, -1), (1,-1))
= 8-6 + 1 = 4 + 6-1 = 4-3-1 = 2+3+1
9\
I
must write
/((1,0), (1,0))
*^® matrix of / in the basis {u^, n^}.
=(r;)'
611
(iii)
-
B^P*AP.
A =
Thus
(ii)
2xi2/i
Find the matrix A of / in the basis {Ui — (1,0), 112 = (1, 1)}. Find the matrix B of / in the basis {vi = (2, 1), V2 = (1, -1)}. Find the transition matrix P from the basis {mi} to the basis that
(i)
=
X2), (yi, 2/2))
ft
Vi
/
's
^^^ matrix of / in the basis {vi, v^}.
and V2
in
ri
V2
terms of the
= =
(2,1)
=
(1,-1)
Mj:
(1,0)
=
+
2(1,0)
(1,1)
-(1,1)
= M1 + M2 = 2Mi-M2
= = = =
3 9
6
{Vi},
and verify
BILINEAR, QUADRATIC
268
^ =
Then
J
12.4.
and so
_j)
(
Q _M
=
P'
AND HERMITIAN FORMS
[CHAP. 12
Thus
.
Prove Theorem 12.1: Let F be a vector space of dimension n over K. Let {^j, ^„} be a basis of the dual space V*. Then {/«: ij" = 1,. .,%} is a basis of B{V) where /« is defined by f^.{u,v) = ^.(m)^.(v). Thus, in particular, dimB(F) = n\ .
.
.
,
.
Let
{e^,
.
.
and suppose
.,
for
ej)
(2aij/ij)(es,
V dual to {^J. claim that /
e„} be the basis of
=
/(ej, e^)
ay.
=
s,t
We
We
l,...,n.
(2 ay/y)(es, et) Hence
as required.
=
1,.
.
12.5.
G B(y) =
/(e^.e,)
have
=
2ay/«(es,
=
2ay«isSjt
is
linearly
0(es, et)
Thus
last step follows as above.
2 Oy ^i(es) ^^(et)
=
ej)
=
"st
=
f(es,et)
Soy/y =
Suppose
independent.
Then
0.
for
= (2 ao/y)(es, Bf) =
{/y} is independent
and hence
a^s is
a basis of B(V).
denote the matrix representation of a bilinear form / on F relative to a basis Show that the mapping / i-» [/] is an isomorphism of B{V) onto the vector space of w-square matrices.
Let
[/]
{ei,
.
.
.,e„) of V.
Since / onto.
completely determined by the scalars show that the mapping / l-> [/]
is
It suffices to
=
[af+bg] However, for
i,j
=
1,.
.
which
is
a restatement of
Prove Theorem
f(e^, ej),
is
a[f]
the
mapping
/*"*[/]
a homomorphism; that
+
is,
is
one-to-one and
that
b[g]
(*)
.,n,
(af
12.6.
show that
to
.,n,
= The
spans B(y). Let /
{/y} spans B{V).
remains to show that {/y}
It s,t
We first show that {/„} = ^ay/y. It suffices
12.2:
+
bg){ei, e^)
(*).
Thus the
Let
P
A
=
+
«/(«{, e^)
bg(ei, Bj)
result is proved.
be the transition matrix from one basis {e,} to another {Ci}, then B = P'^AP is the
basis {gj}. If is the matrix of / in the original basis matrix of / in the new basis {e\}. Let u,v
eV.
Since
P is the transition matrix from {e^} to {e,-}, we = [u]l, PK Thus f{u,v) = [u]lA[v], = [u]l.PtAP[v],.
P[v]e-
=
Since
u and v are arbitrary elements
[v]e-
hence
[u]l
of V, P^ A
P
is
have
P[u]g,
—
[u]g
and
the matrix of / in the basis {e^}.
SYMMETRIC BILINEAR FORMS. QUADRATIC FORMS 12.7.
Find the symmetric matrix which corresponds to each of the following quadratic polynomials: (i)
q{x, y)
(ii)
q{x, y)
= -
4x^
xy
-
6xy
-
7y^
+ y^
The symmetric matrix
A—
(uy)
(iii)
q{x, y, z)
(iv)
q(x, y, z)
= =
+ 4xy - y' + - 2yz + xz
3x^ x^
8xz
—
Qyz
+
z^
representing q(xi, .,x„) has the diagonal entry a^ equal to ajj each equal to half the coefficient of ajjajj-. Thus
the coefficient of xf and has the entries ay and
.
.
CHAP.
BILINEAR, QUADRATIC
12]
AND HERMITIAN FORMS
269
(ii)
12.8.
For each of the following such that P*
AP
A -
(i)
(i)
symmetric matrices A,
real
diagonal and also find
is
its
-:
(ii)
First form the block matrix (A,
find
A =
-
=
R2^^Ri + R2 and iZa -» — 2i2i + iJj -> SCj + C^, and C3 -» — 2Ci + C3
sponding column operations C^ 1
-3 -2
2
1
1
3
1
4
-2
C3-* C2
+ 2C3
and then
1
-» i?2
to
+ ^^3
A
and then the corre-
to obtain 1
-2
1
3
1
4
-2
\o
1/
B3
to (A,/)
/I
0\
Next apply the row operation
P
I):
{A, I)
Apply the row operations
a nonsingular matrix
signature:
1 1
^^^ then the corresponding column operation
to obtain
and then
Now A
has been diagonalized.
The signature S of
(ii)
A
Set
5 =
is
First form the block matrix (A,
P
then
2-1 =
P^AP -
1
I):
(A,/)
=
1
1
1
-2
2
1
2
-1
1 1 1
In order to bring the nonzero diagonal entry —1 into the first diagonal position, apply the row operation Ri <-> R^ and then the corresponding column operation Ci «-> C3 to obtain 1
2
-1
1
-2
2
1
1
1\
1
-*
2
1
2
3
3
1
0/
1
Apply the row operations column operations C^
and then
1
2Ci
i?2 ~*
+ C^
\
1
2
1
2
-2
1
1
1
2Bi + -^2 ^nd JB3 -> iJj + R^ and C3 ^ Ci + C3 to obtain
2
1/
1 1 1
and then the corresponding
/-I
1\ 1
-1
and then \
1
2
3
3
1
1 1
2 1
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
270
Apply the row operation C3
-»
-3C2 + 2C3
—3^2 + 2R3
iJg ->
and then the corresponding column operation
to obtain
/-I
/-I
l\
2
12
3
4
12.9.
'
the difference
is
-3 -4/
1
\l
The signature S of
-14
-3 -4,
2
2\
P =
Set
12
2 \
/O has been diagonalized.
1^
and then
2-3-4/
-7
\
Now A
[CHAP. 12
2
S =
—
1
2
;
=
P'AP =
then
—1.
Suppose 1 + 1 v^ in K. Give a formal algorithm to diagonalize (under congruence) a symmetric matrix A = (an) over K. Case
I:
Apply the row operations
aii¥=0.
corresponding column operations
Cj -*
fij ->
— Oji Cj + an C;
— ajxi?x + OxxiJj, A
to reduce
=
i
to the
2,
.
.
and then the
.,n,
/ill
form
(
0"
^
^0
Case II: a^ = but a^ ^ 0, for some i > 1. Apply the row operation Ri «^i2j and then the corresponding column operation Cj <-> to bring ctjj into the first diagonal position. This reduces the matrix to Case I.
Q
All diagonal entries Oji = 0. Choose i, j such that ay ¥= 0, and apply the row operainto the Rj + Rf and the corresponding column operation Ci-* Cj + Cj to bring 2ay # ith diagonal position. This reduces the matrix to Case II.
Case
III:
i?i ->
tion
In each of the cases,
matrix of order
we can
By
than A.
less
induction
Remark: The hypothesis that
12.10.
1
+1
A
we can
#^
in
/ail
form
(
finally bring
A
to the
K,
is
.
0\ _
)
where
B
is
a symmetric
into diagonal form.
used in Case III where
we
state that 2ay ¥= 0.
Let q be the quadratic form associated with the symmetric bilinear form /. Verify (Assume that the following polar form of /: fiu,v) - ^q{u + v) - q{u) - q{v)).
+ 1^0.)
1
+ v) —
q(u
+
1
If
12.11.
reduce
finally
we can
1 7^ 0,
—
= = =
f(u
+ v,u + v) — f{u, u) — f(v, v) + f{u, v) + f{v, u) + fiy, v) -
f(u, u)
f{u, u)
-
f(v, v)
2f{u,v)
divide by 2 to obtain the required identity.
K
12.4:
+ 1^0). Then V i.e. f{Vi, v,) =
.
matrix, Method
qiv)
(in which Let / be a symmetric bilinear form on V over a diagonal represented by has a basis {vi, .,Vn) in which / is
Prove Theorem 1
q{u)
for
.
i ¥- j.
1.
= = n>
—
then the theorem clearly holds. Hence we can suppose f ¥= Q and - for every v&V, then the polar form of / (see Problem 12.10) implies that / = 0. Hence we can assume there is a vector t?i e V such that f(Vi, v^ ¥= 0. Let consist of those vectors 1; G y for which /(^i, v) = 0. U be the subspace spanned by Vi and let If
dim V
/
or
dim V
if
If
1.
q(v)
=
1,
f(v, v)
W
We (i)
V = U ®W.
claim that
Proof that
UnW = =
uGW,
Since fore u
=
kvi
=
0.
{0}:
f{u,u)
Thus
Suppose
=
uG U nW.
f(kvi,kvi)
=
UnW = {0}.
Since
k^ f(Vi,Vi).
ue^U, u —
But
kv^ for some scalar
/(^i.-Wi) ?^ 0;
hence
k-0
k&K.
and there-
;
CHAP.
V=
Proof that
(il)
U + W:
Then
By
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
12]
Now
f{v^,w)
w G W. By (1), v is (ii), V = U ® W.
Thus and
(i)
vG
Let
/ restricted to
W
the
is
sum
V.
Set
=
f(v„v)
Method
•
- ;^^/(^i'^i) =
of an element of
U
"
V = U + W.
and an element of W. Thus
W
— n. — 1; hence by a symmetric bilinear form on W. But dim such that f(v^, Vj) = for i ¥= j and 2 — i, j — n. But v„} of V for j = 2, .,n. Therefore the basis {v-^,
W
induction there is a basis {^2. . v„} of by the very definition of W, fiv^, Vj) = has the required property that /(-Uj, Vj) = •
271
•
.
for
.
.
.
.
,
i ¥= j.
2.
K
is congruent to a The algorithm in Problem 12.9 shows that every symmetric matrix over diagonal matrix. This is equivalent to the statement that / has a diagonal matrix representation.
12.12.
Let
(i)
(ii)
A =
if
K
if
K
I's,
(i)
Let
is
and
Show
a diagonal matrix over K.
, 1
for any nonzero scalars with diagonal entries aifcf I's
(iii)
^ I
fci,
.
.
.
the complex field C, then O's as diagonal entries;
,
A
that:
fc„
e
If ,
A
is
congruent to a diagonal matrix with only
is
congruent to a diagonal matrix
the real field K, then A is congruent to a diagonal matrix with only —I's and O's as diagonal entries.
P
is
be the diagonal matrix with diagonal entries
ptAP =
^"2 I
11
fc^.
Then
02^2
02
O-nKl
(ii)
Let
P
be the diagonal matrix with diagonal entries
f>i
if «{ '^
fl/Voi
—
"]
_
•«
-.
r,
•
Then P^AP has
the required form.
(iii)
P
be the diagonal matrix with diagonal entries the required form.
Let
Remark. We emphasize that (ii) is no longer true gruence (see Problems 12.40 and 12.41).
12.13.
6j
=
fl/A/hl
if
<
^
if
congruence
Oi^O _
.
Then P^AP has
'
is
replaced by Hermitian con-
Let / be a symmetric bilinear form on V over R. Then there is a basis of V / is represented by a diagonal matrix, and every other diagonal representation of / has the same number of positive entries and the same number of negative entries.
Prove Theorem
12.5:
in
which
u^} of V in which / is represented by a diagonal By Theorem 12.4, there is a basis {itj, ., w„} is another basis of matrix, say, with P positive and negative entries. Now suppose {wi, V in which / is represented by a diagonal matrix, say, with P' positive and N' negative entries. We can assume without loss in generality that the positive entries in each matrix appear first. Since = P' + N', it suffices to prove that P = P'. rank (f) - P + .
N
N
.
.
,
.
.
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
272
Let
W
be the linear span of u^, .., up and let be the linear span of Wp, + 1, for every nonzero v e U, and f{v,v) ^ for every nonzero v
[7
.
>
f{v,v)
[CHAP. 12
UnW = {0}.
.
.
Note that dimU = P and dimW = n- P'. Thus dim{U+W) = dimU + dimW - dim(UnW) = P + {n-P')-0 = P But dim(U+W) ^ dim V = n; hence P-P' + n^n or P ^ P'. Similarly, P' ^ P fore P = P', as required.
Remark. The above theorem and proof depend only on the concept of theorem is true for any subfield K of the real field R.
12.14.
,
.
m)„.
& W.
Then Hence
+ n
P'
and there-
Thus the
positivity.
An nxn real symmetric matrix A is said to be positive definite if X*AX > for every nonzero (column) vector G R", i.e. if A is positive definite viewed as a bilinear form. Let B be any real nonsingular matrix. Show that (i) B*B is symmetric and (ii) B*B is positive definite. {Btpy = ptP" = B<-B\ hence B'JS is symmetric. (i)
X
(ii)
B
Since
is
nonsingular,
BX-BX =
itself,
BX #
X S R".
for any nonzero
(BXY(BX),
BX
with
= (BX)HBX) >
as
Hence the dot product of
Thus XHBtB)X
is positive.
=
{XtBt){BX)
required.
HERMITIAN FORMS 12.15.
Determine which of the following matrices are Hermitian:
2-i
+
4
6
i
i
3
i\
/
(ii)
A
(iii)
Let
A
(ii)
12.16.
A =
matrix
The matrix The matrix The matrix
(i)
(Oj^)
is
not Hermitian, even though
is
Hermitian.
by f(X, a, 6 e C and
all
all
/(aXi + 6X2,
Hence / equal to
its
/
V.
Show
(i)
f{u,v)
(ii)
if
(or:
(i)
it is
Show
that /
iflf
o.^
= 'ajl.
conjugate transpose, symmetric, is
is
Hermitian
and only
if
symmetric.
if it is
a Hermitian form on C« where /
is
Y. X^, X^,
y)
Ye
C",
+ hX^YAY = (aX\+hXl)AY aX\AY + bXlAY = af(Xi,Y) +
= =
(aX^
bf(X2,Y)
Also,
Wa^X = YtA*X =
= xTaY = (XTaYY =
We
Yt
use the fact that X*
AX =
AY
is
a.
f(Y,X) scalar and so
it is
transpose.)
Let
H be the matrix
of / in a basis {d,
.
.
. ,
e„} of
that:
=
[u]lH\v]e tor
u,vGV;
al\
the transition matrix
is
B-Q*HQ
Note that
i.e.
its
In fact, a real matrix
be a Hermitian form on V.
P
equal to
a Hermitian form on C". (Remark.
is
Let
it is
/ is linear in the first variable.
f{X, Y)
12.17.
A*,
is
Y)^X<^A
For
A=
iff
Hermitian, since
be a Hermitian matrix.
defined
Hence
Hermitian
is
(ii)
Let u,v
is
GV u =
from
where Q = P)
is
the complex analog of
new basis {e,'} of V, then B = matrix of / in the new basis {e,'}. the
{ei}
to a
Theorem
12.2.
and suppose ajCi
+
0362
+
•
•
•
+
a„e„
and
v
—
b^ei
+
62^2
+
•
•
•
+
&«««
P*HP
CHAP.
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
12]
Then
=
f{u, v)
+
fia^Ci
•
•
+ a„e„,
•
6161
+
•
+
•
6„e„)
as required.
Since
(ii)
P
\
Thus by
(i),
f(u,v)
hence P*
HP
is
H =
1
=
[u]„ P[v],,
=
{u]\H
\
P
such that
First
and so
[v],
— [u\l,P^HP
[v]^
2i
i
4
i
[v]^,.
[m]^
=
[m]*, P',
[^= P ['^
But u and v are arbitrary elements of V;
—
2
\
3i,a
Hermitian matrix.
Find a nonsingular ma-
1 + Si j P*HP is diagonal.
-2i trix
+
1
-
=
the matrix of / in the basis {el}.
1
Let
"
the transition matrix from {ej to {e^}, then
is
P[u],.
12.18.
273
2
form the block matrix {H,
I):
1
+
1
-
1
2i
i
4
j
-2i
2
+
-
2
3i
7
3i
R2 -* (—1 + 'i)Ri + R2 ^^^ ^3 ^ 2i/2i + i?3 to (A, /) and then the corresponding "Hermitian column operations" (see Problem 12.42) C2 -> (—1 — t)Ci + C2 and C3 -» — 2iCi + Cg to A to obtain
Apply the row operations
and then
Next apply the row operation R^ tion C3 -* hiC^ + 2C3 to obtain
->
— SiBg + Zi^a
and the corresponding Hermitian column opera-
and then
Now
H
has been diagonalized. '1
P =
Set
-1 +
I
+
5
i
and then
P^HP
2
,0
Observe that the signature
9i^
-hi
1
S
of
H
is
jS
=
2
—1 =
1.
MISCELLANEOUS PROBLEMS 12.19.
that any bilinear form / on F is the sum of a symmetric bilinear form and a skew symmetric bilinear form. Set g(v,,v) = ^[f{u,v) + f(v,u)\ and h{u,v) = ^[f(u,v) — f(v,u)\. Then g is symmetric because
Show
g{u,v)
and
/i
is
=
^[f(u,v)
+
f(v,u)]
=
^[f{v,u)
f{u,v)]
=
g(v,u)
-/(«,-!))]
=
-h(v,u)
+
skew symmetric because h{u,v)
Furthermore, f
—
g
+ h.
=
^[f{u,v)
—
f{v,u)]
=
-^[/(v.m)
12.20.
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
274
Then there
Let / be an alternating bilinear form on V. in which / is represented by a matrix of the form
Prove Theorem a basis of
[CHAP. 12
V
12.3:
exists
1 I
-1 I
1-10
1
L_oj
T, To" 1
Moreover, the number of ( \'^ to i[rank (/)]).
= fcifcj f(u, m) = /
If
M^
a
uniquely determined by / (because
then the theorem is obviously true. Also, if dim y = and so / = 0. Accordingly we can assume that dim F > 1
0,
it is
equal
"/
1,
then
and f
fik^u, fcaw)
=
¥- 0.
G 7 such that /(mj, u^) # 0. In fact, multiplying that f{ui, u^ = 1 and so /(m2, "i) = —1- Now Ui and 0. M2 are linearly independent; because if, say, u^ = fewj, then /(mj, u^) = /(mi, ku^) = k f(ui, u^) = Let C7 be the subspace spanned by Ml and M2, i.e. U = L{ux,u^. Note: n 1 Since
/
#
Wj, Wg
there exist (nonzero)
0,
Ml by an appropriate
factor,
we can assume
/
(i)
the matrix representation of the restriction of / to
(ii)
if
uG
U, say u
=
aui
+
W V=U®W.
We claim that V = U+W. Let veV. Since
M
is
=
f{v,
(
_^
^
/(aMi
+
6m2, Ml)
= —6
f(u, M2)
=
f(aui
+
bu2, M2)
=
such that /(w,Mi) f(w, m)
clear that
=
=
for every
UnW = {0},
"'
and /(w,M2)
=
0.
Equivalently,
uGU} and
so
it
remains to show that
u^uy
-
a linear combination of Mi and hence
w =
and
/(«, Mi)m2 Mj,
uGU. We
show
-u that weW.
U)
v
By
(i)
and
(ii),
f(v,Ui);
/(W, Ml)
Similarly, /(m, M2)
=
f(v, Mg)
wG
TF and
and therefore
y=
so, ?7
by
(i),
=
f(v
- U,
Ml)
=
fiv, Ml)
-
/{m. Ml)
=
/(f
- M,
M2)
=
/(v, U2)
-
/(m, M2)
=
and so
f{w, M2)
Then
=
wGV is
is
Set
M =
f(u,U])
/(m. Ml)
= {wGV: It
the basis {mi, Mg}
then
6M2,
Let TF consist of those vectors
?7 in
v
=
= m
+ w where m G
f/
and
w&W.
This shows that
V^U+W\
© TF.
W
there exists is an alternating bilinear form on W. By induction, the restriction ot f to desired form. has the to restricted representing matrix / the in which .,m„ of IF a basis M3, desired form. Thus Mi,M2,M3,. .,M„ is a basis of V in which the matrix representing / has the
Now
.
.
.
W
CHAP.
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
12]
275
Supplementary Problems BILINEAR FORMS 12.21.
12.22.
Let M
=
(xi,
(i)
/(m, V)
(ii)
f(u, v)
(iii)
/(m,v)
x^ and
= = =
-
2a;i2/2
+
xi
v
=
(i/x,
y^. Determine which of the following are bilinear forms on R^: f(u, v)
(iv)
SajjJ/i
3/2
3a;2V2
(v)
f{u, v)
(vi)
/(w, v)
=
A of / in the basis B of / in the basis transition matrix P from
2x12/2
+
(iii)
Find the
{mJ to {vj and verify that
V
be the vector space of 2
A,B&V
where
X 2 matrices over
and "tr" denotes
Find the matrix of / in the basis
-''
f,gGB(V), then
if
f
+ g and
over K.
S-*-
=
{ve^V: f{u,v)
Show (iii)
=
for every
of V,
mSS},
S"""
and S^ are subspaces of V;
that: (i) S"*" {0}-L ={0}T
If / is a bilinear
hence dim
y-*"
=
Show
^ j,
(
that /
is
/O 1^
0^ °
o)'[q
1,
0\ °^
and
f{A,B)
=
a bilinear form on V.
belong to BiV), and so B(V)
=
,p{u) a(v)
is
a subspace
belongs to B{V).
we write
(ii)
= {vGV SjCSg
:
f{v,u)
= S2 c
implies
foreverywGS} and
st
form on V, then
=
f(x,u)
As.
>^
and u{x)=f(u,x).
u and u are each
(ii)
u
(iii)
rank
h*
u and m (/)
rank
(/)
= dim y -
dimy-*"
sj C s}
= dim y - dimy^
M
linear, it
i.e.
For each
;
uGV,
let
m)
-^
K
and
u iV
-*
K
= rank {u K
V
into V*;
tt).
that congruence of matrices is an equivalence relation, i.e. (i) A congruent to B, then B is congruent to A; (iii) if A is congruent to then A is congruent to C.
Show
is
B
congruent to A; (ii) if A and B is congruent to C,
Find the symmetric matrix belonging to each of the following quadratic polynomials: q(x, y, z)
(ii)
q(x, y, z)
(iii)
q(x, y, z)
(iv)
q(x, y, z)
= = — =
- Sxy + y^ - IGxz + — xz + y^ xy + y^ + 4xz + z^ xy + yz. 2x^
x^
be defined by
^ ^ u,u G V*;
are each linear mappings from
= rank (m K
u .V
is
(i)
and
Prove:
SYMMETRIC BILINEAR FORMS. QUADRATIC FORMS 12.29.
let
dim y^.
Let / be a bilinear form on V.
(1)
=
Af
P* A P.
=y.
