INTRODUCTORY LINEAR ALGEBRA
Dr. Balram Dubey
Asian Books Private Limited
INTRODUCTORY LINEAR ALGEBRA
"This page is Intentionally Left Blank"
INTRODUCTORY LINEAR ALGEBRA
Dr. Balram Dubey Department of Mathematics Birla Institute of Technology and Science, Pilani
~ ;4sian 'B""ks 'iJJ'ioaie t..iHliie'J 7128, Mahavir Lane, Vardan House, Ansari Road, Darya Ganj, New Delhi-ll0002
Registered and Editorial office 7/28, Mahavir Lane, Vardan H()Use, Ansari Road, Darya Ganj, New Delhi-I 10 002. E-Mail:
[email protected];
[email protected] World Wide Web: http://ww\... asianbooksindia.com Phones: 23287577, 23282098, 23271887, 23259161 Fax: 91-11-23262021
Sales Offices Baf/galore 103, Swiss Complex No. 33, Race Course Road, Bangalore-560 001 Phones: 22200438, Fax: 91-80-22256583, Email:
[email protected] Chellna: Palani Murugan Building No.21, West Cott Road, Royapettah, Chennai-600 014 Phones: 28486927. 28486928, Email:
[email protected] Delhi 7/28. Mahavir Lane, Vardan House, Ansari Road, Darya Ganj, New Delhi-l 10002 Phones: 23287577,23282098,23271887,23259161, !:ax : 91-11-23262021 Email:
[email protected];
[email protected] Guwahati 6, GN.B. Road, Panbazar, Guwahati, Assam-781 001 Phones: 0361-2513020, 2635729 Email:
[email protected] Hyderabad3-5-1101l11B Hnd Floor, Opp. Blood Bank, Narayanguda, Hyderabad-500 029 Phones: 24754941, 24750951, Fax: 91-40-24751152 Email:
[email protected]@eth.net Kolkata lOA, Hospital Street, Calcutta-700 072 Phones: 22153040, Fax: 91-33-22159899 Email:
[email protected] Mumbai 5, Himalaya House, 79 Palton Road, Mumbai-400 001 Phones: 22619322, 2~623572, Fax: 91-22-2262313 7 Email:
[email protected] Showroom : 3 & 4, Shilpin Centre, 40, GD. Ambekar Marg, Wadala, Mumbai-·~~) 031; Phones: 241576li, 24157612 © Publisher 1st Published: 2007 ISBN: 978-81-8412-019-6 All rights reserved. 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 and/or otherwise, without the prior written permission of the publisher. Published by Kamal Jagasia for Asian Books Pvt. Ltd., 7128, Mahavir Lane, Vaidhan House, Ansari Road, Darya Ganj, New Delhi-110 002 Laser Typeset at P.N. Computers S/:ahdara De!hi-JJO 032 Printed at Gopa/jee Enterprises, Delhi
7n J\1y wifo rama, and our little sons ratsav and ratk.arsh, whose support, encouragement and faithful prayers made this book. possible
"This Page is Intentionally Left Blank"
PREFACE
The main aim of writing this book is to provide a text for the undergraduate students in science and engineering. This book is designed to meet the requirements of syllabus of various Engineering Institutes at the undergraduate level. It is written to facilitate teaching linear algebra in one semester or in a part of it. The presentation of material in this book preassumes the knowledge of determinant of a matrix, gained from class XII. Throughout the text, material is presented in a very smooth and easily understandable manner. In each chapter, theories are illustrated by numerous examples. In each section, exercises of both theoretical and computational nature are included. Theoretical exercises provide experience in constructing logical and deductive arguments while computational exercises illustrate fundamental techniques useful in the field of business, science and engineering. The first chapter deals with the linear system of equations with the help of elementary row operations. The aim of keeping this topic in the first chapter is to provide an efficient development of the basic theory, used in all forth coming chapters. In Chapter 2, vector spaces, linear dependence and independence, basis and dimension are explained. Chapter 3 treats linear transformations, their rank and nullity, inverse of linear transformations and linear functionals. Chapter 4 treats the representation of linear transformation in terms of matrix, representation of matrix in terms of linear transformation, rank and column space of a matrix. Chapter 5 includes a complete treatment of eigenvalues and eigenvectors of matrices, and diagonalization of matrices. Although all possible care has been taken while preparing this text, there may be a possibility of error. I will be thankful to the readers for any such information so that it may be incorporated in the subsequent edition. I also invite constructive suggestions for further improvement. I believe that this book will fulfill the needs of first-year science and engineering students at the undergraduate level in linear algebra.
Balram Dubey
"This Page is Intentionally Left Blank"
ACKNOWLEDGEMENTS
I would like to thank Professor S. Venkateswaran (Former Vice-Chancellor, BITS, Pilani) and Professor L. K. Maheswari (Vice-Chancellor, BITS, Pilani) for inspiring and encouraging to write this book to meet the requirements of undergraduate students. I am very much thankful to the administration of BITS, Pilani for their all round supports. I am very grateful to many of my colleagues, especially to the faculty members of Mathematics Department at BITS, Pilani who directly or indirectly helped me in various ways during the preparation of this book. I would like to convey my sincere appreciation to the students of BITS Pilani, lIT Kanpur and Central University of Tezpur (where I have taught at undergraduate and postgraduate levels) for their useful opinions, suggestions and discussions. I convey my sincere thanks to Professor O. P. Juneja whose lectures (delivered in 1988 when I was a student at lIT Kanpur) helped me a lot. I wish to express my deep gratitude to Professor J. B. Shukla ( Former Professor of Mathematics at lIT Kanpur) for his invaluable suggestions, indefatigable guidance and encouragement throughout my academic career. It is beyond the words to describe what I have received from my wife Dr. Uma S. Dubey, my two little sons Utsav and Utkarsh, my mother Mrs. Raj Kumar Devi and my father Mr. Rama Shankar Dubey, yet I make an eiTort to express my heartiest gratitude to them. I thank Mr. Narendra Saini who typed this book with a great care. Finally, my sincere thanks goes to Mr. R. Taneja (General Manager), Mrs. P. Biswas (Production Manager) and editing staffs of Asian Books Private Limited for gladly accepting this text book for publication.
Balram Dubey
"This Page is Intentionally Left Blank"
CONTENTS
Preface Chapter 1.
vii System of Linear Equations ............... 1-20 1.1. Fields ........................................................................ I 1.2. System of Linear Equations ................................... 3 1.3. Inverse of a Matrix ............................................... 15
Chapter 2.
Vector Spaces ..................................... 21-58 2.1. 2.2. 2.3. 2.4.
Chapter 3.
Linear Transformations ..................... 59-91 3. I. 3.2. 3.3. 3.4. 3.5. 3.6.
Chapter 4.
Vector Spaces ........................................................ 21 Subspaces .............................................................. 26 A Subspace Spanned by a Set ........................... 33 Bases and Dimension ........................................... 39
Linear Transformations ......................................... 59 Range and Null Spaces ........................................ 64 The Algebra fo Linear Transformations ............. 73 Inverse of a Linear Transformation ..................... Tl Idempotent and Nilpotent Operators .................. 83 Linear Functionals ................................................. 86
Matrix Representation of a Linear Transformations ................... 92-109 4.1. Coordinates ............................................................ 92 4.2. Representation of a Linear Transformation in Terms of a Matrix .................. 95 4.3. Representation of a Matrix in Terms of a Linear Transformations ............................... 100 4.4. Row and Column Spaces of a Matrix ............... 104
xii
Chapter 5.