Prove:
u(x)
f{u, v)
S
Let / be a bilinear form on V. For any subset
B=
Prove:
G K, also from V X V into K; k
kf, for
and a are linear functionals on V, then
if
(ii)
V
(i)
1\ /o ^^ 1^ oj' [1
/O (^
o)'\o
Let B{V) be the set of bilinear forms on
Let
B.
trace,
0\ °^
^
a;2?/2
= (1, 1), Mj = (1, 2)}. = (1, —1), V2 = (3, 1)}.
{u^
Let
-
4x2yi
{v^
of the vector space of functions
12.28.
-
Find the matrix
(i)
12.27.
SxiVi
Find the matrix
(ii)
12.26.
0.
(i)
'1
12.25.
j/iJ/2
1
(ii)
tr (At MB)
12.24.
+
X1X2
Let / be the bilinear form on R2 defined by /((«!, X2), (vi, 2/2))
12.23.
= = =
liyz
+
5«2
12.30.
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
276
For each of the following matrices A,
P
matrix
find a nonsingular
[CHAP. 12
such that P'^AP
^ = (3
(i)
A =
(«)
4)'
I
/
! 6
-2
^
-9
\
(12-5-1
A =
(iii)
.
diagonal:
1-2-3^
1
3\
2
is
2 _g -2-5
6 g
9
-3 -1
9
11
,
In each case find the rank and signature.
12.31.
Let q be the quadratic form associated with a symmetric bilinear form alternate polar
12.32.
12.34.
f(u,v)
/:
=
+ v) —
^[q(u
q(u
S{V)
is
a subspace of B(V);
V=
dim
if
(ii)
Let / be the symmetric bilinear form ax^ + bxy + cy^. Show that: is
nondegenerate
if
and only
(i)
/
(ii)
/ is positive definite if
if
and only
n,
Verify the following
/.
— v)].
Let S(V) be the set of symmetric bilinear forms on V. (i)
12.33.
form of
then
associated
Show
that:
=
dim S(y) with
the
— 4ac ¥= 0; and b^ — 4ac < a >
^n(n
+ 1).
quadratic
real
form
qix,y)
=
b^
if
Suppose A is a real symmetric positive P such that A = P«P.
definite matrix.
Consider a real quadratic polynomial
qix^,
Show
0.
that there exists a nonsingular matrix
n 12.35.
.
.
.
,
2=
=
a;„)
ijj
(i)
an
If
'^ 0,
the
ay
=
ttjj.
=
2/n
show that the substitution Xi
yields
where
Oij^i^j, 1
=
(ai22/2
Vi
equation
+
•
+ «-ln2/n).
•
=
«2
••'
2/2.
««
aji
x^)
q{xi
= a^yl +
where
-yVn),
q'iVz,
q'
is
also
quadratic
a
polynomial. (ii)
an =
If
ajg
but, say,
^ 0,
show that the substitution
xi-yi + 2/2. yields the equation
q{xi,...,x„)
=
This method of diagonalizing q
12.36.
Use steps of the type
X2
= yi- V2>
2 Mi^/jis
known
«3
"^^^^^
=
2/3.
^n
•
•
•
^^ 0.
^n
.
i.e.
=
Vn
reduces this case to case
as "completing the square".
problem to reduce each quadratic polynomial Find the rank and signature in each case.
in the preceding
12.29 to diagonal form.
(i).
in
Problem
HERMITIAN FORMS 12.37.
For any complex matrices A, B and any k e C, show (i)
12.38.
ATB = A+B,
(ii)
M^JcA,
(iii)
AB = AB,
For each of the following Hermitian matrices H,
that: (iv)
A*
=
A^.
find a nonsingular
matrix
diagonal:
Find the rank and signature in each
12.39.
Let
A
,
P
such that P*
HP
is
^^.,
.
case.
be any complex nonsingular matrix.
Show that
H = A*A
is
Hermitian and positive
definite.
12.40.
We say that B is Hermitian congruent to A B = Q*AQ. Show that Hermitian congruence
if is
there exists a nonsingular matrix an equivalence relation.
Q
such that
CHAP.
12.41.
BILINEAR, QUADRATIC
12]
AND HERMITIAN FORMS
277
Prove Theorem 12.7: Let / be a Hermitian form on V. Then there exists a basis {e^, .,e„} of V which / is represented by a diagonal matrix, i.e. /(ej, ej) = for t # j. Moreover, every diagonal representation of / has the same number P of positive entries and the same number N of negative entries. (Note that the second part of the theorem does not hold for complex symmetric bilinear forms, as seen by Problem 12.12(ii). However, the proof of Theorem 12.5 in Problem 12.13 does carry over to the Hermitian case.) .
.
in
MISCELLANEOUS PROBLEMS 12.42.
Consider the following elementary row operations:
Bi <^ Rj,
[ai]
[oj
-*
Ri
kRi, k ¥- 0,
The corresponding elementary column operations
Ci^Cj,
Ih] If
K
is
(ii)
Ci'^Ci,
Let
and
Show
that the elementary matrix corresponding to
and
W
W be vector spaces over K.
(i)
f(avi
(ii)
f(v,
A
mapping /
+
w)
bv2,
+ hWi)
owj
a,bQK,
V;
G
= =
af(vi,
w)
af(v, Wj)
[cj
is
the transpose of the elementary
the conjugate transpose of the ele-
VxW-^ K
:
is called
a bilinear form on
V
V, w^
+ bfiv^, w) + bf^v, w^
£ W. Prove
The set B(V, W) of bilinear forms on from into K.
= l,...,n}
= dim F
dim B(V, W)
{Remark. Observe that
•
{aj,
W
a subspace of the vector space of functions
is
...,»„}
/y
a basis of W*, then defined by fij(v,w) -
is
is
{/y 4>i(v)
:
i=l,...,m,
aAw).
Thus
dim W.
V = W,
if
and
B(V,W) where
a basis of
is
the following:
V
V* and
If {^1, ...,0^} is a basis of j
is
[6J
+ Ci
[aj.
VXW
(ii)
+ fij
if:
for every (i)
Ci-*kCj
[cs]
Show that the elementary matrix corresponding to matrix corresponding to [a^].
V
kRj
d^kCj + Ct
[bs]
Cj^feCj, fc^O,
[cj
mentary matrix corresponding to
12.43.
^
are, respectively,
Ci-^kCi,ky^O,
[bzl
jB;
the complex field C, then the corresponding Hermitian column operations are, respectively, [ci]
(i)
[og]
then
we
obtain the space B{V) investigated in this chapter.)
m times 12.44.
Let
y
be a vector space over K. A mapping f: form on V if / is linear in each variable,
m-linear)
f{...,au+bv, where ^ denotes the
=
...)
ith component,
VXVX---xV-*K i.e.
for
=
i
af(...,u, ...)
l,
+
is
called
a multilinear
(or:
...,m,
bf(...,v, ...)
and other components are held
fixed.
An
»i-linear
form /
is
said to be alternating if f{vi,
.
.
. ,
Vf^)
=
whenever
Vj
=
v^,
i^^ k
Prove: (i)
The
set
VXVX (ii)
The
B^(V)
of m-linear forms on into K.
XV
set A,„(y) of alternating m-linear
Remark
1.
If
Remark
2.
If.
m = 2, V=
then
we
V
is
a subspace of the vector space of functions from
forms on
V
is
a subspace of B^{V).
obtain the space B(F) investigated in this chapter.
K'", then the determinant function is a particular alternating m-linear
form on V.
\
Answers 12.21.
(i)
12.22.
(i)
12.23.
(ii)
12.29.
(i)
Yes
A =
(iii)
4
1
7
3
Yes
P =
7
(v)
1
-i^
'1
P -
2
0\
1
3
/^
=02
(iii)
-1 -1
26
3
13
1
P =
(i)
P =
1
P =
1
A
ptHP =
^1
1
P =
'o (iv)
1/
-2 +
i
1
»
3i
PtHP =
-3 + i
0.
°
S =
1.
-38 y
S =
S =
2.
1
-14 '1
i
0'
10
PtHP lO
U .0
1
1
(iii)
2'
7
1
(ii)
5
-2 -2
PtAP =
19 \o
12.38.
°
loo
2/ /l
i
1
S =
-2
PtAP
,
i
2
2
PtAP
3
P =
'0 (iii)
0;
2
(ii)
Yes
(vi)
(iii)
32
10 -i
-3
No
-4 20
(ii)
5,
1
No
B
-4 -8'
1-417
[CHAP. 12
Supplementary Problems
to (iv)
(ii)
1-8
(i)
No
(ii)
2
12.30.
AND HERMITIAN FORMS
BILINEAR, QUADRATIC
278
0-4,
S =
1.
2.
i
o'
1
4
0,
chapter 13
Inner Product Spaces INTRODUCTION The
V R
definition of a vector space
involves an arbitrary field K. In this chapter we either the real field or the complex field C. In the first case we call V a real vector space, and in the second case a complex vector space. restrict
K to be
Recall that the concepts of "length" and "orthogonality" did not appear in the investigation of arbitrary vector spaces (although they did appear in Chapter 1 on the spaces R" and C"). In this chapter we place an additional structure on a vector space V to obtain an
inner product space, and in this context these concepts are defined.
We
V shall denote a vector space of finite dimension unless otherwise In fact, many of the theorems in this chapter are not valid for spaces of infinite dimension. This is illustrated by some of the examples and problems. emphasize that
stated or implied.
INNER PRODUCT SPACES We begin with a definition. Definition:
Let
F
be a (real or complex) vector space over K. Suppose to each pair of there is assigned a scalar {u, v) G K. This mapping is called an inner product in V if it satisfies the following axioms:
vectors u,v
GV
[h]
+ hu2, V) = {u,v) = (v/u)
[la]
{u,
[/i]
{aui,
u) s= 0;
and
The vector space Observe that {u,u) sense.
We
is
V
a(ui, v)
{u, m)
+
=
h{U2, v)
if
and only
if
with an inner product
always real by
[72],
u=
0.
is called
an inner product space.
and so the inequality
relation in [h]
makes
also use the notation
=
||m||
'\/{u,
u)
This nonnegative real number ||m|| is called the norm or length of u. [/2] we obtain (Problem 13.1) the relation {u,
If the base field
K is
real,
avi
+
bv2)
=
d{u, Vi)
+
Also, using
[/i]
and
b{u, V2)
the conjugate signs appearing above and in
[/2]
may
be ignored.
In the language of the preceding chapter, an inner product is a positive definite symmetric bilinear form if the base field is real, and is a positive definite Hermitian form if the base field is complex.
A real inner product space is sometimes called a Euclidean space, and a complex inner product space is sometimes called a unitary space. 279
INNER PRODUCT SPACES
280 Example
13.1
[CHAP. 13
Consider the dot product in R":
:
=
M•v
+
ajfei
+
a262
•
•
+
•
Ok6„
= (6;). This is an inner product on R", and R" with this Although there are usually referred to as Euclidean n-spa-ce. many different ways to define an inner product on R" (see Problem 13.2), we shall assume this inner product on R" unless otherwise stated or implied. where u
—
Example
and v
(aj)
inner product
is
Consider the dot product on C":
13.2:
where u — (Zj) and v = (Wj). As in the real case, this is an inner product on C" and we shall assume this inner product on C" unless otherwise stated or implied.
Example
13.3:
mXn matrices
Let V denote the vector space oi product in V:
=
(A,B)
where
tr stands for trace, the
sum
over R.
of the diagonal elements.
=
{A,B}
Example
13.4:
usual,
Let
V
an inner
is
tr (B*A)
Analogously, if U denotes the vector space of following is an inner product in U:
As
The following
mXn
matrices over C, then the
tr(B*A) of the matrix B.
B* denotes the conjugate transpose
a
be the vector space of real continuous functions on the interval is an inner product on V:
—
t
—
b.
Then the following
=
{f,g)
f m)g(t)dt •'a
Analogously,
if
(real) interval
V a
-
denotes the vector space of complex continuous functions on the t - h, then the following is an inner product on U:
if,
=
9)
f
m^dt
•'a
Example
13.5:
Let
V
be the vector space of infinite sequences of real numbers
2
2,2 + a2
=
of
ai
<
+
(oj, a2,
.
.
.)
satisfying
M
t=i i.e.
the
sum
converges.
Addition and scalar multiplication are defined component-
wise: (tti.aa, ...)
+
(61,62.
fe(ai, a2,
An
inner product
is
defined in
V
.
•
.)
••)
=
=
(011
(.kai,
+
61,02+62,...)
)
ka2,
by
<(«!, tta, ...), (61, 62, ... ))
=
ai6i
+
a2&2
+
' '
The above sum converges absolutely for any pair of points in V (Problem 13.44); hence the inner product is well defined. This inner product space is called J2-space (or:
Remark
1:
If
||f||
=
Hilbert space).
1,
normalized. setting V =
Remark
2:
i.e.
We
{v,v) = 1, then v is called a unit vector or is said to be can be normalized by note that every nonzero vector u
if
&V
m/||%||.
The nonnegative real number d{u,v) = \\v-u\\ is called the distance between u and v; this function does satisfy the axioms of a metric space (see Problem 13.51).
,
CHAP.
INNER PRODUCT SPACES
13]
281
CAUCHY-SCHWARZ INEQUALITY The following formula,
called the
Cauchy-Schwarz
many branches
inequality, is used in
of mathematics.
Theorem
13.1:
For any vectors u,vGV,
(Cauchy-Schwarz):
—
\{u,v)\
Next we examine Example
||m||
this inequality in specific cases.
Consider any complex numbers
13.6:
II^II
aj,
.
.,a„, 6i,
.
.
.
.,
6„
G
C.
Then by the Cauchy-
Schwarz inequality,
(ai6i+---+a„6„P that
=
=
and v
(aj)
+
iWi\^
^
(m-v)2
is,
where u
Example
=^
+
'
KMh]^ +
+
K\^)
Ikll^ll-ulP
(6;).
—
Let / and g be any real continuous functions defined on the unit interval
13.7;
Then by the Cauchy-Schwarz
=
«/,ff»2
Here
V
is
(^f
t
—
1.
inequality,
f{t)g(t)dty
the inner product space of
^^
=s
P(t)dt
Example
^\\t)dt =
\W\\g\\^
13.4.
ORTHOGONALITY Let {u/v) {u,v}
V
= =
e7
GV
V 0.
be an inner product space. The vectors u,v are said to be orthogonal if The relation is clearly symmetric; that is, if u is orthogonal to v, then {v, u) = = and so v is orthogonal to u. We note that e V is orthogonal to every
for {0,v)
Conversely,
u
if
^
GV,
orthogonal to every v
is
=
{Qv,v)
0{v,v)
=
then (u.u)
=
and hence m
=
by
[1 3].
W
Now
suppose is any subset of V. The orthogonal complement of W, denoted by W-^ (read "W perp") consists of those vectors in V which are orthogonal to every w GW: W-'-
We
W^
show that for any a,h
GK
is
Theorem
€.
13i!:
W
{v
GV:
a subspace of V.
and any {au
Thus au + bv
=
(v,w)
=
for every
w GW}
GW^. Now
Clearly,
suppose u,v
G
W-^.
Then
w GW,
+ bv,
w)
=
and therefore
a{u,
W
+
w)
is
w)
h{v,
- a-0 + b'0 =
a subspace of V.
W he a subspace of v^w®w^.
Let
Then
V.
V
is
the direct
=W
sum
of
W
and
W^
I.e.
Now if PF is a subspace of V, then V ® W-^ by the above theorem; hence there a unique projection Ew: F -> 7 with image and kernel W-^ That is, if v and v = w + w', where w GW, w' G W^, then Ew is defined by Ew{v) = w. This mapping Evf is called the orthogonal projection of V onto W.
W
GV
.
Example
13.8:
W be the z axis in B^, W = {(0,0,c): Then W the xy plane, Let
is
W^ =
{{a, 6, 0)
:
i.e.
cSR} i.e.
a, 6
e
R}
W®W W
noted previously, R^ = orthogonal projection E of R^ onto by E{x, y, z) = (0, 0, z).
As
.
is
The given
is
INNER PRODUCT SPACES
282
Example
[CHAP. 13
Consider a homogeneous system of linear equations over R:
13.9:
ffliia;!
+
fflijaig
+
a2iXi
+
022*2
+
ttml*!
+
«m2«'2
+
•
•
•
+
»!„»;„
=
•
•
•
+
a2n«7i
=
AX
W W
= 0. Recall that the solution space may be viewed or in matrix notation as the set of all vectors as the kernel of the linear operator A. We may also vievs^ is the orthogonal V = (xi, ...,«„) vrhich are orthogonal to each row of A. Thus complement of the row space of A. Theorem 13.2 then gives another proof of the fundamental
Remark:
F
If
= n—
rank
(A).
between nonzero vectors
a real inner product space, then the angle
is
GV
u,v
W
dim
result:
W
is
defined
by (u, V)
cos 6
-
\\u\\m
By
^
1 and so the angle 6 always the Cauchy-Schwarz inequality, -1 =^ cos ^ Observe that u and v are orthogonal if and only if they are "perpendicu-
exists.
lar",
i.e.
e
-
7r/2.
ORTHONORMAL SETS A set {Ui} of vectors in V is said to be orthogonal if i.e. if
{Vd, Uj)
=
orthogonal and
if
=
Sij
elements are orthogonal, be orthonormal if it is
for i¥' j
r
=
{Ui, Uj)
its distinct
{Ui} is said to
In particular, the set j. each Ui has length 1, that is, if for i¥=
<
.
[1 for
I
—
.
J
orthonormal set can always be obtained from an orthogonal set of nonzero vectors by normalizing each vector.
An
Example
Consider the usual basis of Euclidean 3-space R^:
13.10:
{ei
=
=
(1, 0, 0), 62
(0. 1. 0). «3
=
(0. 0, 1)}
It is clear that
-
(ei, ej)
(62, 62)
=
(63, 63)
=
and
1
That is, {ei, e^, 63} is an orthonormal basis of of R" or of C" is orthonormal for every n.
Example
Let
13.11:
V
R^.
e^)
=
More
for
i¥=j
generally, the usual basis
be the vector space of real continuous functions on the interval
with inner product defined by cal
(fij,
{f,g)=
I
The following
f(t) g{t) dt.
-v-t-v is
a classi-
example of an orthogonal subset of V: {1,
cos
t,
cos 2«,
.
.
.,
sin
t,
sin 2t,
.
.
.}
The above orthogonal set plays a fundamental role in the theory of Fourier series.
The following properties of an orthonormal
Lemma
13^:
An
orthonormal set
the vector is
^^
(mi,
_
^
set will be used in the next section.
...,Ur}\s linearly independent and, for any
_
^^^ ^^^^^
orthogonal to each of the Uu
_
^^^ ^2)^2
-
-
{v, Ur)Ur
v&V,
CHAP.
INNER PRODUCT SPACES
13]
283
GRAM-SCHMIDT ORTHOGONALIZATION PROCESS Orthonormal bases play an important role in inner product spaces. The next theorem shows that such a basis always exists; its proof uses the celebrated Gram-Schmidt orthogonalization process.
Theorem
{vi, .,Vn} be an arbitrary basis of an inner product space V. Then there exists an orthonormal basis {ui, tt„} of V such that the transition .,n, matrix from {Vi} to {%} is triangular; that is, for i = l,
Let
13.4:
.
.
.
.
.
,
.
Ui
We
Proof.
set Ui
=
By Lemma
13.3,
W2 (and hence
ggf-
=
W3
va
—
OiiVi
is
orthonormal.
then {ui}
Vi/||t;i||;
W2 =
=
V2
—
+
012^2
and
{V2,Ui)Ui
U2) is
+
M2
=
•
+
•
We
next set
W2/||w2||
—
and
{vs,U2)U2
Uz
=
=
Vi+i
—
{Vi+i,ui)ui
— •• —
(Vi+1, 1^)111
and
is
We
orthonormal.
next
Ws/\\ws\\
Again, by Lemma 13.3, Wa (and hence Ua) is orthogonal to Ui and normal. In general, after obtaining {ui, ...,2*1} we set Wi+i
auVi
orthogonal to Ui; then {ui,v^}
{V3,Ui)Ui
.
U2;
then
=
th+i
{ui,
iia,
Ua} is ortho-
Wi+i/\\Wi+i\\
As above, (tti, .,Mi+i} (Note that Wi+i¥'0 because Vi+i€ L{vi, .,ik).) .,Vi) — L{ui, ztn} which is inorthonormal. By induction we obtain an orthonormal set {ui, dependent and hence a basis of V. The specific construction guarantees that the transition matrix is indeed triangular. .
.
.
.
.
is also
Example
.
13.12:
.
.
. ,
Consider the following basis of Euclidean space R^: {^1
=
(1, 1, 1),
V2
=
(0, 1, 1),
V3
=
(0, 0, 1)}
We
use the Gram-Schmidt orthogonalization process to transform {vj into an orthonormal basis {Ui). First we normalize v^, i.e. we set
JVi_
Next we
and then we normalize Wg,
i.e.
~
"^
W3
we
=
^ /J
1_
(0,1,1)
-j=(^j=.^,jl'j we
_1_
=
(-f,|,|
set
W2 Finally
(1,1,1)
set
= .2-<^.%)% =
U,2
^
IKII
_ ~
/
1_ J_ 2 Ve'Ve'Ve
[
set '"a
-
<'>'3.Ml>Ml
-
{'«3,U2)U2
= (0,0,1)-— (^-^,—,-^j--^^--|.-^,-|j = (o,-|,| and then we normalize Wg:
_
Wa
I
11 \/2
The required orthonormal basis of R^
y/2
is
LINEAR FUNCTIONALS AND ADJOINT OPERATORS Let V be an inner product space. Each u&V determines a mapping u:V-^K u{v)
—
{v,u)
defined
by
INNER PRODUCT SPACES
284
Now That sion
G K and any Vi, V2 G V, u(avi + bv2) — {avi + bv2,u) =
[CHAP. 13
for any a,b
a{vi,u}
+
The converse is, M is a linear functional on V. and is an important theorem. Namely,
Theorem
13.5:
b{V2,u)
is also
=
au{vi)
+
bu{v2)
true for spaces of finite dimen-
Let ^ be a linear functional on a finite dimensional inner product space V. such that (j>{v) = (v, u) for every Then there exists a unique vector
uGV
v&V.
We 13.45), is
remark that the above theorem is not valid for spaces of infinite dimension (Problem although some general results in this direction are known. (One such famous result
the Riesz representation theorem.)
We
use the above theorem to prove
Theorem
Let
13.6:
T
be a linear operator on a finite dimensional inner product space V. exists a unique linear operator T* on F such that
Then there
{T{u),v)
=
{u,T*{v))
Moreover, if A is the matrix of T relative to an m, v G V. orthonormal basis {Ci] of V, then the conjugate transpose A* of A is the matrix of T* in the basis {ei}.
for every
We emphasize that no T and T*
if
the basis
is
such simple relationship exists between the matrices representing not orthonormal. Thus we see one useful property of orthonormal
bases.
Definition:
linear operator T on an inner product space V is said to have an adjoint operator T* on V if {T{u),v) = {u,T*{v)) for every u,vGV.
A
Thus Theorem This theorem Example
is
not valid
13.13:
T has an
13.6 states that every operator
Let
if
T be
V has
infinite
adjoint
if
V
has
finite
dimension (Problem 13.78).
the linear operator on C^ defined by T(x, y, z)
=
{2x
+ iv,y- Mz, x + {l-i)y + Sz) Note (Problem
And a similar formula for the adjoint T* of T. matrix of T in the usual basis of C^ is
We
/2
=
[r]
\1
-5i
1-t
3
Theorem
13.6, the
[T*]
=
=
+ z,-ix + y + {l + i)z,5iy + Bz)
Accordingly, T*(x,y,z)
The following theorem summarizes some 13.7:
Let S and
T
(2x
of the properties of the adjoint.
be linear operators on (i)
(ii)
V and
+ T)* = S* + T* (kT)* = kT* (S
7.3)
that the
i
1
Recall that the usual basis is orthonormal. Thus by in this basis is the conjugate transpose of [T]:
Theorem
dimension.
let
kGK.
Then:
(iii)
(ST)*
= T*S*
(iv)
(T*)*
= T
matrix of T*
CHAP.
INNER PRODUCT SPACES
13]
ANALOGY BETWEEN
AND
A{V)
C,
285
SPECIAL OPERATORS
Let A(V) denote the algebra of all linear operators on a finite dimensional inner product space V. The adjoint mapping T^^ T* on A{V) is quite analogous to the conjugation mapping zi-* z on the complex field C. To illustrate this analogy we identify in the following table certain classes of operators T G A(V) whose behavior under the adjoint map imitates the behavior under conjugation of familiar classes of complex numbers.
Class of
Behavior under
complex numbers
conjugation
Unit
circle
(|z|
=
-
z
1)
Behavior under
Class of operators in A(V)
the adjoint
Orthogonal operators (real case) Unitary operators (complex case)
1/z
r*
map
= r-i
Self -ad joint operators
Real axis
=
z
Imaginary axis
z
Positive half axis
Also called: z
Skew-adjoint operators Also called: skew-symmetric (real case) skew-Hermitian (complex case)
— —z
— WW, w
z
(0,«)
The analogy between these
T*
symmetric (real case) Hermitian (complex case)
r*
Positive definite operators
=/=
classes of operators
with
T and complex numbers
= T
=:
-T
T = S*S S nonsingular
z is reflected in
the following theorem.
Theorem
We now belonging to
Proof of
(i)
If
T* = T-\ then
(ii)
If
T* =
(iii)
If
T* = —T, then X
(iv)
If
r = S*S
We
(i):
{v,v) ¥- 0;
Proof of
(ii):
{v, v) ¥- 0;
Proof of
=
{v,v} ¥- 0;
We
1.
is
pure imaginary.
with S nonsingular, then X
=
{XV, v)
=
We
^
{XV, XV)
=1
XX{v,v)
=
=
{XV, V)
hence X
=
X{v, v)
X and so X
is real
and
positive.
or
{v,v):
A.
=
{v,T*T{v))
=
{v,I(v))
=
{v,v)
1.
X{v, v):
{v,T*{v))
=
{v,T{v))
=
{v,Xv)
=
X{v,v)
is real.
= =
=
= =
X{v,v)
{T{v),v)
= —X
|A.|
{T{v),v)
show that
-
{T(v),T{v))
and so
show that
hence A
(iii):
X{v,v)
But
=
T, then k is real.
show that
hence XX
X{v,v)
But
|A.|
V.
prove the above theorem. In each case let v be a nonzero eigenvector of X, that is, T{v) - Xv with v y^ 0; hence {v, v) is positive.
XX{v,v)
But
T on
Let A be an eigenvalue of a linear operator
13.8:
=
—X{v,v):
{v,T*{v))
=
—A, and so A
{v,-T{v)) is
=
{v,-Xv)
pure imaginary.
=
-X{v,v}
T
INNER PRODUCT SPACES
286
Proof of
(iv):
We
positive.
Note
X{v,v)
=
\{v,v)
But
{v,v)
We
and
S{v)¥'0 because S
that
first
show that
=
[CHAP. IS
nonsingular; hence {S{v),S{v))
is
is
{S{v),S(v)y.
{\v,v)
=
=
{T{v),v)
=
{S*S{v),v}
{S{v),S{v))
{S{v),S{v)) are positive; hence A is positive. all the above operators T commute with their adjoint, that operators are called normal operators.
remark that
TT* = T*T. Such
is,
ORTHOGONAL AND UNITARY OPERATORS U
Let above,
then
Z7 is
plex.
he
a.
As
linear operator on a finite dimensional inner product space V.
if
U* =
UU* = U*U =
or equivalently
C/-1
defined
/
said to be orthogonal or unitary according as the underlying field is real or comalternate characterizations of these operators.
The next theorem gives
Theorem
13.9:
Example
The following conditions on an operator
U are equivalent:
UU* ^U*U = I.
= U-\
(i)
[/*
(ii)
U
that is, preserves inner products, i.e. for every v,w {U{v),U{w)) = {v,w)
(iii)
U
preserves lengths,
13.14:
i.e.
for every v
&V,
\[U{v)\\
T RS ^ R3 be the linear operator which rotates each vector about the z axis by a fixed Let
GV, =
||t;||.
^
T(v)
:
angle
e:
T(x, y, z)
-
{x
—
cose
y
X sine
sin e,
+
y cos
9,
z)
Observe that lengths (distances from the oriThus T is an gin) are preserved under T. orthogonal operator.
Example
13.15:
TiV^V
be the linear operator depreserves inner products and fined by r(cii,a2. ••) lengths. However, T is not surjective since, for example, (1, 0, 0, ... ) does not belong Thus we see that Theorem 13.9 is not to the image of T; hence T is not invertible. valid for spaces of infinite dimension.
Let
V
be the ^space of Example
=
(0, ai,
13.5.
Let
•)•
Clearly,
T
one inner product space into another is a bijective mapping which preserves the three basic operations of an inner product space: vector addition, Thus the above mappings (orthogonal and scalar multiplication, and inner products. unitary) may also be characterized as the isomorphisms of V into itself. Note that such a
An isomorphism from
mapping
U
also preserves distances, since
\\U{v)-Uiw)\\
and so
U
is
also called
=
\\U{v-w)\\
=
\\v-w\\
an isometry.
ORTHOGONAL AND UNITARY MATRICES Let C7 be a linear operator on an inner product space V. following result when the base field K is complex.
Theorem
13.10A:
13.6
we
A
obtain the
with complex entries represents a unitary operator (relative to an orthonormal basis) if and only if A* = A'^.
A
matrix
K
is real then A* the other hand, if the base field ing corresponding theorem for real inner product spaces.
On
By Theorem
= A';
hence
we have
U
the follow-
CHAP.
INNER PRODUCT SPACES
13]
Theorem
A
13.10B:
A
matrix
(relative to
with real entries represents an orthogonal operator an orthonormal basis) if and only if A* = A~^.
The above theorems motivate the following
A
Definition:
is
A
Definition:
= A"S
A* — A~^,
matrix A for which an orthogonal matrix.
real
Observe that a unitary matrix with real entries
Example
A =
Suppose
13.16:
)
6i
is
*i\
=
AA* = A*A =
or equivalently
I,
is
orthogonal.
+ +
/ l^il'
\aj6i
62/
A A* = A*A = /,
or equivalently
A A* =
Then
a unitary matrix.
is
"A/^i h^\a2
/"'
AA* =
^
'
(
U
definitions.
complex matrix A for which A* called a unitary matrix.
called
287
+
I
ajftaA
l^zP
ttifti
a262
|6iP+|62l^/
and hence
^
/l
o'
\0
1
^
^
Thus |ai|2
+
laaP
=
1,
|6i|2
+
=
Ibgp
and
1
Accordingly, the rows of A form an orthonormal the columns of A to form an orthonormal set.
The
(i)
(ii)
(iii)
Example
A
The following conditions for a matrix
13.11:
A
A*A =
I
forces
are equivalent:
unitary (orthogonal).
is
The rows of A form an orthonormal set. The columns of A form an orthonormal set.
The matrix
13.17:
Similarly,
set.
example holds true in general; namely,
result in the above
Theorem
+ a^z =
aj)^
basis of R3
A
T
representing the rotation
in
Example
13.14 relative to the usual
is
(cos
— sin B
e
cos
sin e
9
1,
As
A
expected, the rows and the columns of an orthogonal matrix.
A
each form an orthonormal
set;
that
is,
is
CHANGE OF ORTHONORMAL BASIS In view of the special role of orthonormal bases in the theory of inner product spaces, are naturally interested in the properties of the transition matrix from one such basis to another. The following theorem applies.
we
Theorem
13.12:
Let
{ei,
...,««} be an orthonormal basis of an inner product space Y. transition matrix from {ei) into another orthonormal basis is
Then the
unitary (orthogonal). Conversely, if P = (Oij) is a unitary (orthogonal) matrix, then the following is an orthonormal basis: {el
A
=
aiiCi
+ 02162 +
•
•
+ a„ie„
:
i
=
1,
.
.
.
,
n}
and B representing the same linear operator T are similar, i.e. the (nonsingular) transition matrix. On the other hand, if V is an inner product space, we are usually interested in the case when P is unitary (or orthogonal) as suggested by the above theorem. (Recall that P is unitary if P* = P""S and P is orthogonal if P* = p-\) This leads to the following definition. Recall that matrices
B — P~^AP
where
P
is
INNER PRODUCT SPACES
288
Definition:
[CHAP. 13
Complex matrices A and B are unitarily equivalent if there is a unitary matrix P for which B - P*AP. Analogously, real matrices A and B are orthogonally equivalent if there is an orthogonal matrix P for which B = P*AP.
Observe that orthogonally equivalent matrices are necessarily congruent
(see
page 262).
POSITIVE OPERATORS Let
P
semi-definite) if
„ „ P = S*S
said to be positive definite if S characterizations of these operators.
and
is
Theorem
P
be a linear operator on an inner product space V.
13.13 A:
(ii)
(iii)
rt
,
nonsingular.
The next theorems give
P
alternate
are equivalent:
P = T^ for some self -adjoint operator T. P = S*S for some operator -S. P is self -adjoint and (P(m), u)^0 for every uGV.
The corresponding theorem for 13.13B:
is also
said to be positive (or:
some operator &
The following conditions on an operator (i)
Theorem
„
for
is
positive definite operators
The following conditions on an operator (i)
P-T^
(ii)
P - S*S
(iii)
P
P
is
are equivalent:
for some nonsingular self-ad joint operator T. for some nonsingular operator S.
is self -adjoint
and
{P{u),u)
>
w^
for every
in V.
DIAGONALIZATION AND CANONICAL FORMS IN EUCLIDEAN SPACES V over K. Let r be a linear operator on a finite dimensional inner product space
RepT, of eigenvalues and eigenvectors the upon resenting r by a diagonal matrix depends Now (Theorem T 9.6). of A(t) polynomial characteristic and hence upon the roots of the field C, but may not have any ^^t) always factors into linear polynomials over the complex for Euclidean spaces (where situation the linear polynomials over the real field R. Thus
them
K
= C); hence we treat inherently different than that for unitary spaces (where the next spaces unitary and below, spaces separately. We investigate Euclidean
X = R)
is
m
section.
Theorem
13.14:
dimensional Let T be a symmetric (self-ad joint) operator on a real finite basis of V orthonormal an exists there Then V. inner product space by a represented can be T is, that T; of consisting of eigenvectors diagonal matrix relative to an orthonormal basis.
We
give the corresponding statement for matrices.
Alternate
Form
of
Theorem
13.14:
Let A be a real symmetric matrix. Then there exists an orthogonal matrix P such that B = P-^AP = P*AP is
diagonal.
normalized orthogonal eigencan choose the columns of the above matrix P to be eigenvalues. corresponding vectors of A; then the diagonal entries of B are the
We
CHAP.
INNER PRODUCT SPACES
13]
Example
13.18:
2
/
-2\
Let
A =
The
characteristic polynomial A(t) of
We
.
j
(
=
A(t)
find
289
an orthogonal matrix
A
such that
P^AP
is
diagonal.
is
t-2
=
\tI-A\
P
2
(«-6)(t-l)
f-
2
5
The eigenvalues of A are 6 and 1. Substitute « = 6 into the matrix tl obtain the corresponding homogeneous system of linear equations
+
4x
A
nonzero solution
is
=
2y
2x
0,
+
Next substitute (1, —2). homogeneous system
-X + 2y =
-
2x
0,
is (2, 1). As expected by Problem Normalize Vi and V2 to obtain the orthonormal basis
Finally let
P
=
=
-2/V5), M2
=
1
matrix
into the
13.31, v^
/ I/V5
2/V5\
expected, the diagonal entries of
P*AP
Then
/6 (
VO
I/V5/
V-2/X/5
~A
(2/V5, 1/V5)}
P-iAP = PtAP =
and
tl
and V2 are orthogonal.
be the matrix whose columns are u^ and U2 respectively.
P = As
(1/-V/5,
t
=
4y
A nonzero solution
{«!
to
=
2/
=
v^
to find the corresponding
-A
1
are the eigenvalues corresponding to the
columns of P.
We observe that the matrix B - P~^AP = P*AP is also congruent to A. Now if q is a real quadratic form represented by the matrix A, then the above method can be used to diagonalize q under an orthogonal change of coordinates. This is illustrated in the next example. Example
13.19:
Find an orthogonal transformation form q{x, y) = 2x^ — 4xy + 5y^.
of coordinates virhich diagonalizes the quadratic
The symmetric matrix representing Q
/
A =
is
/ I/V5
(Here 6 and
1
.
In the preceding
PtAP =
/fi „
"
(
Vo
1
Thus the required orthogonal transforma-
is
x\ )
this
)
'
1/V5/
are the eigenvalues of A.)
tion of coordinates
Under
2/\/5\ for which
V-2/V/5
-2\
^
example we obtained the orthogonal matrix
P =
2
I
=
X
/x'\
= P[
that
>,')
change of coordinates q
is
a;7\/5
+
^
is,
22/'V5
^
transformed into the diagonal form
q(x',y')
=
6x'^
+
y'^
Note that the diagonal entries of q are the eigenvalues of A.
An orthogonal operator T need not be symmetric, and so it may not be represented by a diagonal matrix relative to an orthonormal basis. However, such an operator T does have a simple canonical representation, as described in the next theorem. Theorem
13.15:
Let
T"
there
form:
be an orthogonal operator on a real inner product space V. Then an orthonormal basis with respect to which T has the following
is
290
INNER PRODUCT SPACES
[CHAP. 13
COS Or
— sin
Or
sin Or
cos
Or
The reader may recognize the above 2 by 2 diagonal blocks as representing rotations in the corresponding two-dimensional subspaces.
DIAGONALIZATION AND CANONICAL FORMS IN UNITARY SPACES We now present the fundamental diagonalization theorem for complex inner spaces,
i.e.
for unitary spaces.
Recall that
an operator T
is
said to be normal
mutes with its adjoint, i.e. if TT* = T* T. Analogously, a complex matrix normal if it commutes with its conjugate transpose, i.e. if AA* = A*A. Example
Let
13.20:
A =
(
.
„,„.).
A*A Thus
A
The following theorem
Theorem
13.16:
is
\1
com-
said to be
S
+ 2iJ\l -'
V^
S-2iJ\i
S-2iJ 3
M
+
3
=
2iJ
+
3 3i
-
3i
14
^"^* '
^
\3 +
3i
14
a normal matrix.
applies.
T be a normal operator on a complex finite dimensional inner product space V. Then there exists an orthonormal basis of V consisting of eigenvectors of T; that is, T can be represented by a diagonal matrix Let
relative to
We
= (^
is
Then 2
\i
A
product
if it
an orthonormal
basis.
give the corresponding statement for matrices.
Alternate
Form
of
Theorem
tary matrix
Let A be a normal matrix. such that B — P~^AP = P*AP
13.16:
P
Then there is
exists a uni-
diagonal.
The next theorem shows that even non-normal operators on unitary spaces have a relatively simple form.
Theorem
13.17:
Let T be an arbitrary operator on a complex finite dimensional inner product space V. Then T can be represented by a triangular matrix relative to an orthonormal basis of V.
CHAP.
INNER PRODUCT SPACES
13]
Form
Alternate
Theorem
of
291
Let A be an arbitrary complex matrix. Then there matrix P such that B = p-^AP = P*AP is triangular.
13.17:
exists a unitary
SPECTRAL THEOREM The Spectral Theorem
Theorem
is
a reformulation of the diagonalization Theorems 13.14 and 13.16.
Theorem): Let T be a normal (symmetric) operator on a complex (real) finite dimensional inner product space V. Then there exist orthogonal projections Ei, ..,Er on V and scalars Xi, ...,\r such that
13.18 (Spectral
.
(i)
T =
+ X2E2 +
XiEi
(ii)
(iii)
EiEj
=
for i¥=
+
•
•
•
E1 + E2+ • -^Er ^
XrEr
I
j.
The next example shows the relationship between a diagonal matrix representation and the corresponding orthogonal projections. '2
Example
13.21
:
A =
Consider a diagonal matrix, say
|
|
The reader can verify that the Ei are (i)
A =
2£7i
+
3^2
+
5^3.
(ii)
projections,
^1
fif
i.e.
+ ^2 + ^3 =
Let
.
=
I,
Bj,
(iii)
and that
E^^ =
Solved Problems
INNER PRODUCTS 13.1.
Verify the relation Using
[12], [Ii]
avi
[I2],
+
bv2)
avi
Verify that the following {u, V)
=
Method
We
XiVi,
-
+ hvi) =
and then {u,
13.2.
{u,
xiy2
-
X2yi
we
d{u, Vi)
+
b{u, V2).
find
=
{avi
=
d <«!,«>
+
6^2, u)
+
=
a{Vi,u)
6 {^2, u)
is
an inner product in
+
3x2y2,
where u =
=
+
b{V2, u)
a(u,Vi)
+
6
R^:
{xi, X2),
v
=
(2/1, 1/2).
1.
verify the three axioms of an inner product.
au
+ bw =
a{xi, Wj)
+
H^i,
Z2)
Letting
=
(ffl«i
+
w= bzi,
(«i, Zz),
aX2
we
+ 623)
find
for i¥= j
INNER PRODUCT SPACES
292
[CHAP. 13
Thus
+ bw,
{au
and so axiom
[/2]
=
—
Method
We
{(axi
=
{ax^
= =
- a;iJ/2 - a!22/i + 3a;2?/2) + a(u,v) + b{w,v) -
j/i^i
= x^-
+
bziivi
—
2/2*^1
so
+
2x1X2 if
13.4.
[/i]
V2))
-
621)2/2
=
0,
Xj a!2
-
+
(ax2
A
Since
holds.
is
—
0,
+ m
i.e.
we can write
is,
=
a'i2/2
a;|
+
=
0.
{u, v) in
~
1;
=
C2
^
Ci
+ C2, we
3222/2)
3a;22/2
=
last
+
2x1
axiom
~
° is satisfied.
[I^]
-l\/2/i^ o m 3/\2/2
Thus we need only show that A is positive ^1 + ^2 *iid then the corresponding ele-
iZg ~*
A
into diagonal
(3, 4)
G
with respect
R'^
(
<(3,4), (3,4)>
=
9
+
16
=
(ii)
||i;||2
=
{v,v}
=
((3,4), (3,4)>
=
9
-
12
-
hence
25;
12
+
48
=
=
l|vll
33;
hence
\\v\\
=
=
12v/||12i;||
=
+
12
+
(—1)2
8 (6, 8,
We
-3).
(6/\/l09, 8/a/109. -3/\/i09
have
<12v, 12t))
)
be the vector space of polynomials with inner product given by
{f,9)
=
(ii)
,/>
=
f{t)
= t+2
and
r
(t
+ 2W-2t-S)dt =
r
(t
+ 2)(t + 2)dt =
•'n
6.
R^:
(2/V6, l/Ve, -1/a/6)
12i; = tear" of fractions: First multiply v by 12 to "clear" 62 + 82 + (—3)2 = 109. Then the ;he required unit vector is
(i)
=
a/sI.
= {hh-i).
ul\\u\\
Let
A
5.
Note {u, u) is the sum of the squares of the entries of u; that is, (u, u) — 2^ Hence divide m by ||m|| = V
Cf{t)g{t)dt.
Thus
to:
Normalize each of the following vectors in Euclidean space
{n)v
.
j
'
the inner product in Problem 13.2.
(ii)
=
(2,l,-l),
form
^
{v,v)
V
{u,v}
matrix notation:
\-l,
transform
=
Let
+
holds.
||i;||2
(ii)
+ 623)2/2
3(aa;2
(x^-Xi)^
Hence the
(a;i,a;2)i
holds.
[/2]
(i)
(i)
+
*22/i
=
2x|
1
=
utAv
symmetric,
the usual inner product,
=
+
6x2)2/1
6(212/1-213/2-222/1
—
«i2/i
2a;ia;2
Applying the elementary row operation
Find the norm of
(i)M
13.5.
(2/1,
+
+ %2*2 =
=
3x1
x-^
argue via matrices. That
mentary column operation is positive definite and [Ig]
(i)
622),
(axi
2.
definite.
13.3.
-
Finally,
{u,v)
and
+
ax^
Also,
2/1X2
and only
if
+ bzi,
a(a;i2/i
is satisfied.
(M,M>
Also,
=
is satisfied.
[/i]
{v,u)
and axiom
V)
19/3
g{t)
t*/4
and
= t^-2t-Z.
- 7t2/2 - 6tT =
||/i|
=
y/Wj)
Find
(i)
-37/4
=
VW3
(f,g)
and
{f,g) (ii)
=
||/||.
CHAP.
13.6.
INNER PRODUCT SPACES
13]
Prove Theorem 13.1 (Cauchy-Schwarz): — If V = 0, the inequality reduces to zz
=
is
any real value:
Set
t
any complex number
(for
|z|2
=
— = =
we
root of both sides,
13.7.
(u, u)
||m|P
=
\\v\\
[N,\. [N,]:
and
[/g],
(v,v)
||m|P-
find
=
\\v\\
from which
,
.
\\kv\\^
if
I
||m||2 H-yH^.
{kv, kv)
+
and only
=
v
if
following axioms:
0.
||i;||
we
inequality,
obtain
+ v,u + v) — {u,u) + {u,v) + (u,v) + {v,v) ||«|p + 2|M||HI + IMP = (IMI + IMIP
= ^
0)112
{u
Taking the square root of both sides yields [A^s]Remark: [Ng] is frequently called the triangle inequality because if we view m + -u as the side of the triangle formed with u and v (as illustrated
Taking the square
= if and only if {v,v) = Q, \\v\\ = ^/l^v/v) — 0. Furthermore, = Q. Thus [Wj] is valid. = kk(v,v) = |fc|2 ||-y||2. Taking the square root of both sides gives [ATg].
Using the Cauchy-Schwarz |m
^
satisfies the
hence if v
0;
=
](m,i>>P
\\v\\.
and only
this holds if
We
—
Using where t
obtain the required inequality.
]H| = |fc||H|. \\u + v\\^\\u\\ +
By and
^ 0;
0.
= (u — {u,v)tv, u — (u,v)tv) — {u, v)t(u, v) — (u, v)t(v, u) + (u, V)(U, V)t^V, V) - 2t\{u,v)\^ + |(M,i;)|2t2||v||2
Prove that the norm in an inner product space [2Vi]:
Now suppose v # ||m —
valid.
is
we expand
{u,v},
ti;||2
\(u ' v)\'^
0^
to find
l/\\v\\^
—
||m
\\u\\ \\v\\.
an d hen ce
and {v,u)
z)
^
\{u,v}\
293
,
on the right), then [Ng\ states that the length of one
sum
side of a triangle is less than or equal to the
of the lengths
of the other two sides.
ORTHOGONALITY 13.8.