Contents
Eigenvalues and Eigenvectors ... 110-132 5.1. Null Space ofa Matrix ....................................... no 5.2. Eigenvalues and Eigenvectors ........................... 112 5.3. Diagonalization of Matrices ............................... 126
Bibliography ................................................................................ 133 Index
....................................................................... 135-136
SYSTEM OF LINEAR EQUATrONS &
One of the important problems in linear algebra is the solution of linear equations. This chapter introduces to the reader some basic concepts of solving a system of linear equations. rn order to s~1ve a system of linear equations we shall use the word "Field" frequently. Hence it is necessary for the readers to know first about a field.
1.1
FIELDS
We assume that F denotes either the set of real numbers 9\ or the set of complex numbers C. Suppose that F has two operations called addition (+) and multiplication C.). The elements of the set Fhave the following properties: I. F is closed with respect to addition:
x +y
E
F for all x, y in F.
2 Addition is associative:
x+(y+z)
=
(x+y)+z
for all x, y and z in F. 3. There is a unique element 0 (zero) in F such that x + 0 = x = 0 + x for all x in F. The element 0 is. known as the additive identity of F. 4. To each x in F, th~re corresponds a unique element C-x) in F such that x + C-x) = 0 = C-x) + x. The element -x is known as the additive inverse of x in F. 5. Addition is commutative: x+y=y+x for all x and y in F. 6. F is closed with respect to multiplication: xy E F for all x and y in F.
Introductory Linear Algebra
2
7. Multiplication is associative: x (yz) = (xy) z for all x, y and z in F. 8. There is a unique non-zero element 1 (one) in F such that
x·l=x=l·x for every x in F. The element I is known as the multiplicative identity of F
9. To each non-zero x in F, there coressponds a unique element in F such that x X-I The element x-I 10. Multiplication is xy = yx for all x 11. Multiplication is
X-I (
or
~)
=I = X-IX.
is known as multiplicative inverse of x in F. commutative: and y in F. distributive over addition:
x(y+ z) = xy+ xz, (x+ y)z
=
xz+ yz,
for x, y and z in F. If the set F, together with these two operations, satisfies all conditions (I) - (ll), then F is called a field.
9\ (the set of real numbers) under the usual addition and multiplication is a field. Example 2. C (the set of complex numbers) under the usual addition and multiplication is a field. Example3. Q (the set of rational numbers) under the usual addition and multiplication is a field. Example 1.
Example 4. The set of positive integers: I, 2, 3, ..... is not a field for several reasons. For example, 0 is not a positive integer; there is no positive integer n for which -n is a positive integer; there is no positive integer n (except I) for which positive integer.
~ n
is a
Example 5. The set of integers: ... , -2, -I, 0, 1, 2, ... , is not a field, because there is no integer n for which or -I.
n
is an integer unless n is 1
3
System of Linear Equations
Now throughout our discussion in this text book, we shall assume that F = 9\ or F = C unless otherwise stated.
1.2 SYSTEM OF LINEAR EQUATIONS A linear equation in the variables xpx2' ..... 'xn is an equation that can be written in the form
where aI' a2 , ••• , an and b are in F (9\ or C), usually known in advance. The equations
5 and 5xI + 2xJ = 0 are both linear. The equations
XI - x
2
=
XI X 2
and.,J-;; + 2X2 == I
are no~inear because of the presence of the term x 1x 2 in the first equation and ..J i in the second equation. A system of linear equations is a collection of more than one linear equations involving the same variables XI' x 2 ' ••••• , xn (say).
Consider the following system of m linear equations in n unknowns:
... (l.l)
all Let
A =
[
l
[XI]
[b ]
~n
b:
... a 2n ] x2 b2 . . , x== . ,b== . .
a:
. a:
1I1
1be matdx B
ln
a 21
~ [:~: :~,:
1I1I
::
l]
is called augmented matrix of the system (l.l).
1I
Introductory Linear Algebra
4
The matrix notation of the linear system of equations given in equation ( l.l) is written as ...(1.2) Ax = b. If b = 0, then equation (1.2) is called homogeneous otherwise inhomogeneous. A solution of the system is a list of numbers (s" $.,2' ••• , sn) that makes each equation (1.1) a true statement when the values sI' s2' .. , snare substituted for xI' x 2 ' ••• , x n' respectively. \ The set of all possible solutions is called the solution set. Two linear systems are caIled equivalent if they have same solution set. That is, each solution of the first system is a solution of the second system and each solution of the second system is a solution of the first. We formulate this result for linear system in the foIlowing form: Equivalent system of linear equations have exactlyfthe same solutions. How to Solve a Linear System? We answer this question with a very simple example. Example 6. Suppose we wish to solve (a) x + y = 1, (b) x + y = 1, (c) x + Y = 1, x - V = o. 2x + 2y =2. x + Y =2.
Fig. 1.1. (a)
Fig. 1.1 (b)
Fig. l.1(c)
We note that system of equations in Ca) are paraIlel straight lines, in
Cb) both equations are straight lines intersecting at a unique point
(i,i)
and in Cc) both equations are straight lines coinciding with each other. This shows that the system of equations in Ca) has no solution, in (b) has exactly one solution, and in (c) there are infinitely many solutions. This example illustrates the following general fact for linear system ofeQuations.
5
System of Linear Equations
A system of linear equations has either no solution or exactly one solution or infinitely many solutions. A system of linear equations is said to be consistent if it has a solution (one solution or infinitely many solutions), a system is inconsistent if it has no solution. Remark: A linear homogeneous system of equations Ax = 0 is always consistent as it has always a trivial solution 0
=
(0, 0, ... , O)T.
We know that there is an exact formula caIIed Cramer's rule which gives the solutions of n equations in n unknowns as a ratio of two n by n determinants. But this formula is a disaster if the order of the matrix is high (even if greater than 4). Hence this formula is very difficult to be used in practical life. In this section, we shaII give a very simple technique to solve a linear system of equations. In order to understand this technique, one must know three elementary row operations on an m x n matrix A over the field F (9\ or C):
1. Multiplication of one row of A by a non-zero scalar
Introductory Linear Algebra
6
2. The zero rows (i.e., rows containing only zero), if any, occur below all the non-zero rows. 3. Let there be r non-zero row. If the leading entry of the ith row occurs in the column k j (i = I, 2, ... , r), then kl < k2 < ... < kr · For example, the matrices
are in row echelon form.
n~ m~ -~ !l
Definition. An m x n matrix A is called in row-repuced echelon form if the following conditions are satisfied:
I. The first non-zero entry in each non-zero row of A is equal to I (this I is called the leading one). 2. If a column contains the leading 1 (that is, first non-zero entry) of any row, then every other entry in that column is zero. 3. The zero rows, if any, occur below all the non-zero rows. 4. Let there be r non-zero rows. If the leading 1 of the ith row occurs in the column ki (i = I. 2, ... , r), then kl < k2 < ... < k,.. That is, in any two consecutive non-zero rows the leading I in the lower row appears farther to the right than leading I in the upper row.