Show
that
to V.
Find a unit vector orthogonal
If
{u, V)
if
u
=
is
orthogonal to
then {ku, v)
=
=
{w,v{)
then every scalar multiple of u
v,
to Vi
= = + y + 2z
k{u, v)
=
X
fc •
=
and V2 =
(1, 1, 2)
as required.
0,
=
and
is
(0, 1, 3)
also orthogonal
in R^.
w = {x, y, z). We want = y + 3z
Let
(w, V2)
Thus we obtain the homogeneous system X Set
z
=
1
to find
y
= —S
and
x
unit vector w' orthogonal to v^ and
13.9.
+ =
+
y
W
=
y
0,
w=
then
1;
^2-
=
2z
+
Sz
= Normalize
(1,-3,1).
w/||w||
=
(l/yfTl,
— 3/-\/ll,
W
be the subspace of R^ spanned by u = (1, 2, 3, -1, Find a basis of the orthogonal complement W^ of W.
Let
We
seek
all
vectors
w=
(x, y, z, s, t)
{w,u) (w,v)
= =
the
=
y, s
and
t.
s-1, t = to find solution W3 = (—17, 0, 5, 0, 1). The
Set y
0,
and v
2)
=
(2, 4, 7, 2,
-1).
2x
+ +
we
2y 4:y
+ +
Sz 7z
+
s
2s
+ -
2t t
= =
find the equivalent system
- s + 2t = z + 4s ~ 5t = Set y = —1, s = 0, t = to obtain the solution Wi = (2, —1, 0, 0, 0). the solution Wj = (13,0,-4,1,0). Set y-Q, 8 = 0, t = l to obtain X
The free variables are
to obtain the required
such that
X
Eliminating x from the second equation,
w
l/-v/li).
+
2y
+
3z
set {wj, W2. Ws) is a basis of
W
.
INNER PRODUCT SPACES
294
13.10.
t;2
=
\M? = =
Thus Ml
=
(l/V2,i/y/2,Q).
=
Vj
—
=
Then
=
=
Vi
—
(1,
i,
and
0)
=
(1
1
2wi
\
i
=
r
+
Suppose aiMi
•
•
•
+
a^u^
=
=
=
Accordingly, {mi, It
.
.
'
.
,
is,
w
is
||vi||
= V2
^_^,^,l-^
-
/l
+
2i
2-t
2
-2i
...,«,} is linearly independent and, for
{t*i,
{V,
is
vig'^'vn
ii2»iii
—
W
18
U^Ut
—
—
•
{V, Ur)Ur
im.
=
2,
Taking the inner product of both
0.
+ a^u^, Ml) ai(Mi,Mi> + a2 + •• + 0^ = + + Oi 1 + a2
= = =
.
.
. ,
sides with respect to Mi,
(aiMi
+
•
•
•
•
•
•
ftr
O-i
'
r,
+
•
•
ai
+ a^M„ Mj) + + Oi
•
•
•
•
+
•
ar(Mr. «{>
=
"i
Uj) is linearly independent.
w
is
orthogonal to each of the
Taking the inner product of
Mj.
w
with
ttj,
is
= =
{V, Ml)
orthogonal to Mi.
w
{V, Ui)Ui
Mi>
= =
(0, Mj)
<«;,Ml>
Thus
=
i
remains to show that
respect to
That
<0,
Similarly, for
0.
—
V
orthogonal to each of the
so
find
i,
Prove Lemma 13.3: An orthonormal set any v e V, the vector yj
We
= l/V2-2i/V2 = (l-2i)/V2
0)>
Thus the required orthonormal basis of
\/l8.
and
2
Vj.
compute
2w2
^2 yT^
13.12.
by
+ 2i, 2 — 2 — 2i). We have = (1 + 2t)(l - 2i) + (2 - 1)(2 + 1) + (2 - 2i)(2 + 2i) =
(2m;i,2wi)
/
or Oi
+ 0-0 =
i'(-i)
l-«), (l/\/2,t/\/2,
<(1,2,
or, equivalently,
||2m)i||2
||2wi|l
first
(v^, Wi>Mi,
First normalize
process.
(i,2.1_^)--^^__,0j
=
.,
Next normalize W2
and
= I'l +
(.'"k'^i)
vi/\\vi\\
To form W2
is
of C^ spanned
(l,2,l-i).
Apply the Gram-Schmidt orthogonalization
13.11.
W
Find an orthonormal basis of the subspace
[CHAP. 13
=
<'U,Mi)
-
orthogonal to Mj for
-
(V, Ml)
•
1
-
Similarly, for <'U,Ml)
=
-
<'y,Mi>
-
r,
1
i
•
•
= •
2,
-
M2KM2, Mi>
.
-
•
.
.
•
•
•
- •• -
•
(V,M^>(M„Mi>
=
,r,
(U,Mi)(Mi,Mj)
-
•
•
•
-
(•«,
M^XMr, Mj)
=
as claimed.
Let TF be a subspace of an inner product space V. Show that there which is part of an orthonormal basis of V. basis of
is
an orthonormal
W
apply choose a basis {v^, ...yV^ioiW and extend it to a basis {vj, ...,v^} of V. We then basis orthonormal obtain an to to {vi,...,v^} process Gram-Schmidt orthogonalization + au^i. Thus 'u^,...,Ur^W and there, w, M; = ai^v^ + M„} of y where, for i = 1,
We the {Ml
fore {mi,
.
.
.
. ,
mJ
is
.
.
an orthonormal basis of W.
•
CHAP.
13.13.
INNER PRODUCT SPACES
13]
Prove Theorem
By Problem
{mi,
=
OjMj
V
.
where
•
•
V
conditions,
=W+W^
.
.
is
.,
.
.,«„} is orthonormal,
.
ajV-i
wGWnW-^,
the other hand, if
The two
Since {mj,
.,m„} of V.
+ + ft„M„ y = W + W"*-.
Accordingly,
On
.
V=W@W-^. u^} of W which
Let W' be a subspace of V; then
13.2:
13.12 there exists an orthonormal basis {ui,
normal basis
295
+
+
•
=
then
and PTn
G W,
a^u^
weW.
Let
Now
W^
®W-^-^
V
V
W
This yields dim TF
{ei,
(i)
for any
(ii)
(iii)
(iv)
(i)
has
= =
...,€„} be
dim
y -
But
dimiy-'--'-.
+ «««« ^
•
V
+
•
•
•
•
•
W = W^-^
when V
Accordingly,
WcW-^-^.
W'^
and, also,
=
at
Substituting
/f 1
«!
+ fcj 62 +
i =
2,
.
=
"^
TF"^
y -
dim
dim W"^
W = W-^-^,
for
Prove:
•
•
+ fen^n-
•
=
for
2
fej
= =
Then by
u = (ii),
+
+
•
+ +
fc2
+
fcj •
(m, e,>ei {u, v)
h
+
kiie^, ej> fci
and
2 =
j
+
(fciCi
+
•
•
•
•
•
(Cj, e^)
=
+•••+(«,
=
(m,
=
•
1
2
1=1
e„>e„
e^Xv, e,)
•
•
•
+
•
fe„
representing
ej,
fc„(e„, ej}
=
•
h fc„e„,
+
•
+
H
fci •
+
+
1
fcie,
for
+
•
•
•
fci
O'^i
and
ej)
•
+
+
•
•
fc„
fc„
=
•
+ fc„e„, we
•
i.i
=
•
•
2=
= i
+
fci(ei, 6;)
6jeA /
i=l
) /
fcjBj
l
2
ajee,
•
fcjej
m=
in the equation
i 7^ j,
ajej,
i=l
1
fcj •
•
•
•
Taking the inner product of u with
fci
2 (\i=l
(ej, e^)
as required.
.,n,
.
We have
(i),
V =
+•••+(«, en>e„; = aibi + chbl + +
•
=
By
=W ®W^.
{u, ei)ei 4- (u, 62)62
{u, gj)
But
{0}.
we remark
W®
by the above; hence
WcW'^-'-
•
Suppose
=
dimension;
+ OnBu, biCi + + 6„e„) Onhn; for any u,v GV, (u, v) = {u, ei){v, ei) + + (u, e„); if T-.V^V is linear, then (r(ej), ei) is the i/-entry of the matrix A T in the given basis {d}. (ttifii
v e.V,
^
hence WnW''-
finite
that
V —
13.2,
dim
and
an orthonormal basis of V.
uGV, u =
Similarly, for
(iii)
"
0;
wSTV^-^.
hence
By Theorem
dim W*"
= =
(ii)
vGW^;
for every
dimension.
finite
has
WcW^^-^, and
that
'
If
Hence
.
Let
=
Then {w,v)
suppose
dim
13.15.
Show
Let W' be a subspace of W. has finite dimension.
+
TF"*-.
give the desired result
{0},
Note that we have proved the theorem only for the case that that the theorem also holds for spaces of arbitrary dimension.
13.14.
i
w=
This yields
0.
=
W'"'-
(ir+i«*r +
part of an ortho-
...,«„£
u^+i,
fej
obtain the desired result.
ai6;
hence, as required.
j;
=
aibi
i^
(u, CzKv, 62)
= +
+
ajbz
+
{v, ei>ei
+
+
•
•
•
•
+
•
•
+
{u, e„){v, e„>
a„6„
(i),
e„>e„
296
INNER PRODUCT SPACES
[CHAP. 13
By(i),
(iv)
r(e,)
=
ei
+
{T(e,), e^je^
+
T{e„)
=
iT(e„), e,)ei
+
{T(e„), 6^)6^
+
A representing T in the basis {e;} hence the v-entry of A is (T(ej), ej.
The matrix efficients;
•
•
•
+
{T{e,), e„>e„
+
(T{e^), e„)e„
the transpose of the above matrix of
is
ADJOINTS 13.16.
T
Let
be the linear operator on C^ defined by
=
T{x, y, z)
(2x
+ (1 - i)y,
+ 2i)x - 4iz, 2ix + (4 - Zi)y - Zz)
(3
¥mdiT*{x,y,z). First find the matrix
A
representing
T
in the usual basis of C^ (see
3
Form
the conjugate transpose
A*
+
-4i
2i
4
2i
\
7.3):
X-i
2
/
A =
Problem
-
-3
3i
of A: 2
/
=
A*
1
+
-
3
-2i
2i
+
4
i
3t
-3
4i
\
Thus T*(x, y,
13.17.
Prove Theorem
13.5:
product space V. every v G.V. Let
{ei,
.
.
.
,
e„} be
=
z)
{2x
(4
+ 3i)z,
=
V
0(ei)ei
+
0(62)62
defined by u(v)
(ej.M)
=
agree on each basis vector,
^ej,
=
0(e7)ei
u =
+
•
•
•
(v, u),
+
•
•
•
+
0(e„)e„
for every v ElV.
+ 0(ije„> =
13.6:
Let
T be
Then for
i
=
1,
.
.
.
,
to,
.
Now suppose m' is another vector in V for which
Prove Theorem
- 3z)
an orthonormal basis of V. Set
w(ei)
13.18.
Aiy
Let ^ be a linear functional on a finite dimensional inner exists a unique % G F such that ^(v) = {v, u) for
Let M be the linear functional on
m and
+ i)x +
(1
Then there
u =
Since
+ (3 - 2i)y - 2iz,
a linear operator on a
for every
and
so (u
vGV.
Then
~u',u- u') =
0.
claimed.
dimensional inner product V such that {T{u), v) = {u, T* (v)), for every u,v GV. Moreover, if A is the matrix representing T in an orthonormal basis {ei} of V, then the conjugate transpose A* of A is the matrix representing T* in {Ci}. space V.
We
Then there
finite
exists a unique linear operator
T* on
the mapping T*. Let v be an arbitrary but fixed element of V. The map a linear functional on V. Hence by Theorem 13.5 there exists a unique element v'&V such that {T{u),v) = {u,v'} for every u&V. We define T*V-^V by T*(v) = v'. Then (T(u), v) = {u, T* (v)) for every u.v&V.
u
first define
h» (T(u), V)
is
CHAP.
INNER PRODUCT SPACES
13]
We
next show that T* T*(av^
(u,
But
+
For any
linear.
is
hv^))
= =
{T(u), av^
=
=
Thus B =
13.19.
=
bT*{v2))
=
{T{ei), e,>
=
o^i
= kT*
(/cT)*
+
For any u,v
(ii)
G
The uniqueness For any u,v
G
+
(S(m)
= =
T{u), V)
of the adjoint implies
(S
=
=
(kT{u), V)
+
{kT)*
=
u,vGV,
that:
(i)
/*
=
{S(T{u)),v}
=
{T*(u),v)
=
7;
u,vGV, u,vGV, = (TT-i)* =
(ii)
0*
(1)
For every
{I(u),v}
(ii)
For every
<0(M),'y)
(iii)
7
Let
r
that
=
{v,
=
= =
0;
uGW^.
If
=
{u,v} {0,v)
wGW,
then T{w)
w
GW
(i)
(r(M),
(ii)
F T
is
i;)
=
= =
for every u,v
G
T.
Set
V
r =
o.
-
T{u).
+
(S*
+
(u, T*{v))
T*){v)}
T*.
=
(u,
and
Then
^
=
kT*iv))
=
W
(kT*)(v))
{u,
(m, (r*S*)(i;)>
{u, T(v))
-
=
then (T-i)*
= T*-\
I.
=
hence 0*
{u,0(v));
0.
= T*-K
W he
and so
W
=
{T(v),u)
hence I*
(u,0)
let
=
T*(S*(v)))
{u,
is invertible,
{u,I{v)};
is
a T-invariant subspace of V.
(w, T*(u))
=
(Tiw), u)
-
0.
Thus
T*(ii)
Show
GW^
invariant under T*.
that each of the following conditions implies
GY;
a complex space, and (r(M),«)
is self -adjoint
T
if
Show
(r(tt),%)
=
Give an example of an operator every m e F but T ^ 0. (i)
S*{v)>
(u,
(u,
T*S*.
=
Hence
.
Let r be a linear operator on V. 0:
=
kT*.
=
(r-i)*r*; hence {T'^)*
orthogonal to every
r=
=
T*{u))
(iii)
be a linear operator on V, and is invariant under T*.
it is
+
k{u, T*{v))
{T(u),S*{v))
W
Let since
/*
S*
=
{T(u), v)
T*(v))
V,
The uniqueness of the adjoint implies (T*)*
Show
=
=
The uniqueness of the adjoint implies (ST)* For any
S*{v)
T)*
k(T(u), v)
of the adjoint implies
{(ST)(u),v)
(iv)
+ +
{S{u), V) {u,
V,
{(kT){u), V)
(iii)
=
T){u), V)
The uniqueness
(iii)
=T
(T*)*
(iv)
For any u,v G V, {{S
13.22.
+
Prove Theorem 13.7: Let S and T be linear operators on a finite dimensional inner product space V and let k G K. Then: (iii) (ST)* = T*S* (i) (5 + 7)* = 5*4-2^*
(i)
13.21.
b{T{u), v^)
aT*(Vi)
= aT*(vi) + bT*(v2). Thus T* is linear. B = (6y) representing T and T* respectively =
r*(e^)>
{ei,
(u,
bv^)
and
and by
e;)
=
K,
A*, as claimed.
(ii)
13.20.
+
+
d{T(u), v^)
b{u, T*(V2)}
(ay)
G
and any a,b
V,
=
bvz)
+
T* {av^
hence
matrices A in the basis {ej are given by Oy = (r(ej), 13.15(iv), the
6y
+
a(u, r*(vi))
uGV;
this is true for every
By Problem
G
u, V;
297
{T{u), T{u)}
=
=
for every
for every
T on
«G
F.
a real space
and hence
T(u)
m G F;
=
V 0,
for which (r(w), m>
for every
uGV.
=
for
Accordingly,
1
1
INNER PRODUCT SPACES
298
By
(ii)
+ w),
(T(v
hypothesis,
and (T{w), w)
=
+ w) =
v
w
arbitrary in
is
-i(T(v),w)
v.wGiV. Expanding and
for any
-
+
Dividing through by
=
{iT(w),v)
and adding
i
=
(T(w), V)
(1)
=
{T(v), iw)
i(T{,v),
w)
=
i{T(w),v),
+
to (1),
we
=
i{T(w), V)
obtain
a
=
{T(w), v)
for
Q
T -Q.
By
=
{T{v),v}
Q
Substituting iw for w, and using
(1).
and {T{iw),v)
w)
-i{T(v), w)
the result holds for the complex case; hence + w),v-\-w) = 0, we again obtain (1). a real space, we have (T(w),v) = (w,T(v)) = {T(v),w). {T(v), w) = Q for any v,w&V. By (i), T - Q.
(iii)
setting
0,
{T(v),
Note
[CHAP. 13
we need Since T
(ii),
Expanding {T(v
v.wGV. By
any
(1)
only consider the real case. is
self-adjoint
and since
Substituting this into
For our example, consider the linear operator T on R2 defined by = for every m G V, but T ¥^ 0.
T(x, y)
=
(1),
(y,
it is
we
obtain
~x).
Then
{T{u), u)
ORTHOGONAL AND UNITARY OPERATORS AND MATRICES 13,23.
Prove Theorem
=
C/-i;
every v
&V.
(i)C/*
Suppose
The following conditions on an operator {Uiv),Uiw)} = {v,w}, for every v.wGV; (iii)
13.9:
(ii)
=
{Uiv),U(w))
Thus
(i)
implies
(ii).
Now
if
(ii)
implies
Suppose
(iii)
(iii).
=
Then for every
Hence
13.24.
and
implies
—
U
w.
is
=
{U{v), V(v))
U*U =
so
U{W) — W; that
nonsingular,
Now
let
V
G W^. Then
Let
A
to
W
.
for
w)
=
Therefore
W
(U(y),
Thus U{v) belongs
{v,vo)
\\v\\
(i).
I.
is,
=
{V, v)
t7*t7
-/
Thus
17*
=
t7-i
let
wG
for any
{I{v), V)
is self -adjoint
C7 be a unitary (orthogonal) operator on V, and under U. Show that W^ is also invariant under U.
Since
13.25.
(iii)
Let
U{w')
-
{v,I(w))
V,
ve.V. But
for every
V*U-I-Q
S
i;
=
(U*U{v), V)
((U*V - I)(v),v) =
=
= V(^> =
V
remains to show that
It
holds.
lem 13.22 we have
{v,U*U{w))
holds, then
(ii)
||C7(i;)||
Hence
are equivalent: = ||t;||, for
||C7(v)||
&V,
Then, for every v,w
holds.
(i)
U
W
(Prove!); then
W
be a subspace invariant
there exists
w'
G
-
(U(v), V(w'))
(ftjj),
then
A* =
(6y)
{V,
=
w')
Show
that:
(i)
the i;-entry of
as required. Also,
<(«(!,
A*A =
= =
2 k=
AA* =
(cy)
where
K— .
.
.
,
ai„), {aji, ...,
UjJ)
=
(Ri, Rj)
where
(dy)
n
Thus
a^.
n
fc—
=
=
where 6y
n
n &ifc«icj l
((«ij,
=
2=
fc
.
.
. ,
a^j), (an,
«kj»fci
=
o-iflii
+ »2j^ -!-•+ a„M^
l
.
.
.
such that
invariant under U.
is
be a matrix with rows Ri and columns d. (ii) the y-entry of A* A is {d, d).
A =
TV
w &W,
any
is {Ri, Rj); If
by Prob-
as claimed.
,
a„i))
=
{Cj, C-)
A A*
CHAP.
13.26.
INNER PRODUCT SPACES
13]
A
(i)
unitary (orthogonal),
columns of Let
AA* =
fij
(cy)
A
Also, if
(Cj, Cj)
of
A
form an orthonormal
(ii).
by the preceding problem. A* A = (dy) where dy = 8y. That is, (i) is equivalent to (iii).
Remark: Since is
set.
and Cj denote the rows and columns of A, respectively. By the preceding problem, where Cy = (Ri,«j>. Thus AA* = I if and only if
equivalent to
13.27.
The following conditions for a matrix A are equivalent: (iii) The (ii) The rows of A form an orthonormal set.
Prove Theorem 13.11: is
299
and
(ii)
A
are equivalent,
(iii)
-
(Cj,
Thus A* A
Q.
unitary (orthogonal)
is
=
and only
if
/
if
and only
if
the transpose
unitary (orthogonal).
A
Find an orthogonal matrix
One such
solution
—
Wg
is
=
W2>
(Ml,
=
W2
First find a nonzero vector
=
whose
+
2y/B
row
which
{x, y, z)
x/3
+
(1/3, 2/3, 2/3).
orthogonal to Mj,
=
2z/3
=
Ui
is
is
+
x
or
W2
Normalize
—1).
(0, 1,
first
2y
2z
for which
=
second
the
obtain
to
i.e.
+
row
of
A,
i.e.
M2=(0,l/vi,-l/\/2).
Next
a nonzero vector Wg
find
= — Set i.e.
W3)
(U2,
Wz)
« = — 1 and find the M3 = (4/Vl8, -l/VlS,
= —
xlZ
—
{x, y, z)
+
2ylZ
+ —
Vlyf2
solution
-l/\/l8).
which 2«/3 zlyl2
Wj = (4, —1, Thus
A =
13.28.
A
emphasize that the above matrix
is
=0
or
=
or
x
2/3
2/3
1/a/2
-1/\/2
-\-
+
2z
2/—
z
2y
= =
«£, i.e.
for which
Q
-l/3\/2y
-l/3\/2
\4/3\/2
We
orthogonal to both Mj and
Normalize W3 and obtain the third row of A,
—1).
1/3
/
is
not unique.
Let {ei, e„} be an orthonormal basis of an inner product space V. Then the transition matrix from {Ci} into another orthonormal basis is unitary (orthogonal). Conversely, if P = (ao) is a unitary (orthogonal) matrix, then the following is an orthonormal basis:
Prove Theorem 13.12:
.
=
{e'i
Suppose
{/j} is
13.15
=
and since
B=
,
+ 02162 +
^ilBl
{/j} is
Sy
Let
.
=
+
61262
+ a„ie„
•
+•••+
=
i
:
1,
.
1=1,
fein^n.
.
,
w}
...,n
(fufj)
=
biibfi
+
6426^
+
•
(Then B*
(1).
+
•
•
is
(ei
{«,'}
is
orthonormal.
By Problem
= auo^ + a^^j +
e'j)
•
•
•
+
(2)
the transition matrix from {ej} to
By Problem 13.25, BB* = (cy) where Cy = 6ii67i + ^12^ + and therefore BB* = /. Accordingly B, and hence B*, are unitary. remains to prove that
U)
bi„i;~
{/J.)
It
.
orthonormal,
be the matrix of coefiicients in
(6y)
•
•
another orthonormal basis and suppose /i
By Problem
auBi
.
•
•
•
+ K^n- By
(2).
"n
Suppose
A
Since
A
is is
orthogonal.
orthogonal,
A A* 1
Therefore
lAI
=
1 or
— 1.
Show that det(A) = = /. Using \A\ = \A*\, =
1/|
=
lAAt|
=
1 or
\A\\At\
=
^H
13.15,
a„ia;;j
=
(Cj, Cj)
where Cj denotes the ith column of the unitary (orthogonal) matrix P = (ay). By Theorem the columns of P are orthonormal; hence (e[, ej) =
13.29.
=
—1.
|Ap
13.11, basis.
INNER PRODUCT SPACES
300
Show
13.30.
/cos
~ sin 9\
6
sm
^
for
„
„
.
y
cos 9
6
Suppose
some
/a
A =
dJ
+
)
.