Examples of row-reduced echelon matrix: l. Om xn: m x n zero matrix 2. In : n x n identity matrix.
3.
[~ ~ -~ ~l :
4'l~ ; ~ ~l where the starred entries (*) may take any values including zero.
System of Linear Equations
5.
*
0
I
0
0
0
0
0
0
0
I
0
0
0
0 0
0
I
0
0
0
0
0 0
0
0
0
0 0
0
0
0
0 0
7
I
* * * * *
0 0
0
* * * * * * * * 0
0 0
0
0 0 0 0
where the starred entries (*) may have any values including zero. Remark 1. Row echelon form of a matrix is not unique, but the row-reduced echelon form of a matrix is unique. Remark 2. The basic difference between row echelon form and row-reduced echelon form of a matrix are the following: (i) The leading entry in row echelon form of a matrix is any non zero number, while in row-reduced echelon form of a matrix the leading entry is always 1. (ii) In row echelon form of a matrix, if a column contains the leading entry of any row, then all entries below the leading entry are zero and all entries above the leading entry may be any number (including zero). But in row-reduced echelon form of a matrix, if a column contains the leading 1, then all other entries in that column are zero. Remark 2. shows that if a matrix is in row-reduced echelon form, then it is also in row echelon form. However, if a matrix is in row echelon form, then it need not be in row-reduced echelon form. Example 7. Obtain the row-reduced echelon form of the matrix
~ [: -5] 2
A
2
6
1 .
8
5
~ [: -;] 2
Solution.
A
6
8
-
[:
2
-4
--6]
6
I
8
5
R, --> R, - R, ,
Introductory Linear Algebra
8
[~ - [~ -
~] ,R2
-4
1~
18
-7
R2 -3Rl
-7
R3 -2Rl
8
-4 18
-6] 19 ,
16
17
[~
-4 2
-6]
16
17
[~ - [~ - [~ - [~
-4
~]I,R2-7R)2
-
R,
2 ,R2 -7 R2 - RI
I
17
16
0
-~] ,R, --> R, +411,
0
1 ,R3 -7 R3 - 16 R2
0 0] R, --> R, + 2R, 1 ,
0 0 1
0
1
n
11, --> R, - R,
Clearly, the final matrix is in the row-reduced echelon form. At this stage we are aware of the row-reduced echelon form of a matrix A. Therefore we are ready to answer the existence and uniqueness question for a linear system of equations.
Existence and Uniqueness Consider a linear equation
ax
=
h.
... (1.3)
System of Linear Equations
9
2. If a = 0 and b = 0, then there are infinitely many solutions of (1.3) as any x satisfies Ox = O. In this case solution exists and it is not unique. 3. If a = 0 and b "# 0, then there is no solution to (1.3). Similar situations may arise for a system of linear equations Ax = b. We ask the following two fundamental questions about a linear system: 1. Is the system consistent? 2. If the system is consistent, is the solution unique?
In order to answer the above two questions, one need to know the row-rank of a matrix. Definition. Number of non-zero rows in a row-reduced echelon form of a matrix A is called row-rank of A. Remark: We shall give a more general definition of row-rank of a matrix in Chapter 4. We know that for a system of In equations in n unknowns Ax = b, ...(1.4) (lA) the matrix B = (A, b) is called the augmented matrix. (i) The system (lA) (1.4) is consistent, i.e., the system (lA) (1.4) has a solution if and only if row-rank of A = row-rank of B = (A, b). Let row-rank of A = row-rank of B = r. (ii) If r = n = the number of unknowns, then solution is unique. (iii) If r <
11,
then there are infinitely many solutions.
(iv) The system (1.4) is inconsistent if and only if row-rank of A 't- rowrank of B.
In the following theorems, we shall establish two important results regarding homogeneous linear system of equations. Theorem 1.2. A homogeneous linear system of m equations in n unknowns always has a nontrivial solution if m < n, that is, if the number of unknowns exceeds the number of equations. Proof: Consider the homogeneous linear system of m equations in n unknowns Ax =0. Let R be the row-reduced echelon form of A. Then A is row equivalent to R. Hence by Theorem 1.1, it follows that Ax = 0 and Rx = 0 have th~ same solution set. Let r = the number of non-zero rows of R.
=> r
.~
m
Introductory Linear Algebra
10
Suppose that m < n. Then r < n. It means that we are solving r equations in n unknowns and can solve for r unknowns in terms of remaining 11 - r unknowns. These n - r unknowns are free variables that can take any real numbers. Ifwe take any free variable to be non-zero, we get a non-zero solution to Rx = 0, and thus a non-zero solution to Ax = o. The following corollary IS an immediate consequence of the above theorem. Corollary: If A is an m x n matrix and Ax m ~ 11.
=
0 has a trivial solution, then
Theorem 1.3. x = 0 is the only soluton to Ax = 0 if and only if row-rank of A equals the number of unknowns. Proof: Let R denotes the row-reduced echelon form of A. Then x = 0 is the only solution to Ax = 0 ¢:;> <:::> x = 0 is the only solution to Rx = 0 ¢:;> <:::> Number of non-zero rows in R = Number of unknowns ¢:;> <:::> Row-rank of R (=row rank of A) = Number of unknowns. Now we summarize the row-reduction process to solve the linear system (lA). (1.4). Using Row Reduction to Solve Linear System Ax = b. I. Write the augmented matrix of the system: B = (A, b)
2. Use elementary row operations on B so that A is transformed into rowreduced echelon form. 3. Decide the consistency or inconsistency of the system. If the system is inconsistent, stop the process, otherwise, go to the next step. 4. Write the system of equations corresponding to the matrix obtained in step 3. 5. Rewrite each non-zero equation from step 4 so that its one basic variable is expressed in terms of any free variables appearing in the equation. Now we shall give some examples on the consistency and inconsistency of linear system of equations. Example 8. Determine whether the following system of equations are consistent or not. If yes, determine all solutions. X I -X2 +X3
=2,
+ x 2 - X3 + 2X2 + 4X3
= 3,
2xI XI
=
7.
System of Linear Equations
11
Solution: The augmented matrix associated with the given system is
B
= (A,
b)
=
-
[~
-I
[~
I
-1
2
4
-1
1
3 -3 3 3
-1
-
-
0
2
5 3 ,R. 1 -1 -3
0 0
~RI+R2
,~~~12
0 0 0
,R2~R2/3
0
0 0
-
~R)-R,
-!l~ -->~-~
-1
0
--> R, -2R,
3 5 ,~ ~~/3 3
-1
0
5 ,R)
1
0
- [:
-~lR' 2
-1
0
~l
0
5 3 2 3 1
,Rz~R2+R)
Thus, ranK of the coefficient matrix = rank of augmented matrix = 3 -' number of unknowns. This implies that the system is consistent and the solution is unique given by
12
Introductory Linear Algebra
5 2 x I = -, X 2 =-, X3
3
3
=1 .
Example 9. Determine whether the following system of equations are consistent or not. If yes, find all possible solutions.
-XI -
5x2 + IOx3 = 7.