Since
=
number A
d2
from det(A)
=
c2
x^
=
We
1.
= 0, the first equation gives 6^ = = —b = ^:l, and the second equation =
^
I)
1
+
=
d^
"
— — r/2, and
e
+
(£2
=
d
fourth equation gives tion gives
b
=:
bU^ +
or
1
or a — —d. —a^ — c2 =
— —c and
a
If 1
a2d2
=
which
=
or
a2
—d,
ad
0,
=
b i
c
If a 7^ 0, the third equation can be solved to give second equation,
and therefore a
is
rows form an orthonormal
and therefore
1
The first alternate has the required form with form with e = ttI2.
62d2/a2
1
-
be
=
of the
form
or
hence
a.
=
and
a¥'0.
±1. Then the fourth equation d = 0. Thus
I
the second alternate has the required
c =^ —bd/a.
(62
set;
1
consider separately the cases
yields
(-:
its
ac+hd =
1,
a c
=
6.
orthogonal,
is
+
62
last equation follows If
det{A)
,
real
b\
{
„2
gives
,
for which
j
^c
The
A
that every 2 by 2 orthogonal matrix
[CHAP. 13
+
aP')d2
=
Substituting this into the
or
a^
a2
=
d2
then the third equation yields c = b and so the Thus a = d. But then the third equa-
impossible.
is
so
Since a^ + c^ = 1, there is a real required form in this case also.
number
9
— c^
/a
=
A
such that
a
=
cos e, c
—
sin «
and hence
A
has the
SYMMETRIC OPERATORS AND CANONICAL FORMS IN EUCLIDEAN SPACES 13.31.
r
Let
r
is
(iii)
be a symmetric operator. Show that: (i) the characteristic polynomial A{t) of a product of linear polynomials (over R); (ii) T has a nonzero eigenvector; eigenvectors of T belonging to distinct eigenvalues are orthogonal.
be a matrix representing T relative to an orthonormal basis of V; then A — A*. Let be the characteristic polynomial of A. Viewing A as a complex self -adjoint operator, A has only real eigenvalues by Theorem 13.8. Thus
A
Let
(i)
A(t)
A(«)
where the
are
Xj
T has
(ii)
By
(iii)
Suppose T(v)
(i),
all real.
13.32.
n\
=
=
is,
(v,
{\v,w}
w)
=
= nw =
a product of linear polynomials over B.
Hence T has a nonzero eigenvector.
where \
{T{v),w)
is
=
¥=
/i.
We
{v,T{w))
show that
=
{v,nw)
X('U,
=
w)
=
ii{v,
w):
ti{v,w)
as claimed.
Let T be a symmetric operator on a real inner product space an orthonormal basis of V consisting of eigenvectors of T; T can be represented by a diagonal matrix relative to an orthonormal basis.
Then there
that
In other words, A(t)
and T(w)
\v
hence
Prove Theorem V.
(i-Xi)(t-X2)---(«-X„)
at least one (real) eigenvalue.
\{v,w}
But \¥'
=
18.14:
exists
is by induction on the dimension of V. If dimV = 1, the theorem trivially holds. suppose dim F = n > 1. By the preceding problem, there exists a nonzero eigenvector v^ of T. Let be the space spanned by v-^, and let Mj be a unit vector in W, e.g. let Mj = i'i/||vi||.
The proof
Now
W
CHAP.
INNER PRODUCT SPACES
13]
Since v^
W^
an eigenvector of T, the subspace TF of
is
Theorem
an orthonormal basis
exists
=
t
A = (
=
.
.
{u^,
.
Thus the restriction T of T to W^ is Hence dim TF"*- = m - 1 since dim W^ ,
.
.
W^
m„} of
Find a
.
2
)
^
.
The characteristic polynomial
=
A(t)
|f/-A|
A
A(t) of
t-
=
-2
1
t-
-2
2x-2y nonzero solution
is
Next substitute
t
—
Vi
(1, 1).
— —l
Finally
As
13.34.
P
let
v^
is
=
(1,
—1).
A —
2t
-
-
=
2j/
P^AP
is
set
diagonal.
=
{2
-
=
3
(t
- 3)(t +
1)
=
t
-2x +
into the matrix tl
3
—A
-
to obtain the
=
2j/
=
to obtain the corresponding
-2a;
0,
—A
(ll\2, l/v2).
homogeneous system
=
2^/
Normalize V2 to find the unit solution u^
=
(1/a/2, —1/^/2).
be the matrix whose columns are Mj and Mj respectively; then
expected, the diagonal entries of
Let
for which
But
1
0,
into the matrix tl
-2x nonzero solution
P
of T.
an orthonormal
is
By
induction, there
T and hence
Normalize v^ to find the unit solution Mi
of linear equations
A
By
1.
is
and thus the eigenvalues of A are 3 and —1. Substitute corresponding homogeneous system of linear equations
A
=
consisting of eigenvectors of
orthogonal matrix
(real)
13.21,
a symmetric operator.
.
,
By Problem
invariant under T.
is
m because Mj G PF-*- Accordingly {%, %,...,«„} 2, eigenvectors of T. Thus the theorem is proved.
for
and consists of
13^3. Let
T.
=W ®W^.
V
13.2,
=
T*
invariant under
is
y
301
\1
2
1
\l
1
2/
Find a
.
P*AP
are the eigenvalues of A.
orthogonal matrix
(real)
P for which P'^AP
is
diagonal.
First find the characteristic polynomial A{t) of A:
-
t
=
A(t)
\tI-A\
=
-1 -1
-1
2
-1 -1
t
-
2
t -
-1
=
(t-l)2(t-4)
2
A are 1 (with multiplicity two) and 4 (with multiplicity matrix tl — A to obtain the corresponding homogeneous system —X — — 2 = 0, —X — y — z = 0, —X — y — z =
Thus the eigenvalues of t
=
into the
1
one).
Substitute
J/
That (1,
is,
—1,
0).
X
+
We
y
+
z
=
0.
The system has two independent
seek a second solution V2
a+ b + For example, Vj
=
(1, 1,
Ml
Now
substitute
—2).
=
c
= (a, 6, c) which = and also
a
Next we normalize f j and V2 (l/\/2,
-I/V2,
0),
Ma
=
One such solution is Vi = orthogonal to v^; that is, such that
solutions.
is also
—
6
=
to obtain the unit orthogonal solutions
(l/\/6, l/\/6,
-2/V^)
= 4 into the matrix tl — A to find the corresponding homogeneous 2x — y — z = 0, -X + 2y - z = 0, -x - y + 2z = t
Find a nonzero solution such as t^s M3 = (l/v3> l/v3, l/vS). Finally, if P
= is
(1, 1, 1),
and normalize v^
to
system
obtain the unit solution
the matrix whose columns are the Wj respectively.
INNER PRODUCT SPACES
302
P =
I
i/\/2
i/Ve
i/VsX
-I/V2
l/Ve
l/Vs
-2/\/6
13.35.
[CHAP. 13
PtAP
and
I/V3/
Find an orthogonal change of coordinates which diagonalizes the real quadratic form q(x, y) = 2x^ + 2xy + 2y^. First find the symmetric matrix '2
A =
representing q and then
and
=
A{t)
A
are 1 and
find the
=
1*7- A|
=
x'^
+
its characteristic
polynomial
A(t):
-1
2
-1
hence the diagonal form of q
3;
q(x', y')
We
t-
1-
2
1
The eigenvalues of
A
-
t
{t-l){t-3) 2
is
Zy'^
corresponding transformation of coordinates by obtaining a corresponding orthonormal A.
set of eigenvectors of
Set
f
=
1
matrix
into the
tl
—A
A
=
nonzero solution is v^ homogeneous system
(1,-1).
A Vi
nonzero solution
=
V2
is
P
The transition matrix
0,
i
—
y
=
3
y
—
into the matrix tl
—A
to find the corresponding
—X + y =
0,
As expected by Problem
(1, 1).
and V2 are orthogonal. Normalize
13.31, v^
P';
that
Then there
=
(l/\/2,
I/V2)} follow:
+ y')/V2 (-x' + y')/^/2 (x'
= P
l/\/2
P
are Mj and
1*2-
We
can also express
x'
and
y' in
terms of x and
=
{x-y)/V2,
y'
=
(«
by
+ j/)/\/2
Let T be an orthogonal operator on a real inner product space an orthonormal basis with respect to which T has the following
13.15: is
j/
is,
x'
Prove Theorem
-l/\/2), M2
and
Note that the columns of
P^i =
(l/\/2,
l/\/2
-l/\/2
using
=
and the required transformation of coordinates
l/^/2
P =
V.
set
—X —
and V2 to obtain the orthonormal basis {ui
13.36.
—
=
y
Now X
homogeneous system
to obtain the corresponding
—X —
form:
1
I
-_j I
-1 -1
-1 ^
1
I
1
Oi
— sm
di
sin di
cos
01
cos
I
I
I
cos
9r
sin 9r
— sm
9r
cos 6r
CHAP.
INNER PRODUCT SPACES
13]
Let S = operator on V.
303
r + r-i = T + T*. Then S* = (T + T*)* = T* + T = S. Thus S By Theorem 13.14, there exists an orthonormal basis of V consisting
Is a symmetric of eigenvectors
denote the distinct eigenvalues of S, then V can be decomposed into the direct Vm where the Vj consists of the eigenvectors of S belonging to Xj. We claim that each Vj is invariant under T. For suppose v e Vj; then S{v) — \v and If Xi,
of S.
.
.
.
,
sum y = Vi
Xjn
©
Va
•
•
•
©
= (T+T-^)T(v) = T(T+T-^){v) =
S(T(v))
TS{v)
=
TiXfV)
=
\iT(v)
That is, T{v) & Fj. Hence Vi is invariant under T. Since the V; are orthogonal to each other, we can restrict our investigation to the way that T acts on each individual 'V^.
On
a given V;, (T
+
=
T-^)v
=
S(v)
Multiplying by T,
\v.
(T2-\T + I){v) =
We
consider the cases
(T
leads to
Xj
± I){v) =
= ±2
or
and
=
T(v)
±2 separately. Thus T restricted
X; ¥=
±v.
=
-
±
±2, then (T I)Hv) If Xj to this Fj is either I or -/.
which
If Xj ¥= ±2, then T has no eigenvectors in Vj since by Theorem 13.8 the only eigenvalues of T be are 1 or —1. Accordingly, for v ¥= the vectors v and T{v) are linearly independent. Let is invariant under T, since the subspace spanned by v and T(v). Then
W
W
=
T(T(v))
By Theorem 13.2, Vj = Thus we can decompose
W © W-^
=
T^v)
\^T(v)
-
v
Furthermore, by Problem 13.24 w'' is also invariant under T. V^ into the direct sum of two dimensional subspaces Wj where the Wj are orthogonal to each other and each Wj is invariant under T. Thus we can now restrict our investigation to the way T acts on each individual Wj.
t^
—
.
T^ — XjT + / = 0, the characteristic polynomial A(t) + 1. Thus the determinant of T is 1, the constant term A representing T acting on Wj relative to any orthonormal — sin e^ 'cos e
T
of
\t
in A(t).
matrix
^
Wj is A(t) = By Problem 13.30, the Wj must be of the form
acting on
Since
basis of
cos e y
sin e
The union of the basis of the Wj gives an orthonormal basis of Vj, and the union of the basis of the Vj gives an orthonormal basis of V in which the matrix representing T is of the desired form.
NORMAL OPERATORS AND CANONICAL FORMS 13.37.
Determine which matrix
<»
-
AA*
¥=
^"'
A*A,
2
1^1
Since
T
2-iJ\l
BB* = B*B,
(i)
Tiv)
(ii)
T—
\I
(iv)
if is
T{v)
=
vectors of
+
the matrix
and only
A =
A
is
if
*
-- g
[2-2i
\2-2i is
^
) ,
(ii)
B =
OG o
I
,
2
+
•
= (-'
not normal.
iJ
B
/
6
6
normal.
Prove: T*{v)
=
0.
normal.
If T{v) = \v, vector of T*. If
2
be a normal operator.
=
(iii)
the matrix
+ iJ\-i 2-iJ
\-i
Let
(i)
= (;:)(-;:) = (-:;)
Since
13.38.
normal:
is
IN UNITARY SPACES
then
T*{v)
= Xv;
hence any eigenvector of
T
is
also
Xiv and T{w) = X2W where A,i ^ A2, then {v,w) = 0; that T belonging to distinct eigenvalues are orthonormal.
an eigenis,
eigen-
.
INNER PRODUCT SPACES
304
We
(i)
show that
=
{T(v), T{v))
{T*(v), T*iv)):
=
(T(v), T(v))
Hence by
show that T —
We
(ii)
=
T(v)
[/g],
Thus T ~\I
(r-X/)*(i>)
We
(iv)
show that
0.
13.39.
Xj ¥= X2;
Prove Theorem
{T-{v), T'-(v))
0.
its adjoint:
=
hence
{v,
Now T -
0.
=
hence
0;
Xl
normal by
is
(ii);
therefore,
by
(i),
-Xv.
T*(v)
X2
=
(Xiv.w)
=
w)
Let
13.16:
=
Xl){v)
[T* -Xl)(v)
is,
=
w)
Xi{v,
-
(T
That
Xi{v,w}
But
=
TTHv)) =
{V,
- \i)* = (r-x7)(r*-x/) = rr* - XT* - xr + xx/ _ T*T -XT - XT* + XXI = {T* -Xl){T - XI) = (T - XI)*{T - XI)
\i)(,T
then
XV,
=
=
normal.
is
=
T(v)
If
(iii)
T*(v)
if
commutes with
\I
{T -
T*T{v))
(V,
and only
if
[CHAP. 13
{T(v),w)
=
-
{v,T*(w))
{v.X^w)
=
X
0.
T
be a normal operator on a complex finite dimensional exists an orthonormal basis of V consisting of can be represented by a diagonal matrix relative to an
Then there
inner product space V. eigenvectors of T; that
T
is,
orthonormal basis. If dim V = 1, then the theorem trivially is by induction on the dimension of V. suppose dim V — \. Since V is a complex vector space, T has at least one eigenvalue and hence a nonzero eigenvector v. Let be the subspace of V spanned by v and let u^ be a
The proof
n>
Now
holds.
unit vector in
W
W.
Since v is an eigenvector of T, the subspace eigenvector of T* by the preceding problem; hence
W W
invariant under T.
is
W
is invariant under T** = T. The remainder of the proof the proof of Theorem 13.14 (Problem 13.32).
13.40.
Prove Theorem
However, v is also an under T*. By Problem 13.21,
also invariant
is
is identical
with the latter part of
T
be an arbitrary operator on a complex finite dimensional inner product space V. Then T can be represented by a triangular matrix relative .,n, to an orthonormal basis {Ui, U2, that is, for i = l, ., Wn} 13.17:
Let
.
;
.
—
T{ui)
.
+
OiiMi
+
ai2U2
•
•
+
•
.
aiiUi
If dim V = 1, then the theorem trivially is by induction on the dimension of V. suppose dim V = n > 1. Since V is a complex vector space, T has at least one eigenbe the subspace of V spanned by v and value and hence at least one nonzero eigenvector v. Let let Ml be a unit vector in W. Then itj is an eigenvector of T and, say, T{ui) = a^Ui.
The proof
Now
holds.
W
By Theorem
W
Clearly {^2,
.
.
.
,
is
M„} of
13.2,
V
=W ®W^.
Let
E
denote the orthogonal projection of
invariant under the operator ET.
W
such that, for
i
=
ET{Ui)
2,
=
.
.
By
W
OJ2M2
+
ttiaMg
+
•
•
+
for
i
=
2,...,n.
This with
T(ui)
=
=
W'''
djiitj
E
is
the orthogonal projection of
;
T(Ui)
into
.,n,
(Note that {ui,U2, ...,u„} is an orthonormal basis of V.) But hence we must have
onto
V
induction, there exists an orthonormal basis
ajiMi
+
ai2M2
+
•
•
+
a^Ui gives us the desired
fliiM;
result.
V
CHAP.
INNER PRODUCT SPACES
13]
305
MISCELLANEOUS PROBLEMS 13.41.
Prove Theorem 13.13A:
(ii)
(iii)
P
and
is self-adjoint
P
The following conditions on an operator
P = T^ for some self-adjoint operator P — S*S for some operator S.
(i)
{P(u),
u}^
are equivalent:
T.
GV.
for every u
Suppose (i) holds, that is, P = T^ where T = T*. Then P = TT = T*T and so (i) implies Now suppose (ii) holds. Then P* = (S*S)* = S*S** = S*S = P and so P is self-adjoint. Furthermore, (ii).
=
{P(u),u}
Thus
implies
(ii)
Now
and so
(iii),
it
=
{S*S(u),u)
remains to prove that
^
(S(m), S(m)>
implies
(iii)
(i).
Since P is self-adjoint, there exists an orthonormal basis {u^, of V consisting of eigenvectors of P; say, P^.u^) = XiMj. By Theorem 13.8, the Xj are real. (iii), we show that the Xj are nonnegative. We have, for each i,
suppose
holds.
(iii)
Thus
-
-
represented by a real diagonal matrix relative to the orthonormal basis {u^, Moreover, for each i,
T
Xj
Accordingly,
as claimed.
0,
by
=
for
VXiMi,
T^Ui) Since T^ and
P
Remark:
=
^/\^ is
t=l,
agree on a basis of y,
a real number.
\{u.i
=
is the unique positive operator such that square root of P.
Show
T
the
sum
be the
is self-
P(ud
P=
The above operator T
is
m„}
proved.
13.93); it is called the positive
that any operator
.,
Using
...,n
= V\iT{ut) = V^^^/\^u^ = P = T^. Thus the theorem is
r(vTiMi)
.
\(Ui, Mi>
is
adjoint.
13.42.
=
T
forces
r(Mj)
T
(XjMj, Mi)
Let
linear operator defined
Since
=
.
T^
(Problem
of a self-adjoint operator and skew-adjoint
operator.
S^^{T+T*)
Set
and
S*
and i.e.
13.43.
U*
S
self-adjoint
is
and
U = ^(T-T*).
= {^{T+T*))* = {^(T-T*))*
U
is
skew
S+U
Then T =
:=
:^(T*
=;
^(T*
+ -
where
= 1{T* + T) = S = -^(T - T*) = -U
T**) T)
adjoint.
Prove: Let T be an arbitrary linear operator on a finite dimensional inner product space V. Then T is a product of a unitary (orthogonal) operator U and a unique positive operator P, that is, T = UP. Furthermore, if T is invertible, then U is also uniquely determined.
By Theorem
P
13.13,
r*r
is
a positive operator and hence there exists a (unique) positive operator
= T*T (Problem 13.93). Observe that ||P(t;)||2 = {P{v),P(v)} = (PHv),v} = {T*T(v),v} =
such that P2
We now If
T
U* Thus
consider separately the cases
U
is invertible,
(Pr-i)* is
then
we
set
- T-i*P* =
unitary.
We
next set
when T
is
invertible
U = Pr-i. We (r*)-ip
U =U-K
To prove uniqueness, we assume T
Then
= UqP^
show that
U
U is
(i)
unitary:
(T*)-^PPT-i
is also
where U^
\\T{v)\\2
and non-invertible.
U*U =
and
=
{T(v), T(v))
is
unitary and
= (T*)-^T*TT-^ = T = UP
unitary and Pg
I
as required.
is positive.
Then
T*T = PtutUaPo = PoIPo = ^o But the positive square root of T*T is unique (Problem 13.93); hence Pq = P. (Note that the invertibility of T is not used to prove the uniqueness of P.) Now if T is invertible, then P is also by (1). Multiplying U^P = UP on the right by P-i yields Uo = U. Thus U is also unique when
T
is invertible.
INNER PRODUCT SPACES
306
Now
suppose
T
W
Let
not invertible.
is
Ui-.W-^V by
-
Ui(w)
[CHAP. 13
be the image of P,
= ImP. We
define
=w
where P(v)
T(v)
W
i.e.
(2)
We
must show that Ui is well defined, that is, that P(v) - P{v') implies T(v) = T(v'). This follows from the fact that P{v — v') = is equivalent to ||P(d — i;')li = which forces ||r(i) — v')l| = by (1). Thus U^ is well defined. We next define Ui-W^V. Note by (1) that P and T have the = same kernels. Hence the images of P and T have the same dimension, i.e. dim (Im P) — dim dim(Imr). Consequently, and (Im 7^-'- also have the same dimension. We let C/j be any
W
W
isomorphism between W-^ and (Im T)
We
U = UiQ
next set
weW.w'eW^,
then U(v) w, then by (2)
— Thus T = UP and P(v)
It
U
(Here
U^.
=
.
+ =
Ui(w) T(v)
defined as follows: it
is
Now U
U^iw').)
Ui(w)
=
=
U(w)
remains to show that
U
U(x))
7(a;),
13.44.
also used the fact that
Let
(tti,
a2,
.
and v
— w + w'
where v
GV
UPiv)
as required.
Now
is unitary.
every vector
x
GV =
X = P(v) + w' where w'GPT-^. Then U{x) = UPiv) + U2{w') by definition of U^. Also,
(We
GV
v
linear (Problem 13.121) and, if
is
and
)
. .
= = = =
{P(v), P{v)}
.
+ +
^(^;)
(1).
+
where
?72(w')
<7'(i;),
Thus
+ U^(w'))
U^iW), T(v)
{T(v), T(v))
(Ui(w'), U^iw'))
=
+ w',
P{v)
+ w')
(X, X)
=
(P{v),w')
(6i, &2,
+
{T{v)
can be written in the form
. .
)
U
Thus
0.)
unitary and the theorem
is
is
proved.
be any pair of points in i2-space of Example
13.5.
00
Show
that the
By Problem
sum ^(hhi =
+
azbz
+
•
•
•
converges absolutely.
(Cauchy-Schwarz inequality),
1.16
K6,l
aibi
+
•••
+
KM
^
2
A
«i
J2
\2«fA
-
6?
which holds for every n. Thus the (monotonic) sequence of sums S„ = l^i^il bounded, and therefore converges. Hence the infinite sum converges absolutely.
13.45.
V
Let ,
fl')
be the vector space of polynomials over
=
f{t) g{t) dt.
I
Theorem {f)
=
for every
i.e.
-Pif)
f(t).
•
+
•
la„6„|
is
V
for which
there does not exist a polynomial h{t) for which
feV.
Let ^ y -> R be defined by ^(/) = /(O), that is, ^ evaluates constant term. Suppose a polynomial h(t) exists for which
for every polynomial
•
with inner product defined by
:
its
+
Give an example of a linear functional ^ on
13.5 does not hold,
R
S^-i
m
=
-
f(t)
at
and hence maps
C f{t)h(t)dt
Observe that ^ maps the polynomial
f{t) into
(1)
tf(t)
into 0; hence
by
(1),
-1
tf(t)h(t)dt
=
(2)
'0
for every polynomial
f{t).
In particular,
(2)
must hold for
C.m^t)dt
f(t)
=
th{t),
that
is.
=
This integral forces h(t) to be the zero polynomial; hence 4>(f) = (f.h) = ,0) = for every polynomial f(t). This contradicts the fact that is not the zero functional; hence the polynomial h(t) does not exist.
CHAP.
INNER PRODUCT SPACES
13]
307
Supplementary Problems INNER PRODUCTS 13.46.
Verify that (ajMj
More
+ a^u^,
+ h^^ =
fei'Wi
generally, prove that
m
I
\ i=X
13.47.