Solution: The augmented matrix associated with the given system is I
3]
2 B "(A, b)" [-: -3 4 2 -1 -5 10 7 1
2 6
3]5
-[i
-2
-[i
-2 6 5 0 o 0, R3
-
-
[:
-4
,R2~R2+RI
12 10 ,R3 2
~
R3
+ R,
3]
2
31
0
0
0
0
5
~
R3 - 2R2
-3 -~ 2 ' R2 ~-R/2 2
0 0 0
-3 0
11
2 ,R, ~ RI -R2 5
2 0
Thus, rank of coefficient matrix = rank of augmented matrix = 2 < number of unknowns.
=> The system is consistent and it has infinitely many solutions given by
13
System of Linear Equations
XI
=-1 (11- 10k), x 2 =-1 (6k 2
2
5), X3
=k, k E 9\
is arbitrary (i.e., x3 is a free variable). Example 10. Determine whether the following system of equations are consistent or not. If yes, find all possible solutions. XI -2X2 -x] +3x4 =0, -2xl + 4X2 + 5X3 - 5x4 =3,
3xI -6x2 -6x] +8x2 =2. Solution: The augmented matrix associated with the given system is
B
~
(A, b)
~
H
-1
-2
3
5 -5
4
-6 -6
8
-2
-1
- [i
0
3
1
0 -3
-1
[~
-2
-I
0
3
0
0 0
-
-2
- [:
3
-I
3
0
1 3 0
0 0 -2
~
3
0
0
(l
0
0
0
~l
~lR'
--> II, +2R,
2 ,R] -7 R] -3RI
~t ]11,
--> II, + R,
-->R,D
10
-
,RI -7 RI +R2 3 1 3 0 5
Thus, rank of coefficient matrix = 2, rank of augmented matrix = 3. Since rank of coefficient matrix"# rank of augmented matrix, the system is inconsistent.
Introductory Linear Algebra
14
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ __ 1. Which of the following matrices are in row-reduced echelon form?
2. Dl!termine the row-reduced echelon form for each of the matrices given below.
~ -~ ~ -~] (l)
(Ii,) (11,)
-1
r
1
2
4
3
2 -1
2
r~ j ;~]
3. Determine the solution set of the following system of homogeneous equations:
(l) 4xI + 2X2 - 5x3 =0, 3xI + 6x2 + X3 = 0, 2xI +8x2 +5x3 =0.
(it)
+X2 +X3 +X4 =0, x2 + X3 + x 4 = 0, +X2 +3X3 +3x4 =0. -x2 +5x3 +5x4 =0.
-XI
XI -XI XI
System of Linear Equations
15
4. Determine whether the following system of equations are consistent or not. If yes, determine all possible solutions. (I)
(iii) (v)
(vii)
XI + X2 + 2X3 = 3, -xl -3x2 +4X3 =2, -XI - 5x2 + JOX3 IOx3 =I I.
(ii) 9x, + 3x2 + xJ = 0, 3x, +xJ = 0, XI + x 2 + X3 =0, . -6x2 +xJ =9.
XI + 3x2 + 4X3 = 7, 3x, +9x2 + 7X3 = 6.
(iv)
3x, - 4X2 + 2X3 = 0, -9x, + 12x2 - 6X3 = 0, -6x, + 8x2 - 4xJ = 0.
(vi)
x 2 + 2xJ + 3x4 = 1, 2x, + 2X2 + 0,X3 + 2X4 = 1, 4xI + x 2 - xJ - x 4 = 1, XI + 2X2 + 3xJ + 0,x4=1.
5. Let A =
XI -
(vii)
XI + 4X2 + 0,x3 =7, 2x, + 7X2 + 0,x3 =10.
x, -7X2 + O.xJ + 6x4 = 5, O.X, + 0,X2 +x3 - 2X4 =-3, -x, + 7X2 - 4xJ + 2X4 = 7.
x,-3x2+X3 -x4 =7, 2x, +4X2 -XJ +6x4 =-6, 2x, +X2 +O,XJ +X4 =O.
[~ ~ ~ ~ ~l 00012
Determine all b for which the system AX = b is consistent. 6. Let RI and R2 be 2 x 3 real matrices that are in row reduced echelon form and the system RIX = 0 and R~ = 0 are equivalent. Without making use of any result, show that RI Rl = R2 • 7. Let AX = b be a consistent non-homogeneous system. Prove or disprove: If Xl XI and X 2 are solutions of the system AX = b, then so is Xl + X 2 • 8. If Xl is a solution of the non homogeneous system AX = band Yl is a solution of the corresponding homogeneous system AX = 0, then XI + Yl is a solution of the system AX = b. Moreover, if Xo show that Xl is any solution of the system AX = b, then there is a solution Y Yoo of the system AX = 0 such that Xo = Xl XI + Yo' 9. Prove or disprove: If two matrices have the same row-rank, then they are row equivalent.
1.3 THE INVERSE OF A MATRIX Defintion: An n x n matrix A is is said to be invertible (also called nonsingular) if there exists an n x n matrix C such that AC = 1= CA,
Introductory Linear Algebra
16
where I = In' the n x 11 identity matrix. In this case C is called inverse of A. Note that C, the inverse of A, is always unique, because if B is another inverse of A, then AB = 1= BA, and B = BI = B (AC) = (BA) C = IC = C . Thus, inverse of a matrix A is unique and is denoted by A-I so that AA-I = = 1= I = A-lA A-IA
...(1.5)
A matrix that is not invertible is called singular matrix. Now let us motivate the concept of finding the inverse of an n x n matrix A. For this purpose we start from A and consider the matrix equation Ax = y, ...(1.6) where x and yyare are column vectors with n components XI' x 2' ..• , xn and Yl' Y2'''' Yn' respectively. I f the inverse matrix A-I exists, we may pre-multiply equation (1.6) by A-I and have A-I (Ax) = A-Iy => (A- I A) xX = A-Iy
=> Ix = A-Iy =>
X
= A-Iy
The above equation expresses x in terms of y, and it involves A-I. For a given y, equation (1.6) may be regarded as a system of 11 linear equations in n unknowns x" x 2 , ••• , x" and we know from previous section that it has a unique solution if and only if row-rank of A = n, the greatest posible rank for an n x n matrix. Thus, we can state the following theorem. Theorem 1.4. (Existence of Inverse) For an n x 11 matrix A, the inverse A-I exists if and only if row-rank of 11. Hence A is non singular if row-rank of A = n, and is singular if rowrank of A < 11. From Theorem 1.3, we note that the homogeneous linear system of equations
A=
Ax
=0
has a trivial solution if and only if row-rank of A equals the number of unknowns. If A is 11 X 11 matrix, then from Theorem 1.4 lA it follows that A is non singular <=> row-rank of A equals n <=> Ax = 0 has a trivial solution. Thus, we can state the following theorem.
Theorem 1.5. If A is n x n matrix, the homogeneous linear system of equations Ax = 0 has a nontrivial solution if and only if A is singular.
l
System of Linear Equations
17
Algorithm for FindingA-1: I. If A is n x n matrix, take an n x n identity matrix 1. 2. Place A and 1 side by side to form an augmented- matrix. B = [A 1]. 3. Perform elementary row operations on B. 4. If [A I] is transformed into [I C], then C = A-I; otherwise A does not have an inverse.
Remark 1. Note that by performing elementary row operations on
B = [A
1]
if it is possible to find row-reduced echelon form of A as I,
then row-rank of A = n and so A is non singular. If row-reduced echelon form of A I by performing elementary row operations on B = [A 1], then row-rank of A < n, and so A is singular.