Let u (i)
=
(xi, x^)
=
and v
(j/j,
For what values of k For what values of
a, 6,
13.49.
Find the norm of v = (1, product in Problem 13.47(i).
=
£R
c,d
=
2)
Let
u
(i)
Verify that the following
(zi, Z2)
and v
€
For what values of
13.51.
d{u,v) d(u, v)
[Dg]
d{u, v)
— = —
0;
a, b, c,
d
= eC
and d(u,v)
Si)
=
if
d{u,
w)
+
d{w,
13.54.
13.55.
13.56.
(u,v}
BxiVz
a^ij/i
+
6*12/2
+
ZiWi is
= =
u
(iii)
/(()
(iv)
A —
-
SXiVi
+
(i)
ca;23/i
aziWi
(1
- 2t, 3 + _ 2t +
t2
)
V3 -4/ [
+
da;2J/2
— i)22'"'l +
bziW2
+
CZ2W1
with respect to
C^
=
+
||v
— m||, if
where
u=
the inner
(ii)
SZ2W2
(i)
+
dz2W2
the usual inner product,
u,v&V,
(ii)
satisfies the following
the
axiom
v.
Hm
ijII
+
||M-'y||
=
2||m||
+
2||d||.
(u, v):
mX
ti
matrices over R.
i,
3
2
- 5i) G
Show that {f,g)=
C^,
in the space of
in the space of
Problem
Problem
13.55,
13.54.
(complex case).
Show that {A,B) =
(|,-Jt,^,^)GR4,
=
kx2y2
the usual inner product,
+ v\\2-l\\u-v\\2 (real case); = J^||M + a'||2-|||M-i)||2+i||M + i'u||2-i||M-w||2
(1
+
C^:
+ i)ZiW2 +
Find the norm of each of the following vectors: V
5x2y2
the following an inner product on C^?
= G
(1
Let V be the vector space of polynomials over R. inner product in V.
(ii)
+
l\\u
Let V be the vector space of product in V.
(i)
ZxzVi
the following an inner product on R^?
and only
+
Verify the following polar forms for
(ii)
-
v).
13.53.
==
v^
d{v, u).
Verify the Parallelogram Law:
{u,v)
-
XiVi
-
d(u,v)
13.52.
(i)
^^^{u^,
R^:
2XiV2
an inner product on
is
Show that the distance function of a metric space: [D2]
+
belong to C^.
(w^, W2)
Find the norm of v = (l — 2i,2 + inner product in Problem 13.49(i).
[I>j]
-
XiVi
= is
f(u, v)
13.50.
a^-^{u^, v-^
i,i
with respect to
R2
f{u, V) (ii)
/
+
the following an inner product on R2?
is
f{u, v)
13.48.
i=l
=
f(u, v) (iii)
\
an inner product on
is
f(u, V) (ii)
n
a.i62(Wi, v^)
belong to R2.
y^)
Verify that the following
+
a^^ifi^, v{)
tr(B'A)
I
defines
f{t) g(t)
dt
an inner
defines
an
INNER PRODUCT SPACES
308
13.57.
Show
that:
13.58.
Let
a, 6, c
result ||tM
13.59.
13.60.
to
sum of two inner products an inner product.
the
(i)
inner product
is
[CHAP. 13
an inner product;
is
a positive multiple of an
(ii)
e R be such that at^ + bt + cfor every t e R. Show that 62 _ 4^0 ^ 0. Use this prove the Cauchy-Schwarz inequality for real inner product spaces by expanding
+ i;||2 ^
0.
Suppose |! = ||m|| H^yll. (That is, the Cauchy-Schwarz inequality reduces to an equality.) that u and v are linearly independent. Find the cosine of the angle (i)
u =
(ii)
u
(ill)
—
(1,
-3,
2),
2t
— l,
V
V
=
=
e
between u and v
Problem
in the space of
/2 1\ M=(_l,v
=
if:
in RS;
(2, 1, 5)
t^
Show
13.55;
/O -1\ (
)
in the space of
Problem
13.54.
ORTHOGONALITY
W of R* orthogonal to
13.61.
Find a basis of the subspace
13.62.
Find an orthonormal basis for the subspace
13.63.
Let
V
W of
u^
=
(1,
C^ spanned by Wj
R
be the vector space of polynomials over
of degree
—
%=
and
—2,3,4)
=
(1, i, 1)
—5,
(3.
7, 8).
%=
and
(1
2 with inner product
+ i, 0, 2). ,
g)
=
Cf{t)g{t)dt.
Find a basis of the subspace
(i)
Let
to
h{t)
Apply the Gram-Schmidt orthogonalization process normal basis {%(*), U2(t), u^(t)} of V.
(ii)
13.64.
W orthogonal
y
Show
(i)
that the following
is
2t
+ 1.
to the basis {1,
t,
i^}
an ortho-
to obtain
with inner product defined by
—
{A,B)
tr(B*A).
an orthonormal basis of V:
0\ /O 07' Vo
'1
1\
/O
o;-
(,0
Find a basis for the orthogonal complement of
(a)
,0 (ii)
R
be the vector space of 2 X 2 matrices over
=
/O 0\ \)' Vo
ON \,
the diagonal matrices,
the symmetric
(6)
matrices.
13.65.
Let If be a subset (not necessarily subspace) of V.
W
dimension, then
13.66.
—
(Here UyV)
I'(W).
is
Prove:
(i)
W
=
•E'(PF);
if
(ii)
V
has
finite
the space spanned by W.)
W be the subspace spanned by a nonzero vector w in Y, and w) of V onto W. Prove B(d) = —t- w. We call E(v) the projection Let
let
E
be the orthogonal projection
{v,
tj
of v along w.
\\w\\^
13.67.
Find the projection of v along
if:
V
=
(ii)
V
(iii)
V
-(l-i,2 + 3i),w = (2~-i,S) = 2t — l, w = t^ in the space
v
(iv)
13.68.
w
(i)
=
Suppose
(1,
-1,
2),
w=
/I
2\
I
q)''*'~(i
{mj,
.
.
.,mJ
.
.
.,a>„_r} is
in R3;
in C2;
of Problem 13.55;
/O -1\
is
independent set of n {^1,
(0, 1, 1)
9)^"
*^® space of Problem 13.54.
a basis of a subspace vectors such that
—r
W
of
V
(Uj, Vj)
a basis of the orthogonal complement
where
=
W
dim
for .
V=
each
n. i
Let
{vi,
and each
.
.
.,i;„_,} be
j.
Show
an
that
CHAP.
13.69.
INNER PRODUCT SPACES
13]
Suppose
{mi,
mapping
..,u^) is
.
„,
=
,
E(V)
Show
that
E
{V,
+
Mi>Mi
V
the orthogonal projection of
Is
W of
an orthonormal basis for a subspace
defined by
,
(V,
M2>M2
+
309
+
,
,
,
'
'
•
Let
V.
E :V -^V
be the linear
,
{V,
ll^U,
W.
onto
r
13.70.
Let
known
(This Is
13.71.
Let
y
13.72.
be an orthonormal subset of V. as Bessel's Inequality.)
= + 1^112 =
(I)
||m|| ||ti
ll'i'll
Let
U and
U-^
nW-^;
and only
+
||m||2
{u
If
any v€.V,
2 "
K^'.Wi)!^
-
ll^'ll^-
that:
+ v,u — v) =
and only
If
11^112
W be subspaces of a (UnW)-^ =
(11)
:
Let
If
that, for
Q;
=
(u,v)
If
0.
by counterexamples that the above statements are not true for, say, C^.
ADJOINT OPERATOR Let r R3 ^ R3 13.73. 13.74.
Show
be a real inner product space.
(II)
Show
Show
mJ
{tti
r
:
C3
^
dimensional inner product space V.
finite
Show
that:
=
(U+W)
{i)
+ W^.
U-^
be defined by Tix,
y, z)
=
(x
+ 2y,
Zx
- Az,
3a!
+ (3 - i)z,
Find T* (x, y,
y).
«).
03 be defined by
=
T(,x, y, z)
(ix
+ (2 + Zi)y,
(2
- hi)y + iz)
Find T*{x,y,z).
13.75.
For each of the following linear functions ^ on every v G V: (i)
R3
->
R
defined by
{x,
y,z)
(ii)
C3
-»
C
defined by
{x,
y, z)
(iii)
13.76.
^
Suppose
y -» R V
has
kernel of T,
i.e.
defined
finite
Show that T*T =
13.78.
Let
—
/(I)
=
(Ker
implies
T)-^
7=
.
uG.V
find a vector
+ 2y — 32. + (2 + St)y + (1 - 2i)z. where V is the vector space
such that
(v)
=
{v, u)
for
x
ix
Prove that the image of T*
dimension.
Im T*
13.77.
by
— -
F
Hence rank(r)
=
is
of Problem 13.63.
the orthogonal complement of the
rank(r*).
0.
V be the vector space of polynomials over R with
inner product defined by
Let D be the derivative operator on V, i.e. D{f) = df/dt. such that (D(f),g) = {f,D*(g)) for every f,g &V. That
Show is,
D
(/, ff>
=
I
f(t) g(t) dt.
that there is no operator D* on has no adjoint.
V
UNITARY AND ORTHOGONAL OPERATORS AND MATRICES row
13.79.
Find an orthogonal matrix whose
13.80.
Find a symmetric orthogonal matrix whose
first
is:
(1)
(l/VS, 2/v'5);
first
row
is
(il)
a multiple of (1,1,1).
(1/3,2/3,2/3).
(Compare with Problem
13.27.)
row
— t);
13.81.
Find a unitary matrix whose
13.82.
Prove: The product and inverses of orthogonal matrices are orthogonal. matrices form a group under multiplication called the orthogonal group.)
13.83.
Prove: The product and Inverses of unitary matrices are unitary. form a group under multiplication called the unitary group.)
13.84.
Show that
if
first
is:
(1)
an orthogonal (unitary) matrix
a multiple of
is
(1, 1
triangular, then
it is
(ii)
(\, \i,
^—
Ji)
(Thus the orthogonal
(Thus the unitary matrices
diagonal.
INNER PRODUCT SPACES
310
13.85.
[CHAP. 13
Recall that the complex matrices A and B are unitarily equivalent if there exists a unitary matrix such that B = P*AP. Show that this relation is an equivalence relation.
P 13.86.
Recall that the real matrices A and B are orthogonally equivalent if there exists an orthogonal P such that B — P*AP. Show that this relation is an equivalence relation.
matrix
13.87.
Let
W be a subspace of V.
v&V
For any
V=W®W^.)
Let unique because a self-adjoint unitary operator on V. is
13.88.
let
= w + w'
v
T-.V^V
V be an inner product space, and suppose U and preserves inner products, i.e. {U(v), U(w)) linear and hence unitary. Let
:
13.91.
13.93.
=w-w'.
operators
E
W
an orthogonal projection onto some subspace
is
A;
a=
(A;
>
13.95.
An « X M
P
of V.
on
K»
is
=
Prove that kl
(T{u), v).
+E
is
Show
positive
(The corresponding Theorem
.,«, in the proof of Theorem 18^3A 1, the only positive operator for which T^ - P.
=
i
is
Prove that
both positive and unitary.
is
is
0).
Consider the operator T defined by T(Ui) = VTjMi, (Problem 13.41). Show that T is positive and that it
Suppose
U
is positive (positive definite).
Prove Theorem 13.13B, page 288, on positive definite operators. 13.13A for positive operators is proved in Problem 13.41.)
13.94.
.
Prove that
r
Suppose
(Such a sum Show that T is
surjective (onto)
is
be a linear operator on V and let f-.V^-V-^K be defined by f{u, v) that / is itself an inner product on V if and only if T is positive definite.
Let
(positive definite) if
13.92.
T(v)
V -* V (not necessarily linear) = {u,w) for every v,w&V.
POSITIVE AND POSITIVE DEFINITE OPERATORS Show that the sum of two positive (positive definite) 13.89. 13.90.
wGWyW'e W^
where
be defined by
.
P=
.
/.
matrix A = (ajj) is said to be positive if A viewed as a linear operator (An analogous definition defines a positive definite matrix.) Prove A is positive if and only if ay = a^ and
(real or complex)
positive.
(positive definite)
n
2=
i,3
for every
13.96.
(a;i,
(>0)
definite):
Determine which of the following matrices are positive (positive
A =
a,
d and ad
— be
)
[
is
positive if
Prove that a diagonal matrix A is positive (positive a nonnegative (positive) real number.
Suppose
THv)
=
T
is
and only
if
(i)
A= A*,
and
are nonnegative real numbers. definite) if
and only
SELF-ADJOINT AND SYMMETRIC OPERATORS For any operator T, show that T + T* is self -adjoint and T - T* 13.99. 13.100.
(vi)
(v)
(iv)
(iii)
(ii)
Prove that a 2 X 2 complex matrix (ii)
13.98.
-
...,«;„) in K^.
(i)
13.97.
ayiCjSc] 1
self-adjoint.
also implies
T(v)
Show
=
that
lor
w >
THv) 0.
=
implies
is
T(v)
if
every diagonal entry
is
skew-adjoint.
=
0.
Use
this
to
prove
that
CHAP.
13]
13.101.
Let
INNER PRODUCT SPACES F
Suppose
be a complex inner product space.
311
(T(v), v) is real for
every
v
G
Show
V.
that
T
is self-adjoint.
ST
13.102.
Suppose S and T are self-adjoint. i.e. ST = TS.
and only
if
S and T commute,
13.103.
For each of the following symmetric matrices A, And an orthogonal matrix
P
for which
Show
that
is
self-adjoint if
P'AP
is
diagonal:
13.104.
Find an orthogonal transformation of coordinates which diagonalizes each quadratic form: (i) q{x, y) = 2x^ — 6xy + lOy^, (ii) q{x, y) = x'^ -\- Sxy — 5y^
13.105.
Find
an
=
q(x, y, z)
orthogonal transformation 2xy + 2xz + 2yz.
of
which
coordinates
diagonalizes
the
quadratic
form
NORMAL OPERATORS AND MATRICES 13.106.
find
/2
A =
Verify that
P*AP.
I
i\ 1
.
normal.
is
Find a unitary matrix
^.*
13.107.
Show that
13.108.
Prove that if T is normal on V, then ||r('y)|| holds in complex inner product spaces.
13.109.
Show
13.110.
Suppose T (i)
(ii)
(iii)
P
such that
P*AP
is
diagonal, and
''^
a triangular matrix
is
normal
if
and only
=
if it is
diagonal.
vGY.
for every
||r*('u)||
Prove that the converse
that self-adjoint, skew-adjoint and unitary (orthogonal) operators are normal.
T T T
is
normal.
Prove that:
and only
is
self-adjoint
is
unitary
if
and only
if its
eigenvalues have absolute value
is
positive if
and only
if its
eigenvalues are nonnegative real numbers.
if
if its
eigenvalues are real. 1.
13.111.
Show that
13.112.
Suppose
S and T
13.113.
Suppose
T
13.114.
Prove: Let S and T be normal operators on a complex finite dimensional vector space V. there exists an orthonormal basis of V consisting of eigenvectors of both S and T. (That is, T can be simultaneously diagonalized.)
T
if
is
is
T and T* have
normal, then
the same kernel and the same image.
are normal and commute.
Show
that
S+T
and
normal and commutes with S.
Show
that
T
commutes with S*.
also
ST
are also normal.
Then
S and
ISOMORPHISM PROBLEMS 13.115.
13.116.
Let {ei, e„} be an orthonormal basis of an inner product space V over K. V ^[v]g is an (inner product space) isomorphism between V and X". (Here ordinate vector of v in the basis {cj}.) .
Show
.
.
,
V
that inner product spaces
and
W over K
are isomorphic
if
and only
same dimension. 13.117.
Suppose
{ej,
be the linear
...,ej and {ei
map
defined
by
.
.
.
,
e^} are
T{ei)
=
e(,
W
if
Show that [v]^
V
the map denotes the co-
and
W
orthonormal bases of V and respectively. Let for each i. Show that T is an isomorphism.
have the
T V^ :
W
INNER PRODUCT SPACES
312
13.118.
13.119.
V
Let
uG V
be an inner product space. Recall (pag:e 283) that each
u
in the
M
H-
M
V
Consider the inner product space
Show
of Problem 13.54.
determines a linear functional
for every
u)
{v,
and nonsingular, and hence an isomorphism from
linear
is
=
u (v)
dual space V* by the definition
[CHAP. 13
that
V V
v E.V.
Show that
the
map
onto V*.
is
isomorphic to R"*" under the
mapping
=
where Ri
(ctii,
ajj,
.
«12
\^m\
^m2
••
•
•
«ln
•
\
^mnl
•
row of A.
the fth
.,«i„),
.
/«U
MISCELLANEOUS PROBLEMS 13.120.
that there exists an orthonormal basis {mj, .,m„} of V consisting of eigenvectors of there exist orthogonal projections Ei,...,E^ and scalars \i,...,\ such that: + X^^; (ii) Ei+ + Er = I; (iii) EiEj = for i j. \iEi +
Show only
•
13.121.
13.122.
.
.
if
U®W
V =
Ti ©
T2
then
T(v)
=
+
T.^(u)
TiiU^V
and suppose (Here
is also linear.
U
Suppose
T
is
and
defined as follows: if
t)
an orthogonal operator on R3 with positive determinant.
is
a
13.47.
(ii)
/c
13.48.
(i)
VE,
13.50.
(i)
3V2,
13.56.
(i)
IJMll
13.60.
(i)
cos
13.61.
{vi
=
(1,2,1,0), v^
13.62.
{vi
=
(1, i,
13.63.
(i)
{/i it)
(ii)
{Mj(t)
9;
(iii)
>
Show
that
0,
ad
||i;||
=
2\/ll,
>
6c
5V2
(ii)
=
-
>
d
0,
Supplementary Problems
to
VH
=V65/12, e
and
T =
TgC^)-)
Answers
(ii)
if
T^:W-^V are linear. Show that T = e V and v = u + w where uG:U,wG.W,
rotation or a reflection through a plane.
>
(i)
¥
•
Suppose
T
(ii)
9/\/420,
(ii)
=
1)/VS, V2
= =
7t2
- 5«,
1, M2(t)
cos e
=
(iii)
\/l5/6,
\\f{t)\\
cos
(iii)
= e
^/83/15,
-
(i)
13.73.
r*(a;, y, z)
-
(«
13.74.
T*{«,y,z)
=
(-iae
||A||
= V30
2/y/2\a
(4,4,0,1)}
=
(2i, 1
- Si, 3 - i)/V24}
= 12*2 _ 5} = (2t - l)/\/3, M3(t) = fzit)
(6*2
- 6f + 1)/a/5 }
(ii)
+ 82/,
2a;
+ Sj/,
(26
+ x, (2
+ 7t,
27
+ 24t)/V14,
(iii)
V5
-4j/)
- 3t)a; +
(2
+ 5i)z,
(3
+
/
r-
,
(0,l/\/2, 1/^/2),
13.67.
(iv)
%-
iz)
tVS,
(iv) i
7/V6
_
r-
...^
U
is either
a
CHAP.
13.75.
INNER PRODUCT SPACES
13]
= ^(ei)ei + = (1, 2, -3),
Let u (i)
u
^
13.80.
2/3
2/3
-2/3
,2/3
1/3
/
13.96.
1.103.
Only
p
(i)
X
(i)
13.105.
a;
13.106.
P =
=
+ 0(e„)e„ where {ej} is an orthonormal basis. u = i-i, 2 - 3i, 1 + 20, (iii) u = (18t2 -8t + 13)/15
(ii)
1/3
and
=.
=
(l-0/V3\
(v)
x'Jy/l
y -1/f 2/V^/
- j/')/\/lO,
v
=
+ y'lyfl + 2'/v/6,
[llyfi -l/\/2\
V
\1/V2
/
are positive.
2/^
(
(3a;'
-1/V6 -l/x/e/
-2/3/
l/x/3
(i)
\2/\/6
2/3^
-l/\/5
13.104.
•
/
'l/3 1
•
313
V
.
1/a/2/'
(a:'
y
Moreover,
(ii)
p =
a;'/v/3
positive definite.
(v) is
-i/^y (iii)
f 2/f \-l/\/5
+ 3j/')/\/l0, =
*"*''
*^
*
(ii)
x
=
(2a:'
- j/'/^/B + z't^f^,
/2 + ^*^P = ('^
i'
2
-
i
p =
z
- 2/')/\/5, =
I
^'^ -^'^
\-l/v^
2/\/5/
a;'/\/3
y
=
(x'
- ^z'l^T^
+ 2j/')/a/5
3/\/T0
A
Appendix
and Relations
Sets SETS,
ELEMENTS
Any
well defined list or collection of objects
set are called its elements or
p If every element of
a subset of
B
or
is
We
members.
GA
if
p
an element
is
A
A
A
A=B
AcB
GA,
p
if
same elements; that
and only
if
AcB
A—B
and
G
B, then
A
is called
BdA
or
sets are equal if they both contain the
The negations of
in the set
also belongs to a set B, i.e. if a; G implies x said to be contained in B; this is denoted by
AgB Two
a set; the objects comprising the
is called
write
and
is,
BcA
A^B
p ^ A,
are written
and
A¥'B
respectively.
We specify a particular set by either listing its elements or characterize the elements in the set. For example,
A means
A
is
(1,3,5,7,9}
the set consisting of the numbers
B —
(a;
:
a;
is
by stating properties which
7 and
9;
and
a prime number, x
<
15}
1, 3, 5,
B is the set of prime numbers less than 15. We also use special symbols to denote sets which occur very often in the text. Unless otherwise specified:
means that
We
N =
the set of positive integers:
Z =
the set of integers: ...,—2,-1,0,1,2,...;
Q =
the set of rational numbers;
R = C =
the set of real numbers;
1, 2, 3, ...
;
the set of complex numbers.
to denote the empty or null set, i.e. the set which contains no elements; this assumed to be a subset of every other set. Frequently the members of a set are sets themselves. For example, each line in a set of lines is a set of points. To help clarify these situations, we use the words cUiss, collection and family synonymously with set. The words subclass, subcoUection and subfamily have meanings analogous to subset. also use
set is
Example A.l
:
The
sets
A and B above A = {a; G N :
9GA
Observe that 3 GB, and 6
The
sets of
Example A.3
Let
C =
Example A.4
The members of the
a;
is
but 9
€ A and
Example A.2
can also be written as
6
g
odd,
a;
g B, B.
numbers are related as {x
:
x^
=
4,
X
is
odd}.
<
and
11GB
follows:
Then
B =
and
10}
but
11
{2,3,5,7,11,13}
€ A;
NcZcQcRcC.
C=
0, that
is,
C
is
the empty
class {{2, 3}, {2}, {5, 6}} are the sets {2, 3}, {2}
315
3GA
whereas
and
set.
{5, 6}.
and
SETS AND RELATIONS
316
The following theorem
Theorem
applies.
Let A, B and C be any then A = B; and (iii) if
A.l:
[APPENDIX A
sets.
AcB
Then:
and
Ac A;
(i)
BcC,
emphasize that does not exclude the possibility but A¥' B, then we say that A is a proper subset of B. symbol c for a subset and the symbol c only for a proper subset.)
AcB
if
AcB
and B(zA,
AcC. that A = B.