*
Remark 2. If we wish to check whether the given n x n matrix is singular or non-singular, then perform elementary row operations only on A. Ifrowreduced echelon form of A is an n x n identity matrix I i.e., if row-rank of A is equal to n, then A is nonsingular; otherwise A is singular. Example 11. Let us consider the matrix
A
~ [~ -~ -~]
Note that the matrix A is the coefficient matrix of Example 8. We wish to find A-I, if it exists. We will write
[I -I 11
~]
0
B = [A
I]
=
2
1 -1 0
1
2
4 0 0
- [~
-1 3
3
- [~
-I
I
I
0
3 -3 -2
1
-I
0 1
~l1l, -->11,-211, 1 ,R) 0
1 -1 -2/3 113
-113
~R)-Rl
0]
o
,R2~R2/3
0 1/3 ,R) ~ R) 13
Introductory Linear Algebra
18
-
[~
1/3 0 0 113 -1 -2/3 1/3 1 1/3 -1/3 0 2
-
[~
1/3 0 0 1/3 -1 -2/3 1/3 -1/6 1/6 0
-
[~
0 0 1/3 1 0 -1/2 1/ 6 1/6, Rz ~ R2 + R3 0 116 -1/6 1/6
1/~ ,R3 ~ R3 -R2
~JI<,
-> 1<,12
113 0]
A-I
Thus,
O]'R' -> R, +R,
=
lI~l
["3 113 -1/2 1/6
1/6 -1/6 1/6 Example 12. Let us consider the matrix
A
=
[I -I 2
1
1
2
n
In order to find A-I, if it exists, we write
B= [A
~ ~ [~
-1
- [~
-1
-
[~
0 3
2 3
1 0I 00]
o o
0
0 3 3 -2 3 3 -1
-1 0
1
1
o
0]
1 0 ,Rz ~ Rz - 2R, o 1 ,R3~R3-R,
o
0]
1 0 3 3 -2 0 0 -1 1, R3
~
R3 -
Rz
At this stage, we note that row-rank of A < 3 = the order of the matrix A. Hence by Theorem 1.4, it follows that A is singular implying that A-I does r.ot exist.
System of Linear Equations
19
We may further note that the above augmented matrix is equivalent
1 - [00 -1
o
-
[~
~ -2/~ I1~ ~], ~ Rz
0
1
-1
to
Rz 13
1
01 1/3 1/3 O]'Rl~Rl+Rz 1 1 -2/3 113 0
o
0
1
-1
1
At this stage, we also note that the row-reduced echelon form of A :t= I, and hence A-I does not exist.
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ l. Let a, b, c, d are all nonzero and the entries not written are all O's. Find
the inverses of the matrices, if they exist, given below.
2. For the following matrices A, determine whether A-I exists. If A-I exists, find it.
(/)
[-~
-1 3
-7 -2 0
(iii)
~] 0 0
[: a
3
a aZ
a
(ii)
~l
[~
-1
0
4
0
0
5
0
-6
-2
(iv)
H -:] 5
-4
j]
Introductory Linear Algebra
20
(,) [: 0 ;]
-: -;]
(,;,{~ (a)
[~ -i] 0
(,,) [j ~ ~l (w)
0
1 -3 (viii) [-1 -13 21]4 (x)
[! ~ ;]
3. Let A be an n x n matrix. Prove the following: (I) (t) If A is invertible and AB = 0 for some matrix B, then B =
o.
(ii) If A is not invertible, there exist an n x m matrix B such that AB = 0, but B "* O.
4. If A and B are invertible, are A + B, A - B and -A invertible? Justify your answer.
~2 2.1
11
VECTOR SPACES
VECTOR SPACES
Vector space (or linear space) is the central part of linear algebra. Linear algebra is that branch of mathematics which deals with the common properties of algebraic system which consists of a set together with reasonable notion of linear combination of elements in the set. The set V of all vectors together with two algebraic operations of vector addition and multiplication of vectors by scalars defined on V motivates an algebraic structure which is called a vector space (or linear space) which occur quite often in mathematics and its application. Vector space is basic in functional analysis which has applications to differential equations, numerical analysis and several other fields of practical interest to the engineer. First of all, we shall give the formal definition of a vector space. Definition. Let V be a non-empty set of objects, called vectors and F be , a field of scalars. On the elements of V, we define two operations (+) and ( . ), called vector addition and multiplication of vectors by scalars. Then V is called a vector space if the following ten axioms (or rules) are satisfied: I. V is closed under addition, i.e., u + V E V for all u and v in V. 2. Addition is commutative, i.e.,
u + v =v + u for all u and v in V. 3. Addition is associative, i.e., (u+v)+w=u+(v+w) for allu, v and w in V. 4. Existence of additive identity, i.e., there is a unique vector 0v in V, called the zero vector, such that u + Ov =
II
for all u in V.
5. Existence of additive inverse, i.e., for each vector u in V there is a unique vector -u in V such that lI+(-u)=Ov'
22
Introductory Linear Algebra
6. For each u in V and for each a in F, au (called product of a and u) is in V 7. a (u + v) = au +uv for each a in F and for all u, v in V. 8. (a + p) U = au".J- pu for each u in V and for all a, p in F. 9. a (pu) = (a[3) u for each u in V and for all a, p in F. 10. 1u = u for all u in V. Remark: As stated in chapter 1, the field F = 9t or F = C. If F = 9t, V is called a real vector space; and if F = C, V is caned a complex vector space. Example 1. Consider the n-tuple space Vn (=9t n) whose all elements are ntuples of the form x = (XI.X2, ... ,xn), where all x/s are in 9t.
Let
Y = (YI'Y2'"'' Yn) E Vn with Y; Define + and . in Vn as follows:
E
9t.
x + Y = (XI + YI' x 2 + Y2 , ... , xn + Yn) E
v"
ax = (<XXI' (axl' <XX2 ax2 ,, ... ... ,, <xxJ axJ E Vn <XX Then Vn is a vector space over the field of real numbers. Solution: I. It follows from the definition that X + Y E Vn for all X and Y in Vn.
2. x + Y
=
(XI
+ YI' x2 + Y2,· .. , xn + Yn)
= (Y I + xI' Y2 = =
+ x 2 , ..., Yn + xn), addition is commutative in 9t
(Y I , Y2'"'' Yn)+(xl'x 2"",+xn)
Y + X for all
X
and Y in Vn
=> Addition is commutative in Vw 3. Let z = (zl'z2,,,,,zn) E Vn,z; (x + y) +
Z =
E
9t Then
(XI + YI'x2 + Y2""'X" + Y,,)+(ZI' Z2""'Z,,)
= (XI
+ YI)+ zl' (x2+ Y2)+ zw'" (x" + y,,)+ z,,)
= (XI +(YI +ZI),X2 +(Y2+Z2),,,,,Xn +(Yn +zn)),
since addition is associative in 9t
23
Vector Spaces
= (xpxz,···,x/l)+ (YI + ZPYz + ZZ'···'Yn + Zn) x,y,z in Vn. = x+(y+z) for all x,Y,z ~
Addition is associative in Vn . 4. There is an unique vector 0v = (0,0, ... ,0) in Vnn, called the zero vector of VII such that (XI +0,X1 +O, ... ,xn+O) ,xn +0) x+ov= (xl+O,XZ+O,
=
(X 1,X2 ,···,x,,)
= x for all x in VI/" VI/' 5. For each vector x in VII' there is a unique vector -x = (-x (-x,,-x1, ... ,-x/l) ,-xn) in Vn such that P-x2 , ••• x + (-x) = (XI + (-X (-XI),X + (-x1),,,,,x (-xJ) 2), ... ,xn + (-x.) I ),X1 2 +(-x =
(0,0, ... ,0)
=Ov· =Ov' <XX2' ... , <XX,,) E Vn ' as given in the definition. 6. ax = (axl' (axl'<XX1'''''<XXn)
7. a (x + y) = a (XI + Y"x YPX21 + Y1,,,,,xn Y2'·.·'X" + Yn) Y,,) = (a (XI + yyl),a(x y1),···,a(x .. ·,a(xn + Yn» l ),a(x1 2 + y1), = = (<XXI +aYp<XXl +aY"<XX2+aYz'''',<XX +aYl'···'<XXn n +aYn) = .. ·,<xxn) + (ay"aYz, .. ·,aYn) = (<xx,,<xx1, (<XXp<XXz'·.·'<XX/I) (aYp aY2,···,aYn) =
a
(Xl'xZ' (XI'X '+Xn) + a (Y"Yz, (YPY2, ... ,Yn) 2, .....·,+X
= = <xx+ay ';;f x, YEVn,aE9t. 8. (a+~)x = (a+~)(x"x2, (a+~)(xpx2' .....·,xJ 'x.)
I
~)xp (a+~) xz, xl, ... ,(a+~) xn) = «a+ ~)x" = (<XXI + ~x" ~xp <XXl + ~l'''''<XXn ~l' ... '<XXn + ~xn) ~x,,)
= (<XXI , <XX1'· .. ,<XXn) + (~I ,~x2'''''~n) ,~xl'···'~n) =
a
(xPxl, ... ,xn)+~ (x"x1, (xpx1, ... ,xn) (x"xl,,,,,xn)+~
= = <xx+~';;fa,~E9t,xEv'" <xx+~';;fa,~E9t,xEv',. 9. a (~) = a (~"~l""'~n) (~P~2' ... '~n) = (a(~I),a (~l), .. ·,a (~n» (~2),···,a (~n))
= «a~) xI> ... ,(a~) xn) xl' (a~) xl, x2,···,(a~) = (a~)(xl,xl,,,,,xn) (a~)(xI,xl, ... ,xn) = (a~)x ';;f a, ~ E 9t, x E V Vn. n.
24
Introductory Linear Algebra
=
(X 1,X2 ,···,X,,)
=
x V
X E
V". v".
Thus, all ten axioms of a vector space are satisfied. Hence vector space.
VII
is a real
Example 2. Let V be the set of pairs (x I' x 2) of real number, and let F be the field of real numbers. Define. (xl' (x" x 2 ) + (YI' Y2) = (x) + YI' 0)
a (xl' x 2 ) = (ax «(XXI' p 0) Is V, with these operations, a vector space? Solution. No, as I (xI' x 2) = (IxI' 0) =(xi' 0) #- (xi' x 2 ) if x 2 #- 0. Example 3. For n. Let
11 ~
0, let P II be the set of polynomials of degree at most
p (x) = a o +a)x+a2 x 2 + ... +a"x"
E Pn,
...(2.1)
where the coefficients ao' al' a" ... , an and the variable x are real numbers. The degree of p is the highest power of x in (2.1) whose coefficient is not zero. If p (x) = ao #- 0, then the degree of pis zero. If all the coefficients are zero, p is called zero polynomial. The zero polynomial is assumed to be included in Pn' even though its degree, for technical reason, is not defined. Let p (x) be given by (2.1) and q (x) = bo +b)x+b2 x2 + ... +bnxn E Pn. For any real number a define the two operations + and in P n as follows: (p+q)(x) = p(x)+q(x) = (a o +bo) +(a) + b)x+ ... + (a" +b,,)x" (ap)(x) = apex) = aGo + (aG)x+ ... + (aa,,)x".
Clearly axioms I and 6 given in the definition of vector space are satisfied, because (p+q)(x) E Pn and (ap)(x) E Pn as being polynomials of degree less than or equal to n. The zero polynomial acts as a zero vector of Pn which satisfies Axiom 4, and (-1· p) (x) =-p(x) acts as the additive inverse which satisfies Axiom 5. Other Axioms 2, 3, 7:.10 follow from the properties
25
Vector Spaces
of real numbers (details can bt;verified by the readers). Thus Pn is a vector space. Example 4. Let Wbe the se~ofall points in 9\2 of the form (3x, 2 + 5x), where
x E 9t Is Wa vector space with usual addition of vectors and multiplication of vectors by scalars? Solution. No, for eaxmple, take x = 1 so that u = (3, 7) E W. For a = 3, we have au = (9, 21). If au were in W, then If au = (9, 21) = (3x, 2 + 5x). That would imply x = 3 and x = 19/5 which is impossible, and hence au is not in W. Thus, W is not a vector space. Several other examples are given in Exercises. Readers are advised to verify all ten axioms of a vector space. In the following proposition, we shall list some important properties of a vector space. These properties can easily be verified using ten axioms given in the definition of a vector space. These are left as an exercise for the readers (see Exercise 6). Proposition 2.1. For each u in a vector space V and scalar a in F, the following statements are true (Ov denotes the zero vector in V): I. Ou= 0v' Qv' 2. aOv=Ov'
3. (-I)u=-u.
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ 1. Show that the set 9\ of real numbers is a vector space over the field 9\ under usual addition of real numbers and multiplication of real numbers by real numbers. 2. Let e be the field of complex numbers and n-tuples of the elements of C of the form where ui '
Vi
e"
be the set of ordered
u =(u 2,UZ,. .. , un)' V =(V2' V2'"'' vn)' are in C. Define + and . in as follows:
en
u+v =(u) +v),u2+v2, ... ,un +Vn) au =(aul'au2, ... ,aun),a e C. Then show that (!' is a vector space over C. In particular, C is a vector space over C.
26
Introductory Linear Algebra
3. Let C [a, b] be the set of all real-valued continuous functions on the interval [a,b].For l,gEC[a,b] andaE 9\,define+and·in C[a,b] as follows:
(f + g) (x) == I(x)+ g(x), (a!) (x) == al (x), for all x in [a, b]. Show that-under these operations C [a, b] is a vector space over the field of real numbers.
4. Show that the set M of all
111 x
n matrices forms a vector space over
the field 9\ of real numbers under the usual operations of addition of matrices and multiplication of matrices by real numbers. 5. Let V be the set of all pairs (x, y) of real numbers and F be the field of real numbers. Define (x,y) + (x"y) = (x+ x"y+ Y 1) a. (x,y) == (ax,y), a. E 9\.
Show that V, with these operations, is not a vector space over the field of real numbers. 6. Prove all three properties stated in Proposition 2.1. 7. Let V be the first quadrant in the xy - plane; that is, let V
=
{(x,y):x~O,y~O}.
Is Va vector space over 9\ under usual vector addition and mUltiplication of vectors by scalars? 8. Let V be the union of the first and third quadrant in the xy - plane; that is, let
V=={(x,y):xy~O}.