AcB
We
(ii)
then
However, if (Some authors use the
When we
speak of an indexed set {at: i G I}, or simply {Oi}, we mean that there is a mapping ^ from the set / to a set A and that the image 4>{i) oi i & I is denoted (M. The set are said to be indexed by /. / is called the indexing set and the elements a, (the range of A set (tti, a2, } indexed by the positive integers N is called a sequence. An indexed class of sets {Ai i G I), or simply (Ai), has an analogous meaning except that now the map 4, assigns to each i G I a set Ai rather than an element a,. )
.
.
.
:
SET OPERATIONS Let A and B be arbitrary
The union of A and B, written AuB, is the set of sets. elements belonging to A or to B; and the intersection of A and B, written AnB, is the set of elements belonging to both A and B:
AUB If
AnB = 0,
that
{x: is,
xG A or xGB} and AnB = {x: x GA and x G B} if A and B do not have any elements in common, then A and B
are
said to be disjoint.
assume that all our sets are subsets of a fixed universal set (denoted here by Then the complement of A, written A'=, is the set of elements which do not belong to A:
We
A<=
ExamplA
AJ5:
=
{X
U).
gU: x^A)
The following diagrams, called Venn diagrams, illustrate the above set operations. Here sets are represented by simple plane areas and U, the universal set, by the area in the entire rectangle.
AuB
is
shaded
AnB
is
shaded
QD A<^ is
shaded
SETS AND RELATIONS
APPENDIX A]
317
Sets under the above operations satisfy various laws or identities which are listed in the table below. In fact, we state
Theorem
Sets satisfy the laws in Table
A.2:
1.
LAWS OF THE ALGEBRA OF SETS Idempotent Laws la.
AuA = A
2a.
(AuB)uC - Au(BuC)
3a.
AuB - BuA
4a.
Au(BnC) = (AuB)n(AuC)
lb.
AnA = A
Laws
Associative
2b.
(AnB)nC = An(BnC)
Commutative Laws 3b.
Laws
Distributive
4b.
Identity 5a. 6a.
AnB = BnA An(BuC) = (AnB)u(AnC)
Laws
Au0 = A AdU = U
5b.
6b.
AnU An0
= A =
Complement Laws 8a.
AuA= = U (A':)<= = A
9a.
(A\jBY
7a.
7b.
AnA<; =
8b.
U<:
=
=
0, 0c
[7
De Morgan's Laws
=
A'ni?»
9b.
Table
Remark:
Each
(Here
we
=
[x:
xGA
and x
A<'uB<=
1
from an analogous
of the above laws follows
AnB
{AnBY =
G B} =
{x:
logical law.
xGB
and x G A)
For example,
=
BnA
use the fact that the composite statement "p and q", written p a "q and p", i.e. q a p.)
g, is
logically equivalent to the composite statement
The
relationship between set inclusion and the above set operations follows.
Theorem
A.3:
Each of the following conditions (i)
(ii)
AnB = A AuB = i?
is
AcB:
equivalent to
(iii)
B<^cA<^
(iv)
AnB' =
We
(v)BuA==C7
generalize the above set operations as follows. Let [Ai i S 7} be any family of Ai, written U^^,A^ (or simply UjAi), is the set of elements each belonging to at least one of the Ai; and the intersection of the At, written n^^^A^ or simply n i Ai, is the set of elements each belonging to every Ai. sets.
:
Then the union of the
PRODUCT SETS Let A and B be two ordered pairs
(a, 6)
sets.
where
The product set and b G B:
AxB The product
A
of
and B, denoted hy
aG A
of a set with itself, say
=
{{a, b):
Ax A,
is
aGA, bGB}
denoted by A".
AxB,
consists of all
SETS AND RELATIONS
318
Example
[APPENDIX A
The reader is familiar with the cartesian plane R^ = R x R as shown below. Here each point P represents an ordered pair {a, b) of real numbers, and vice versa.
A.6:
•P
-3
2^3
1
---1
Example
A =
Let
A.7:
and
{1, 2, 3}
B = {a, 6}. = {(1, a),
AXB Remark:
The ordered pair definition, the
a
if
=
=
(1, 6), (2, a), (2, 6), (3, a), (3, b)}
defined rigorously
"order" property
and b
c
(a, 6) is
Then
may
by
=
{a, b)
be proven; that
is,
{{a}, {a, b}}. {a, b)
=
(c,
d)
From if
this
and only
d.
The concept of product set is extended to any finite number of sets in a natural way. The product set of the sets Ai, x Am, is the set consisting of Am, written Ai x A2 x all TO-tuples (ci, a2, fflm) where ai G A. for each i. .
.
.
.
.
•
,
•
•
. ,
RELATIONS
A
binary relation or simply relation R from a set A to a set G B exactly one of the following statements:
pair {a,b)
Example
A.8:
"a
is
related to b", written
(ii)
"a
is
not related to b", written
Set inclusion
R
ot
a^b.
a relation in A.
a relation in any class of or A ():B.
sets.
For, given any pair of sets
R from A to B uniquely defines R = {(a, b): aRb}
Observe that any relation
Conversely, any subset
is called
is
A cB
either
if
and only
if
(a, b)
A relation i? from A to B is
a subset of
EQUIVALENCE RELATIONS A relation in a set A is called an equivalence [El] Every a € A is related to itself. related to
If
[Es]
If a is related to b
is
In general, a relation
and transitive it is reflexive,
then b
6,
and b
is
is
is
R
of
A
and B,
Ax B as follows:
to
B
as follows:
A
to
B
GR
In view of the above correspondence between relations from we redefine a relation as follows:
[E2]
a subset
AxB defines a relation from A aRb
a
assigns to each ordered
aRb,
(i)
A relation from a set A to the same set A
Definition:
B
Ax
and subsets of
A x B,
A x 5.
relation if
it satisfies
the following axioms:
related to a.
related to
c,
then a
said to be reflexive if
if it satisfies [£"3].
is
related to
it satisfies [Ei],
In other words, a relation
symmetric and transitive.
c.
symmetric if it satisfies [Ez], is an equivalence relation if
APPENDIX
SETS AND RELATIONS
A]
Example
and
On Example
C
Consider the relation
A.9:
A.IO:
a
AqB
By Theorem A.l, That is, C is both not symmetric, since A c B and
of set inclusion.
and B
the other hand,
C
319
is
Ac
C.
A cA
for every set A;
reflexive
A
¥=
B
and
transitive.
implies
B cjiA.
In Euclidean geometry, similarity of triangles is an equivalence relation. For p and V are any triangles, then: (i) a is similar to itself; (ii) if a is similar to then p is similar to a; and (iii) if a is similar to p and /8 is similar to y, then a similar to y. a,
If
R
is
denoted by
an equivalence relation in A, then the equivalence [a], is the set of elements to which a is related: [a]
The
=
[x:
collection of equivalence classes, denoted
AIR = The fundamental property
Theorem
A.4:
class of
if yS,
is
any element a G A,
aRx)
by A/R,
{[a]:
a
is called
the quotient of
A
by R:
G A}
of equivalence relations follows:
R
Let be an equivalence relation in A. Then the quotient set A/R is a partition of A, i.e. each belongs to a member of A/R, and the members of A/R are pairwise disjoint.
aGA
Example A.11
:
Let R5 be the relation in Z, the set of integers defined by
X = y (mod 5) which reads "x is congruent to y modulo 5" and which means "x - y is divisible by 5". Then ^5 is an equivalence relation in Z. There are exactly five distinct equivalence classes in Z/R^:
Ao Ai
A2 A3 A4
==
= = = =
{...,-10,-5,0,5,10}
{...,-9,-4,1,6,11} {...,-8,-3,2,7,12} {...,-7,-2,3,8,13}
{...,-6,-1,4,9,14}
Now each integer x is uniquely expressible in the form x = 5q + r where - r < 5; observe that x G E^ where r is the remainder. Note that the equivalence classes are pairwise disjoint and that Z = A0UA1UA2UA3UA4.
B
Appendix
Algebraic Structures INTRODUCTION We define here
algebraic structures which occur in almost all branches of mathematics. In particular we will define a field which appears in the definition of a vector space. We begin with the definition of a group, which is a relatively simple algebraic structure with only one operation and is used as a building block for many other algebraic systems.
GROUPS Let G be there
is
GG
a nonempty set with a binary operation, i.e. to each pair of elements a,b assigned an element ab G G. Then G is called a group if the following axioms hold:
For any a,b,c G G, we have
[Gi]
There exists an element
[G2]
every a aa~^
=
a~^a
GG =
called the identity element, such that ae
=
ea
=a
for
a'^GG,
there exists an element
called the inverse of a, such that
e.
A ab =
GG,
(the associative law).
a{bc)
GG.
For each a
[Ga]
e
=
{ab)c
group G is said to be abelian ha for every a,h GG.
(or:
commutative)
if
the commutative law holds,
i.e.
if
the binary operation is denoted by juxtaposition as above, the group G is said written multiplicatively. Sometimes, when G is abelian, the binary operation is deto be noted by + and G is said to be written additively. In such case the identity element is denoted by and is called the zero element; and the inverse is denoted by —a and is called negative the of a.
When
If
A
B are AB =
and
G then we write aGA, bGB}, or A + B =
subsets of a group {ab:
We also write a for A subset H of a
group
G
is called
a subgroup of
^ is a subgroup of G and aGG,
Definition:
is called
a
left coset of
B.1:
b
G B}
G
if
H itself
then the set
forms a group under the
Ha
is
called a right coset
H.
if
aH = Ha
Note that every subgroup of an abelian group
Example
G A,
a
A subgroup H of G is called a normal subgroup if a-^HacH Equivalently, H is normal left cosets of H coincide.
Theorem
+ b:
{a}.
operation of G. If of and the set aH
H
{a
is
for every
aGG,
for every aGG. the right and
i.e. if
normal.
Let f? be a normal subgroup of G. Then the cosets of i? in G form a group under coset multiplication. This group is called the quotient group and is denoted by G/H. B.1
:
The
set
Z
of integers forms an abelian group under addition. (We remark that the denote the Z but the odd integers do not.) Let
H
even integers form a subgroup of set of multiples of 5,
i.e.
(necessarily normal) of Z.
H =
{.
The
320
.
.,
-10, -5,
cosets of
H
0, 5, 10,
in
Z
.
.
follow:
.}.
Then
H
is
a subgroup
APPENDIX
ALGEBRAIC STRUCTURES
Bj
1
2
3 4
= = = = =
+ + + + +
1
2 3 4,
H = H = = H = H = H = i?
321
{...,-10,-5,0,5,10,...} -9, -4,
1, 6, 11,
.
.
.}
{..., -8, -3, 2, 7, 12,
.
.
.}
.
.
.}
{.
{.
.,
.
.,
.
-7, -2,
3, 8, 13,
{...,-6,-1,4,9,14,
...}
For any other integer w S Z, n = w + H coincides with one of the above cosets. Thus by the above theorem, Z/H = {0, 1, 2, 3, 4} forms a group under coset addition; its
addition table follows:
+
T
2
3
4 4
T
2
3
T
T
2
3
4
2
2
3
4
3
3
4
4
4
1
I
2
2
3
1
This quotient group Z/H is referred to as the integers modulo 5 and is frequently denoted by Z5. Analogously, for any positive integer n, there exists the quotient group Z„ called the integers modulo n.
Example B^:
The permutations of n symbols (see page 171) form a group under composition of mappings; it is called the symmetric group of degree n and is denoted by S„. We investigate S3 here; its elements are "2
"a
<'i
Here
n f
.
2 .
}
i
(3
—
02
a
\)
(III)
3\ 3 ,
k
1
is
maps
the permutation which
1 -»
t,
2
The mul-
I-* j,
tiplication table of S3 is e
<'l
«'2
6
e
"1
"2
"1
"1
€
01
"2
"2
H
e
01
''S
"1
"3
"3
01
02
e
"1
"2
"3
G
''S
•'a
"2
02
e
"2
"3
<'l
€
01
into a
is ab.)
H
{01, 02}
iH
tfs}
02-^
{02,
group G'
The
set
H=
{e,
ffj
is
a sub-
Left Cosets
{e,
Observe that the right cosets and the subgroup of S3.
a.bGG.
02
•Pi
= H^l = H2 =
for every
01 "2
02
Right Cosets
mapping / from a group
"3
02
01
H
A
02
2
(The element in the ath row and 6th column group of S3; its right and left cosets are
/(a)/(b)
01
= — =
{e,„,} {^j,
org}
{02><'2}
left cosets are distinct;
is
called a
hence
H is not a normal
homomorphism
if
/(a6)
=
one-to-one and onto, then / is called an isomorphism and G and G' are said to be isomorphic.) If f:G-*G' is a homomorphism, then the feerraei of / is the set of elements of G which map into the identity element e' e G': (If / is also bijective, i.e.
kernel of /
(As usual, /(G)
is called
= {aGG:
f(a)
the image of the mapping
/:
=
e'}
G^G'.)
The following theorem
applies.
Theorem
B.2:
X
Let /: G-» G' be a homomorphism with kernel K. Then is a normal subgroup of G, and the quotient group GIK is isomorphic to the image of /.
ALGEBRAIC STRUCTURES
322
[APPENDIX B
Let G be the group of real numbers under addition, and let G' be the group of positive real numbers under multiplication. The mapping f G -* G' defined by /(a) — 2" is a homomorphism because
Example B^:
:
f(a+b) In particular, /
Example
is bijective;
=
hence
=
2° + "
G and
2''2i'
=
f{a)f(b)
G' are isomorphic.
Let G be the group of nonzero complex numbers under multiplication, and be the group of nonzero real numbers under multiplication. The mapping f defined by f(z) — \z\ is a homomorphism because
B.4:
:
/(ziZa)
K
The kernel which real
RINGS, INTEGRAL
\z\
=
=
|ziZ2|
=
|zi| [zal
Thus G/K
numbers under
is
isomorphic to
/(^i)
numbers the image of
of f consists of those complex
1.
=
z
let
G -*
G' G'
f(H)
on the unit circle, i.e. for to the group of positive
/, i.e.
multiplication.
DOMAINS AND FIELDS
Let i? be a nonempty set with two binary operations, an operation of addition (denoted by +) and an operation of multiplication (denoted by juxtaposition). Then R is called a ring if the following axioms are satisfied: [Ri] For any a,b,e G R, we have {a + h) + c = a + (6 + c). [Ri] There exists an element G /?, called the zero element, such that a + = + a = a for every
aGR.
For each a G
[Rs\
+ (—a)
a
=
J?
(—a)
G
[R^ For any a,b
there exists an element
+a = R,
we have a + b =
For any a,b,cG R, we have For any a,b,c G R, we have:
[Rs]
[Re]
(i)
a{b
+ c) =
ab
+ ac, and
is
defined
R, called the negative of
a,
such that
It
can be shown is called
A
nonempty subset S of
b
+ a.
=
a{bc).
+ e)a =ba + ca. through [Rt] may be summarized by (b
iniJby a —
R
(see
{ab)c
(ii)
Observe that the axioms [Ri] abelian group under addition. Subtraction
—a G
0.
b
saying that
R
is
an
= a + (— &).
Problem B.25) that a-
=
•
a
=
for every a
G
R.
a commutative ring if ab — ba for every a,b G R. We also say that R is a ring with a unit element if there exists a nonzero element 1 G R such that o • 1 = 1 • a = o for every a G R. a subring of R if S itself forms a ring under the a subring of R if and only if a, & G S implies a-b
i? is called
We note that S is
operations of R.
GS
and ab G S. A nonempty subset / of jB is called a left ideal in R if: (i) a — 6 G / whenever a,b G I, and (ii) ra G I whenever r GR, aG I. Note that a left ideal I in R is also a subring of R. Similarly we can define a right ideal and a two-sided ideal. Clearly all ideals in commutative rings are two-sided. The term ideal shall mean two-sided ideal unless otherwise specified.
Theorem
Now (a)
=
B.3:
R
he a commutative ring with a unit element. For any aGR, the set is an ideal; it is called the principal ideal generated by a. If every ideal a principal ideal, then R is called a principal ideal ring. let
{ra: r
in iZ is
Let / be a (two-sided) ideal in a ring R. Then the cosets {a + I: aGR} form a ring under coset addition and coset multiplication. This ring is denoted by R/I and is called the quotient ring.
Definition:
G R}
A
commutative ring has no zero divisors,
R
with a unit element is called an integral domain ab — implies a = or b = 0.
i.e. if
it
R
APPENDIX
ALGEBRAIC STRUCTURES
B]
A
commutative ring R with a unit element is called a field if every nonzero a E R has a multiplicative inverse, i.e. there exists an element a~^ E R such that aa~i = a~^a — 1.
Definition:
A field
323
is
necessarily an integral domain; for b
= I'b =
a-^ab
=
if
ab
-
a-^'O
and a
t^ 0,
then
=
We
remark that a field may also be viewed as a commutative ring in which the nonzero elements form a group under multiplication. Example B5:
The
Z
of integers with the usual operations of addition and multiplication is the example of an integral domain with a unit element. Every ideal / in Z is a principal ideal, i.e. / = (m) for some integer n. The quotient ring Z„ = Z/(ji) is calle^ the ring of integers modulo n. If n is prime, then Z„ is a field. On the other hand, if n is not prime then Z„ has zero divisors. For example, in the ring Zg, 2 3=0 and 2^0 and 3 # 0. set
classical
Example
Q
The rational numbers
B.6:
and the
to the usual operations of addition
Example
Let
B.7:
C
real
numbers
R
each form a
field
with respect
and multiplication.
denote the set of ordered pairs of real numbers with addition and multiplica-
tion defined
by (a, 6)
+
(c,
(a, 6) '(c, d)
Then C
satisfies all the
complex numbers
(see
d)
=
=
(a
{ae
—bd, ad +
+ e,b + d)
required properties of a
page
field.
be)
In fact,
C
is
just the field of
4).
M
Example
B.8:
The set of all 2 by 2 matrices with real entries forms a noncommutative ring with zero divisors under the operations of matrix addition and matrix multiplication.
Example
B.9:
Let R be any ring. Then the set jB[a;] of all polynomials over R forms a ring with respect to the usual operations of addition and multiplication of polynomials. Moreover, if R is an integral domain then R[x] is also an integ^ral domain.
Now let D be an integral domain. We say that b divides a in D if a = bc for some G D. An element u G D ia called a unit if u divides 1, i.e. if u has a multiplicative inverse. An element b GD is called an associate of a G D if b = ua for some unit uG D. A nonunit p G D is said to be irreducible if p = ab implies a or 6 is a unit. c
An integral domain D is called a unique factorization domain if every nonunit a G D can be written uniquely (up to associates and order) as a product of irreducible elements. Example BJO:
The ring Z of integers is the classical example of a unique factorization domain. The units of Z are 1 and —1. The only associates of n G. Z are n and —n. The irreducible elements of
Example
B.ll
:
The
set
Z
D = {a+ b^/JS
are ±1, 18
are the prime numbers.
:
irreducible in D.
is an integral domain. The units of D The elements 2, 3 - Vl3 and -3 - Vl3 are 2 • 2 = (3 - Vl3 )(— 3 - Vi3 ). Thus D is not
a, b integers}
± 5^13 and -18 ±
5\/l3.
Observe that 4
a unique factorization domain.
=
(See Problem B.40.)
MODULES
M
M
be a nonempty set and let Rhe a. ring with a unit element. Then is said to be a R-module if is an additive abelian group and there exists a mapping RxM-* which satisfies the following axioms: Let
(left)
M
M
ALGEBRAIC STRUCTURES
324
[APPENDIX B
+ mz) — rwi + rm2 (r + s)m = rm + sm {rs)m = r{sm) I'm — m
[Ml] r(mi [Mz] [Ma]
[M4]
for any r,s
GR
and any mi G M.
We
emphasize that an JR-module is a generalization of a vector space where the scalars to come from a ring rather than a field. Example
B.12:
Let G be any additive abelian group. integers by defining
We make G
into a
we
allow
module over the ring Z of
n times
=
nff
where n
is
ff
any positive
+ ff+---+ff,
Oflr
=
{-n)ff
0,
= -ng
integer.
Example
B.13:
Let
Example
B.14:
be a vector space over a field and let T y -» V be a linear mapping. into a module over the ring K[x\ of polynomials over by defining f(x)v = f(T) (v). The reader should check that a scalar multiplication has been
iJ
be a ring and
/ be an ideal in R.
let
Then
/
may
K
Let
V
We
make V
be viewed as a module over R.
:
K
defined.
Let it
An additive subgroup AT of iW is called a submodule of M G N. (Note that N is then a module over R.) /2-modules. A mapping T M-* M' is called a homomorphism (or:
be a module over R. imply ku
iJf
uGN and kGR Let M and M' be
R-homomorphism or (i)
for every u,v
:
R-linear)
if
T{u + v)
=
G M and
T{u)
+
and
T(v)
(ii)
T{ku)
=
kT{u)
kGR.
every
Problems GROUPS BJ.
Determine whether each of the following systems forms a group G: (i)
(ii)
(iii)
(iv)
(v)
B.2.
Show (i)
B.3.
G G G G G
— set of integers, operation subtraction; = {1, —1}, operation multiplication; = set of nonzero rational numbers, operation division; = set of nonsingular nXn matrices, operation matrix = {a+bi: a,b e Z}, operation addition.
that in a group G:
G
the identity element of
(ii)
each a
(iii)
(o-i)-i
(iv)
0.6
=
SG
ac
=
a,
is
=
and (ab)-^
implies
b
=
c,
a"
(iii)
unique;
has a unique inverse a~^
=
GG
e,
G
G;
fe-ia-i;
and 6a
In a group G, the powers of a
Show
multiplication;
a"
=
ca implies
6
=
c.
are defined by
=
aa"~i, o~"
=
(a")~i,
where
that the following formulas hold for any integers r,8,tS Z:
(«>•+«)«
=
a'-t+st.
nGN (i)
a^'a^
—
a^+'^
(ii)
(a"")'
=
a",
APPENDIX
ALGEBRAIC STRUCTURES
B]
B.4.
Show
B.5.
Suppose
B.6.
Suppose if empty, and
B.7.
B.8.
B.IO.
B.ll.
G
is
is
is
an abelian group, then
a group such that
{ab)^
=
=
(a&)»
a^bn
for any
for every
a^b^
H
a subset of a group G. Show that a,b G implies o6~i G H.
a, 6
Show that
G
a, 6
and any Integer «
Show that G
G.
a subgroup of
is
GG
G
Z.
abelian.
is
and only
if
G
if
(i)
H
is
non-
H
(ii)
Prove that the intersection of any number of subgroups of
by B.9.
G
that if
325
the set of all powers of a
GG
is
G
is also
a subgroup of G;
it is
a subgroup of G.
called the cyclic
group generated
a.
A group G is said to be cyclic if every subgroup of a cyclic group
G
is
aG G,
generated by some
G=
i.e.
Suppose G is a cyclic subgroup. Show that G is isomorphic to the set or to the set Z„ (of the integers modulo n) under addition. Let
H be a subgroup of
Show
G.
{a."
nG
:
Z}.
Show that
is cyclic.
that the right
(left)
cosets of
H
Z
of integers under addition
partition
G
into mutually disjoint
subsets. B.12.
The order of a group G, denoted by |G|, is the number of elements of G. Prove Lagrange's theorem: H is a subgroup of a finite group G, then \H\ divides \G\.
If
B.13.
B.14,
Suppose
—
\G\
p where p
is
prime.
Show
that
G
is cyclic.
H and N are subgroups of G with N normal. Show that HN is a subgroup of G and HnN is a normal subgroup of G. Let H be a subgroup of G with only two right (left) cosets. Show that H is a normal subgroup of G. Suppose
(i)
(ii)
B.15. B.16.