Va
Is vector space over 9\ under usual vector addition and multiplication of vectors by scalars? 9. Let
V={A=[: ~lad-bc:;t:O}. Is Va vector space under usual
addition and scalar multiplication of matrices?
2.2
SUBSPACES
In this section, readers will be introduced with some basic concepts of vector spaces.
Vector Spaces
27
Definition. Let V be a vector space over the, field F and W be a non-empty subsct of V. Then W is called a subspace of V if W is itself a vector space under the samc operations of vector addition and scalar multiplications as in V. The above dctinition shows that if we want to check whether a nonempty subset IV of a vector space V is a subs subspace pace of V or not, we will have to check all ten axioms of a vector space for the elements of W; keeping both operations same as in V. But the following theorem makes all computations simple. Theorem 2.1. A non-empty subset W of a vector space V over the field F is a subspace of V if and only if (i) u + v E W for all u and v in W,
(ii) au E TV for all u in W, and a in F.
Proof. Let W is a subspace of V. ~
W is itself a vector space under the same operations as in V and hence (i) and (ii) are satisfied. Conversely, let (i) and (ii) are satisfied. Then we wish to show that W is a sllbspace subspace of V, that is, W is itself a vector space under the same operations as in V. Thus, we need to check that elements of W satisfy all eight axioms (two are already satisfied as in the hypothesis) of a vector space. Since
au E W for all u in W, and
a in F, therefore, in particular for
a = 0, we have
Ou = 0v
E
W.
Thus W has additive identity. Further for a = -1, we have
-Iu =-u E W for all u in W. Thus W has additive inverse. Other six axioms are automatically satisfied because of being vectors of V. Thus W is a vector space, and hence a sllbspace subspace of V. From Theorem 2.1, it is evident that W is a subspace of V if and only if (I) W is non-empty, that is, zero of V is the zero of W, (il) W is closed under vector addition, and (iil) (iii) W is closed under multiplication of vectors by scalars. The following theorem shows that we can simplify the properties of subspace still further.
Introductory Linear Algebra
28
Theorem 2.2. A non-empty set W of a vector space V over the field F is a subspace of V if and only if alt all + v E W for all
1I, It,
v in Wand for each a a in F.
Proof: Let W is a substance of V :::::> all E W for all
1I It
in Wand a in F, (by Theorem 2.1)
Let v E W be any element. Then again by Theorem 2.1, all
+ v E W for all
ll, v in Wand for each ex Cl in F.
Conversely, let W is a non-empty subset of V such that au + v E W for all It, u, v in Wand for each Cl ex in F. In particular, for Cl ex = I, we have :::::>
l,u+vEWfor allll, v in W l·lt+vEWfor + v E W for all 1I, It, v in W.
U
Since W is non-empty :::::>
There exists a vector p in W such that (-I)p + PEW
:::::>
01' EW.
Therefore, for any
II 11
au = all + Ov
in Wand Cl ex in F, we have
W subspace pace of V. Thus, W is a subs E
Example 5. The set consisting of only zero vector in a vector space V is a subs subspace pace of V. That is, Vo = {Ov} is a subspace of V, where 0v is the zero pace of V, called zero subs pace of V. vector in V. Vo is the smallest subs subspace subspace Example 6. If V is a vector space, then V is itself a subspace of V; which is the largest subspace. Example 7. It is easy to check that any straight line passing through the 2 origin is a subspace of V2 (= !H m2). ). However, any straight line not passing subspace pace of Vz as it does not contain the zero through the origin is not a subs vector. Example 8. It is easy to check that any plane in V3 (= 9\3) passing through the origin is a subspace of V3. However, any plane in V3 not passing through the origin is not a subspace of V3 as it does not contain the zero vector. Example 9. Let P be the set of all polynomials with real coefficients. For each n ~ 0, let Pn be the set of all polynomials of degree less than or equal to n. Then Pn is a subspace of P as
29
Vector Spaces
(i) P n contains the zero polynomial.
(ii) The sum of any two polynomials in Pn is again in Pn"
P,!, (iii) The scalar multiple of any polynomial in Pn is again in P'I' Example 10. The vector space V2 (= 9\2) is not a subs subspace pace of V3 (= 9\3) as V2 is not even a subset of V3; since the vectors in V3 all have three entries. whereas the vectors in V2 have only two entries. Example 11. Consider the set
W={(x p x 2 ,0):x, and x 2 are real numbers} Then W is a subs subspace pace of V3 as (I) (O,O,O)eW (ii) Let x
= (xI'
x 2, 0)
E
11'
andy = (YI' Y2' 0) e W Then for any real scalar a, we have
ax + Yy = a(x" x 2 ' 0) + (yp Y2' 0) = (axp ax2 , O)+(YpP Y2' 0) = (ax, + y"ax 2 + y 2 ,0) E W. Example 12. Let W be the set of points inside and on the unit circle in the xy - plane, that,
IV = {(x,y): X2 + y2 u
=
au
=
::;
I} . Then W is not a subspace of V2 as
(~2'2~) E Wand for a (2, 2)
~
= 4.
W.
In the following theorem we shall establish an important property of a subspace. Theorem 2.3. Let V be a vector space over the field F. The intersection of any collection of subspaces of V is a subspace of V. Proof: Let {W) be the collection of all possible subspaces of V and let
W= nWi Then we wish to show that W is a subspace of V. (i) Since W Wii is a subspace of V for all i
30
Introductory Linear Algebra 01' 01'
E
W; for all
En W;
Let u, v E W =
i
=W
nw,
and a is any scalar.
=>
W; for all i au + v E W; for all i as each W; Wi is a subspace of V
=>
aU+VEnWj=W
=>
u, V E U,
This proves that W is a subs subspace pace of V. Remark: Union of two subspaces need not be a subspace.
To prove this, we let ~ ={(O, 0, z):z E9\} and ~ ={(O, Y, O):YE 9\}
Then it is easy to check that
~
and W2 both are subspaces of V3 .
Let W= W WI1 U W2 . Then (0, 0, 8) E Wand (0, 5, 0) E W, but (0, 0, 8) + (0, 5, 0) = (0, 5, 8) E Was neither (0, 5, 8) E W WI1 nor (0, 5, 8) E W2 • Thus W is not closed under vector addition and hence is not a subspace of V3 . Example 13. Let V be the vector space of all functions from 9\ into 9\; let Ve be the subset of even functions, fe-x) = f(x); let Vo be the subset of odd functions, f(-x)=-f(x}. Prove that
(/) Ve and Vo are subspaces ot (ii)
(iii) (hi)
Y.
v;. + Vo = V. v;. n Vo = {Ov }.
Solution. (t) Let I\> = 01" a constant zero function, then we note that 1\>(-x)=I\>(x)=O, the zero element of 9\.
=> zero function is even, that is, 01' Let f, gE Vc
=>
f(-x)=f(x) and g(-x)=g(x)
Then for any scalar a, we have
E
V.