Prove Theorem
G/H under B.17.
Suppose
B.18.
Let f
B.l:
Let
H
he
a.
normal subgroup of G. Then the cosets of
G
is
an abelian group. Show that any factor group
G -* G' be a group homomorphism. Show that: = e' where e and e' are the identity elements /(a-i) = /(a)-i for any a G G.
(ii)
in
G form
a group
G/H
K
a normal
is also abelian.
:
/(e)
(i)
H
coset multiplication.
of
G
and G' respectively;
B.19.
Prove Theorem B.2: Let f G -* G' be a group homomorphism with kernel K. Then subgroup of G, and the quotient group G/K is isomorphic to the image of /.
B.20.
Let G be the multiplicative group of complex numbers z such that group of real numbers. Prove that G is isomorphic to R/Z.
B.21.
:
For a
G B.22.
B.23.
B.24.
G
G,
let
g
:
G^G
be defined by g(a)
=
g-^ag.
Show
and
1,
that
G
G
G
G
For a
be an abelian group.
wG
Z,
show that the map a
G
with
N
normal.
=
-ab,
fixed
l-»
a"
H
and
is
N
are subgroups of isomorphic to HN/N.
Prove that
HnN
RINGS
B.26.
be the additive
an isomorphism of
Show o
is
a homomorphism
into G.
Suppose
(i)
is
B
be the multiplicative group of n X w nonsingular matrices over R. Show that the mapping is a homomorphism of G into the multiplicative g:roup of nonzero real numbers.
H/{HnN)
B.25.
let
h* |A|
Let of
fir
=
onto G.
Let
A
fixed
\z\
is
•
Show
that in a ring R:
=
• a.
=
0,
that in a ring
(ii)
o(-6)
R with
=
(-a)b
a unit element:
(i)
(iii)
(-o)(-6)
(— l)a
=
—a,
= (ii)
ab.
(— 1)(— 1)
=
1.
is
normal
in
H
and
ALGEBRAIC STRUCTURES
326
B.27.
B.28.
Suppose a^ = a for every a Boolean ring.) Let
e
Prove that
i?.
We make R
be a ring with a unit element.
i?
=
and^ U'b
ab
+ a+b.
(i)
i? is
Verify that
a commutative ring.
into another ring
a ring,
fi is
[APPENDIX B
R
(Such a ring
by defining
is called
a
a®b = a+b + l
Determine the 0-element and 1-element
(ii)
of R.
B.29.
Let
G
be any (additive) abelian group. G into a ring.
makes
Define a multiplication in
G
by a-b
=
0.
Show that
B.30.
Prove Theorem B.3: Let / be a (two-sided) ideal in a ring R. Then the cosets {a + I:. a a ring under coset addition and coset multiplication.
B.31.
Let
/j
and
B.32.
Let
R
and R' be
be ideals in R. Prove that
I^
rings. (i)
an
mapping f
f(a
a,bGR. Prove
for every is
A
ideal in R.
(The set
+ b) =
+ 1^
and hnlz are also ideals
G R} form
R.
in
R ^ R' is called a homomorphism (or: ring homomorphism) if + f{b) and (ii) f(ab) = /(a)/(6), R ^ R' is a homomorphism, then the set K = {r G R f{r) = 0} :
f(a)
that if f
K is
/j
this
:
:
called the kernel of
/.)
INTEGRAL DOMAINS AND FIELDS B.33.
Prove that
in
B.34.
Prove that
F=
B.35.
Prove that
D = {a+ 6\/2
B.36.
Prove that a
B.37.
Show
B.38.
A
B.39.
ab
=
ae,
a, b rational}
is
a
field.
a, b integers}
is
an integral domain but not a
an integral domain D, {a
+ byji
:
:
domain
D
that the only ideals in a field
K
finite integral
if
is
a
D
¥= 0,
then
b
=
c.
field.
field.
are {0} and K.
complex number a + bi where a, b are integers of Gaussian integers is an integral domain. Also Let
a
is
called a
Gaussian integer.
show that the units
G
in
are
Show that the ±1 and ±i.
set
G
be an integral domain and let / be an ideal in D. Prove that the factor ring D/I is an integral if and only if / is a prime ideal. (An ideal / is prime if ab G I implies aG I or b&I.)
domain B.40.
Consider the integral domain D = {a + b\/l3 define N(a) = a^-lSb^. Prove: (1) NiaP)
we
(iii)
the units of
D
are ±1, 18
±
5-\/l3
a, b integers}
:
=
and -18
N{a)N(fi);
±
5\/l3
;
(see
Example
a
a unit
(ii)
(iv)
is
the numbers
B.ll).
if 2,
If
and only 3
- a/13
— a+ by/Ts N(a) = ±1; and -3 - y/Ts a
if
are irreducible.
MODULES B.41.
B.42.
B.43.
M
Let be an iJ-module and submodules of M.
A
and
B
be submodules of M.
M
Show
that
A+B
and
AnB
are also
Let be an iZ-module with submodule N. Show that the cosets {u + N u G M} form an iJ-module under coset addition and scalar multiplication defined by r{u + N) = ru + N. (This module is denoted by M/N and is called the quotient module.)
Let
M
:
M
M
-* M' and M' be B-modules and let f be an iZ-homomorphism. f(u) = 0} is a submodule of /. (The set K is called the kernel of
K = {uGM: B.44.
let
:
Show
that the set
/.)
M
Let be an i?-module and let E(M) denote the set of all fi-homomorphism of into itself. Define the appropriate operations of addition and multiplication in E{M) so that E(M) becomes a ring.
Appendix
C
Polynomials over a Field INTRODUCTION We will investigate polynomials
K
and show that they have many properties over a field integers. These results play an important role of the to properties are analogous which T on a vector space V over K. linear operator for a forms canonical in obtaining
RING OF POLYNOMIALS
X be a field. Formally, a polynomial / over K is K in which all except a finite number of them are 0: from
an
Let
=
/
(
.
.
.
0, On,
,
.
. ,
.
ai,
infinite
sequence of elements
tto)
so that it extends to the left instead of to the right.) The entry ak the kth coefficient of /. If n is the largest integer for which a„ ¥- 0, then we say that the degree of / is n, written
(We write the sequence is called
deg/ = n
We
also call a„ the leading coefficient of
the other hand, if every coefficient of / is = 0. The degree of the zero polynomial
/
Now
if
g
is
is, if
sum
m—n
f+g then
-
+g =
(
Furthermore, the product fg
the kth coefl5cient Cfc
=
.
is
=
fg is,
{..
.,0,bm,
.
.
.,bi,bo)
the polynomial obtained by adding corresponding coefficients.
is
f
that
is
another polynomial over K, say
g then the
On if a„ = 1 we call / a monic polynomial. then / is called the zero polynomial, written not defined.
and
/,
Ck
.
,
.
.
.
.
,
+ 6m,
a™
.
.
.
,
ai
+ 6i,
ao
+ bo)
the polynomial (
.
of fg
2^
0, a„,
That
.
.
,
0, anbm,
.
.
. ,
aibo
+ aobi, Oobo)
is
ttibk-i
=
aobk
+
aibk-i
+
•
•
•
+
akbo
i=0
The following theorem
Theorem
C.l:
applies.
P of polynomials over a field K under the above operations of addiand multiplication forms a commutative ring with a unit element and with no zero divisors, i.e. an integral domain. If / and g are nonzero polynomials in P, then deg (fg) = (deg /)(deg g). The
set
tion
327
POLYNOMIALS OVER A FIELD
328
NOTATION We identify
GX
the scalar ao
with the polynomial
=
ao
We
also choose a symbol, say
t,
call
the symbol
=
t'
=
{..
Thus the above polynomial
.,
the symbol
is
t
,
.
0,
tto)
Multiplying
=
t'
0, 1, 0, 0),
(.
.
with
t
.,
itself,
=
Unt"
+
•
+
•
ait
we
+
obtain
...
0, 1, 0, 0, 0),
/ can be written uniquely in the usual /
When
.
(...,0,1,0)
an indeterminant.
t
.
(
to denote the polynomial t
We
[APPENDIX C
form
ao
selected as the indeterminant, the ring of polynomials over
K
is
denoted by
and a polynomial /
We
also
frequently denoted by
is
view the
field
X as a subset of K[t]
(..., 0,6o)
=
{...,0, ao
(..., 0,ao) •(..., 0, 6o)
-
(...,0, aobo)
+
remark that the nonzero elements of
We
This is posare preserved
identification.
K
identification:
(...,0, ao)
We
under the above
and multiplication of elements of
sible since the operations of addition
under this
f(t).
K are
+ bo)
the units of the ring K[t].
remark that every nonzero polynomial is an associate of a unique monic polyHence if d and d' are monic polynomials for which d divides d' and d' divides d, then d = d'. (A polynomial g divides a polynomial / if there is a polynomial k such that also
nomial. /
=
hg.)
DIVISIBILITY The following theorem formalizes the process known as "long
Theorem
C.2 (Division Algorithm)
K
with g ¥=0. Let / and g be polynomials over a field Then there exist polynomials q and r such that
:
/
where Proof:
If f
—
or
if
deg /
<
deg
g,
f
Now
suppose deg / /
where
a„,
bm
?^
=
—
deg
Unt"
and n
+
either r
we have
then
=
=
Og
+
=
qg
+
or deg r
r
< deg
g.
the required representation
f
say
g, •
division".
•
•
+ait +
— m. We
ao
and
=
g
hmt^
+
+
hit
+
bo
form the polynomial Om
Then deg /i < deg /.
By
induction, there exist polynomials qi /i
=
qig
+
r
and r such that
APPENDIX
POLYNOMIALS OVER A FIELD
C]
where either r
=
or degr
<
deg
Substituting this into
= (qi+^t^-Ag +
/
which
g.
329
{1)
and solving for
/,
r
the desired representation.
is
Theorem
The ring K[t] of polynomials over a field X is a principal ideal ring. If / is an ideal in K[t], then there exists a unique monic polynomial d which gen-
C.3:
erates
/,
that
is,
such that d divides every polynomial /
G 7.
Proof. Let d be a polynomial of lowest degree in 7. Since we can multiply d by a nonzero scalar and still remain in 7, we can assume without loss in generality that d is a monic polynomial. Now suppose / G 7. By Theorem C.2 there exist polynomials q and r such that
=
/
qd
+r
where either r
=
or deg r
<
deg d
f,d G I implies qd G I and hence r = f — qd E I. But d is a polynomial of lowest degree in 7. Accordingly, r = and / = qd, that is, d divides /. It remains to show that is d unique. If d' is another monic polynomial which generates I, then d divides d' and d' divides d. This implies that d = d', because d and d' are monic. Thus the theorem is proved.
Now
Theorem
Let / and g be nonzero polynomials in K[t]. Then there exists a unique monic polynomial d such that: (i) d divides / and g; and (ii) if d' divides / and g, then d' divides d.
C.4:
Definition:
is called the greatest common divisor of / and then / and g are said to be relatively prime.
The above polynomial d d
—
1,
g.
If
Proof of Theorem CA. The set 7 = {mf + ng m,nG K[t]} is an ideal. Let d be the monic polynomial which generates 7. Note f,g G I; hence d divides / and g. Now suppose d' divides / and g. Let / be the ideal generated by d'. Then f,g GJ and hence Icj. Accordingly, d Gj and so d' divides d as claimed. It remains to show that d is unique. If di is another (monic) greatest common divisor of / and g, then d divides di and di divides d. This implies that d — di because d and di are monic. Thus the theorem is proved. :
Corollary C.5:
Let d be the greatest common divisor of the polynomials / and g. Then there exist polynomials and n such that d = mf + ng. In particular, if and n such and are relatively prime then there exist polynomials / g that mf + ng = 1.
m
The
corollary follows directly
7
m
from the fact that d generates the
ideal
= {mf + ng:m,nGK[t]}
FACTORIZATION
A / or
gr
polynomial p is a scalar.
Lemma
C.6:
G
K[t]
Suppose p f,g
G
of positive degree is said to be irreducible
G K[t]
K[t], then
is
irreducible.
p divides
product of n polynomials
p
f or
/1/2.
.
if
p
— fg
implies
p divides the product fg of polynomials More generally, if p divides the then p divides one of them.
If
divides g.
.fn,
Proof. Suppose p divides fg but not /. Since p is irreducible, the polynomials / and must then be relatively prime. Thus there exist polynomials m,nG K[t] such that p — 1. Multiplying this equation by g, we obtain m,fg + npg = g. But p divides fg + np mf and so mfg, and p divides npg; hence p divides the sum g = mfg + npg.
?-i«i./i>'—--r--
.Tii?-«'-TJ?»7^T^-i--
T
"""" "'??^'?2?CT5^-';fTr.js"'w«-5.a
POLYNOMIALS OVER A FIELD
830
[APPENDIX C
Now suppose p divides /1/2. ./«. If p divides /i, then we are through. If not, then by the above result p divides the product /2. ./«. By induction on n, p divides one of the polynomials A, Thus the lemma is proved. /«. .
.
.
.
. ,
Theorem C.7 (Unique Factorization Theorem): Let / be a nonzero polynomial Then / can be written uniquely (except for order) as a product
=
/
GK
where k
and the
kpiP2.
.
in K[t].
.Pn
are monic irreducible polynomials in K[t].
Pi
Proof: We prove the existence of such a product first. If / is irreducible or if / e K, then such a product clearly exists. On the other hand, suppose f = gh where / and g are nonscalars. Then g and h have degrees less than that of /. By induction, we can assume
-
g
where
is
ki,
kiGK
and the
gi
and
hj
/
=
h —
and
kigig2...gr
kihihi.
.
.hs
are monic irreducible polynomials. {kik%)gig2.
.
.grhjii.
Accordingly,
.hs
.
our desired representation.
We next prove uniqueness
(except for order) of such a product for
/
kpiP2.
.
.
.
kpi.
By
induction,
the
qi.
.
.Pn
=
k'qiQi.
.
Suppose
.Qm
.,qm are monic irreducible polynomials. Now pi ., Pn, qi, Since pi is irreducible it must divide one of the qi by the above lemma. Since pi and qi are both irreducible and monic, pi = qi. Accordingly,
where k,k' E.K and the Pu divides k'qi . . .qm. Say pi divides qi.
=
/.
we have
n=
that
We also have that
k
=
m
k'.
.
.
.pn
=
k'qi.
.
.qm
and P2 = 92, for some rearrangement of ., Pn = qm Thus the theorem is proved. .
.
K
If the field is the complex field C, then we have the following result which is as the fundamental theorem of algebra; its proof lies beyond the scope of this text.
Theorem
C.8
known
(Fundamental Theorem of Algebra): Let /(<) be a nonzero polynomial over the complex field C. Then f{t) can be written uniquely (except for order) as a product /(*)
where
k,
In the case of the real
Theorem
C.9:
nG
C,
field
i.e.
B we
= k{t-ri){t-r2)---it-rn)
as a product of linear polynomials.
have the following
result.
Let f{t) be a nonzero polynomial over the real field R. written uniquely (except for order) as a product f{t)
where or two.
kGK
and the
Pi{t)
=
kpi{t)p2{t)
Then
f{f)
can be
Pm{t)
are monic irreducible polynomials of degree one
INDEX Column, of a matrix, 35
Abelian group, 320 Absolute value, 4
rank, 90 space, 67 vector, 36
Addition, in R", 2 of linear mappings, 128 of matrices, 36
Companion matrix, 228 Complex numbers, 4
Adjoint, classical,
Components, 2 Composition of mappings, 121 Congruent matrices, 262 Conjugate complex number, 4
176
operator, 284
Algebra, isomorphism, 169
Consistent linear equations, 31 Convex, 260 Coordinate, 2 vector, 92 Coset, 229
of linear operators, 129 of square matrices, 43
Algebraic multiplicity, 203 Alternating, bilinear forms, 262 multilinear forms, 178, 277 Angle between vectors, 282 Annihilator, 227, 251 Anti-symmetric bilinear form, 263 operator, 285 Augmented matrix, 40
Cramer's
rule,
177
Cyclic group, 325 Cyclic subspaces, 227
Decomposition, direct sum, 224 primary, 225
Binary
Degenerate bilinear form, 262 Dependent vectors, 86 Determinant, 171 Determinantal rank, 195 Diagonal matrix, 43 of a matrix, 43
C,4
Diagonalization, Euclidean spaces, 288 unitary spaces, 290 vector spaces, 155, 199
Basis, 88
change
of,
153
Bessel's inequality, 309
Bijective mapping, 123
Bilinear form, 261, 277 relation, 318 Block matrix, 45 Bounded function, 65
Dimension, 88 Direct sum, 69, 82, 224 Disjoint, 316 Distance, 3, 280 Distinguished elements, 41 Division algorithm, 328
C», 5
Cayley-Hamilton theorem, 201, 211 Canonical forms in Euclidean spaces, 288 unitary spaces, 290 vector spaces, 222 Cauchy-Schwarz inequality, 4, 10, 281
Domain, integral, 322
Cells, 45
Change of
basis,
of a mapping, 121 Dot product,
153
Characteristic,
in C", 6
equation, 200
in R", 3
matrix, 200 polynomial, 200, 203, 210 value, 198 vector, 198 Classical adjoint, 176 Co-domain, 121 Coefficient matrix, 40 Cofactor, 174
Dual basis,
250
space, 249
Echelon form, linear equations, 21
matrices, 41
331
INDEX
332
Echelon matrix, 41 Eigenspace, 198, 205 Eigenvalue, 198 Eigenvector, 198
Element, 315 Elementary,
column operation, 61 divisors, 229 matrix, 56 row operation, 41 Elimination, 20
Empty
set,
315
Equality of matrices, 36 of vectors, 2
Equations (see Linear equations) Equivalence relation, 318 Equivalent matrices, 61 Euclidean space, 3, 279
Integral domain, 322 Intersection of sets, 316 Invariant subspace, 223 Inverse,
mapping, 123 matrix, 44, 176 Invertible,
linear operator, 130
matrix, 44 Irreducible, 323, 329
Isomorphism of algebras, 169 groups, 321 inner product spaces, 286, 311 vector spaces, 93, 124
Jordan canonical form, 226 Kernel, 123, 321, 326
Even function, 83
permutation, 171 External direct sum, 82 Field, 323
Free variable, 21 Function, 121 Functional, 249
Gaussian integers, 326 Generate, 66
Geometric multiplicity, 203
Gram-Schmidt orthogonalization, 283 Greatest common divisor, 329 Group, 320
Hermitian, form, 266 matrix, 266 Hilbert space, 280 Hom (V, U), 128
Homogeneous linear equations, 19 Homomorphism, 123 Hyperplane, 14 Ideal, 322
Identity,
element, 320
mapping, 123 matrix, 43 permutation, 172 Image, 121, 125 Inclusion mapping, 146
Independent subspaces, 244 vectors, 86
Index of nilpotency, 225 set,
316
22-space, 280
Line segment, 14, 260 Linear combination of equations, 30 of vectors, 66
Linear dependence, 86 in R", 28 Linear equations, 18, 127, 176, 251, 282 Linear functional, 249 Linear independence, 86 in R", 28 Linear mapping, 123 matrix of, 160 rank of, 126 Linear operators, 129 Linear span, 66
Mapping, 121 linear, 123 Matrices, 35 addition, 36
augmented, 40 block, 45
change of
basis, 153
40 column, 35 congruent, 262 determinant, 171 diagonal, 43 echelon, 41 coefficient,
equivalent, 61
Hermitian, 266 identity, 43 multiplication, 39
normal, 290 rank, 90 row, 35 row canonical form, 42, 68
row row
equivalent, 41
space, 60
Infective mapping, 123 Inner product, 279
scalar, 43
Inner product space, 279 Integers modulo re, 323
similar, 155
scalar multiplication, 36 size,
35
INDEX Matrices
333
Primary decomposition theorem, 225 Prime ideal, 326
(cont.)
square, 43
Principal ideal, 322 Principal minor, 219
symmetric, 65, 288 transition, 153
Product
transpose, 39 triangular, 43 zero, 37 Matrix representation, bilinear forms, 262 linear mappings, 150
Proper subset, 316
value, 198
Maximal independent set, 89 Minimal polynomial, 202, 212 Minkowski's inequality, 10 Minor, 174 Module, 323 Monic polynomial, 201 Multilinear, 178, 277
vector, 198
Q
(rational numbers), 315 Quadratic form, 264
Quotient,
group, 320 module, 326 ring, 322 set, 319 space, 229
Multiplication of matrices, 37, 39
N
317
set,
Projection operator, 243, 308 orthogonal, 281
(positive integers), 315
n-space, 2 n-tuple, 2
K
Nilpotent, 225
R", 2
Nonnegative semi-definite, 266
Rank,
linear mapping, 127
matrix, 130 Norm, 279 in R«, 4
290, 303
Normalized vector, 280 Null set, 315 Nullity, 126 Odd, function, 73
permutation, 171 One-to-one mappings, 123 Onto mappings, 123 Operations with linear mappings, 128 Operators (see Linear operators) Ordered pair, 318
Orthogonal complement, 281 matrix, 287 operator, 286 vectors, 3, 280 Orthogonally equivalent, 288 Orthonormal, 282
Parallelogram law, 307 Parity, 171 Partition, 319 Permutations, 171 Polar form, 264, 307 Polynomials, 327 Positive
matrix, 310 operator, 288 Positive definite,
315
bilinear form, 262
Nonsingular,
Normal operator, 286, Normal subgroup, 320
(real field),
linear mapping, 126 matrix, 90, 195 Rational canonical form, 228 Relation, 318 Relatively prime, 329 Ring, 322
Row, canonical form, 42 equivalent matrices, 41 of a matrix, 35 operations, 41 rank, 90 reduced echelon form, 41 reduction, 42 vector, 36
Scalar, 2, 63
mapping, 219 matrix, 43 Scalar multiplication, 69 of linear mappings, 128 of matrices, 36 Second dual space, 251 Self -adjoint operator, 285 Set, 315 Sgn, 171 Sign of a permutation, 171 Signature, 265, 266 Similar matrices, 155 Singular mappings, 127 Size of a matrix, So Skew-adjoint operator, 285 Skew-symmetric bilinear form, 263 Solution, of linear equations, 18, 23
space, 65
bilinear form, 265
Span, 66
matrix, 272, 310 operator, 288
Spectral theorem, 291 Square matrices, 43
J
334
INDEX
Subgroup, 320 Subring, 322 Subset, 315
Subspace
(of
a vector
Unique factorization, 323 Unit vector, 280 space), 65
sum of, 68 Surjective mapping, 123 Sylvester's theorem, 265 Symmetric, bilinear form, 263 matrix, 65 operator, 285, 288, 300 System of linear equations, 19
Unitarily equivalent, 288 Unitary, matrix, 287 operator, 286 space, 279 Universal set, 316 Upper triangular matrix, 43 Usual basis, 88, 89
Vector, 63
Trace, 155 Transition matrix, 153 Transpose, of a linear mapping, 252 of a matrix, 39 Transposition, 172 Triangle inequality, 293 Triangular, form, 222 matrix, 43 Trivial solution, 19
Union of
sets,
316
in C", 5
in R", 2
Vector space, 63
Venn diagram, 316
Z
(integers), 315
Z„ (ring of integers modulo Zero,
mapping, 124 matrix, 37 of a polynomial, 44 solution, 19 vector, 3, 63
w),
323
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