Vector Spaces
31
g(-x) (af + g)(-x) = (af) (-x) + gC-x) g(-x) = af(-x) + gC-x)
=af(x)+g(x) =(af + g)(x)
=> af+g EV" Hence Ve is a subspace of V. It is left as an exercise for readers to show that Vo is also a subspace of V (see exercise 4).
v;, + Vo = V
(ii) To show that
Let
f
V be any element. Then for any x in 9\ we may write
E
I "2I (f(x) + fC-x»+"2(f(x)fe-x»~
f(x) =
= ~(x)+
where
hi (x)
.
h2 (x),
I -(f(x)+ fe-x»~,
=
2
It is easy to see that hi (-x)
=> hi
E
Ve and h2
E
Hence
f
=~
V" + Vo .
Thus V k
+ h2
v;, + Vo '
certainly ''-: +
E
=
1 2
~(x)=-(f(x)- fe-x»~.
hi (x) and h2 (-x) = -h2 (x)
Vo· Vu·
and
r:, k
V.
Therefore, V = V" + Vo . (iii) To show that V" 11 II Vo = {Ov} {Or}
Since Ve and Vo are subspaces of V
=>
v;, and 0v E v;, 11 II Vo
=>
{Ov} k V"
=>
0v Or
E
Now let
f
E
II 11
0v
E
Vo
Vo.
V. 11 II Vo
=>
f
=>
fe-x) = f(x) and fe-x) = - f(x) for all x in 9\
E
V" and
f
E
Vo
Introductory Linear Algebra
32
=> 2f(x) = 0 for all x in 9L =>
f=Ov E{OJ'}
Thus
v" + Vo ~ {O,,} {Ov}
and hence V. + Vo = {Ov} .
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ 1. Let W be a non-empty subset of a vector space V. Show that W is a subspace of V if and only if all au +
Bv E W
for all u,
vE
2. Let W = {(x,2x+ I):x E
~H}
Wand all scalars
(1,
p.
. Is W is a subspace of V2?
3. Let W = {(x,mx+c):xE9\;mandc are fixed real numbers}. Is Wa subspace of V2? 4. Show that Vo Va defined in example 13 is a subspace of V. 5. Let W= {(XI'X2'X3,X4)EV4:3xl-X2+X3+2x4=O}. Is W a subspace of V4 ? 6. Which of the following sets are subspaces of Vn? (i) {(X I'X 2' .•••.. ,xn ):xl ~O} (ii) {(XI'X2, ..... ,xlI ):xl =O} (iii) {(XI'X2, .... ,x,J:XI +x2 =4x3}
(iv) {(XI'X2' .... 'Xn):XI2 =x2}
2
(v) {(XI'X 2, .... ,Xn):XI +xi + ..... +x~ =
I}
2
(vi) {(XI' x 2' .... ,xll ) :X1 +xi + ..... +x~ ~ I}
(vii) {(Xl' x 2'·· .. , x n): XIX2 = O} (viii) {(X I'X2' •..•.. x,,):x2 is rational}
7. Which of the following sets are subspaces of P? (i) {p
E
P : degree of p = 7}
(ii) {p E P : degree of p
~
7}
P : degree of p
~
7}
(iii) {p
E
(iv) {pEP:p(l)=O}
Vector Spaces
33
8. Determine if the given set is a subspace of P n for appropriate value of n. Justify your answer. (/) The set of all polynomials of the form p(x) = ax 2 , where a is in
9L (ii) The set of all polynomials of the form p (x) = a + x 2, where a is in (iii) The set of all polynomials of degrees less than or equal to 4, with integers as coefficients.
9\.
9. Which of the following set Ware subspace of the indicated vector space?
(i) W = {(x" x 2' xJ~
~ :x +X2 = I} 1
(ii) W={(X"X2,X3)E~ :X3 =x1xJ (iii) W = {(x" xl' Xl' x4 )
(iv) W w = {[;,
(v)
:] E
EV4 : x2 = 0 =x4 }
M 2X2 : l!,b,c E
9\}
W={[~ ~]EM2X2:~,bE9\}
(vi) W ={aD +al x+a2x 2 E P,. I: al
=0; ao' a2 E 9t}.
10. Let W WII and W2 be subspaces of a vectors space V such that W WII + W2 = Vand W WII n W2 = {Qv}. {Ov}. Prove that for each vector u in V there are WI and u2 in W2 such that 11II = III + 112 • (In such unique vectors 1I, in WI a situation, V is said to be the direct sum of the subspaces W WII and W2 and it is written as V = W WII Ei3 W2 ).
It,
11. Determine if the set W of all matrices oft,he form
[~ ~]
is a subspace
of M2 x 2? , 12. Let W WII and W2 be two subspaces of a vector space V. Prove that W WII u W2 is a subspace of V if and only if W WII ~ W2 or W2 ~ W WI· I•
2.3
A SUBSPACE SPANNED BY A SET
To understand a subs pace spanned by a set, one should be aware of the subspace term 'linear combination'. Let
34
Introductory Linear Algebra
be a subset of a vector space V, and let a p a z , •••.•• , an be n scalars. Then'
is called linear combination of vectors up uz, .... , un' It is sometimes also called linear combination of the set S. Remark: Note that u is a linear combination of finite set S, and hence u is also known as finite linear combination. In case S is an infinite set, we shall take a finite subset A of S. Then form a linear combination of the set A, which is referred to as a finite linear combination of the set S. Now we can define the span of a set. Definition: Let S be a subset of a vector space V. Then span of S, denoted
by [S], is the set of all finite linear combinations of S. That is, if S = {upuz, ..... ,un }, then
[S] = {alul + azu azuzz + ..... + anu anUn : a i E 9\}. Remark 1: If
S = {upuz, ..... ,un} , then span of S is also written as
[S] = [u\, [ul' 1I2' tl2' ... , un]' Remark 2:
S = {upuz, ..... ,uJ is a finite set, but [8]
=
{alul + azu azuzz + ....... + anu anUn : a i E 9\}
is an infinite set as each
is any arbitrary real number.
(Xi
Remark 3: By convention, we take [
] = Vo . Example 14: Suppose that V = V3 and S = {(l, 0, 0), (0,0, I)} and we wish to find [S]. Then
[S] = {a(1, 0, 0)+ P(O, 0, 1): a,p E 9\} = {(a, 0, p):a, p E 9\}
In this example, we note the following about [S]. (I) (0,0,0) E [S]
(ii) Let
u = (XI' 0, Xz)
E
[S]
and v=(YI,O,Yz)E[S] Then for any scalars e, we have eu +v = (ex l , 0, CXZ)+(YI' 0, Yz) = (exli + YI'O, exz + Yz) E [S]
35
Vector Spaces
This shows that [S] is a subspace of V3 . In fact, in the following theorem we show that [S] is always a subspace of a vector space V. Theorem 2.4. Let S be a non-empty subset of a vector space V. Then span of S, written as [SI, [S), is a subspace of V. Proof: Note that 0v E [S] and hence [';] Let u, v E [S]
* $.
.,
=>
U
=
L u j uj
for some u, E S and for some scalar a i
;=1
n
and
v
=
LP
;Vj
for some Vj E S and for some scalar Pj .
;=1
Then for each scalar a, we have III
II 11
au + v = L(auJuj + 1=1
L P;Vj j=1
=> UU+VE[S] => [S] is a subspace of V. Theorem 2.5. If S is a non-empty subset of a vector space V, then [S] is the smallest subspace of V containing S. Proof: From Theorem 2.4, [S] is a subspace of V. First of all, we shall show that S~[S].
To prove this, let
U
E S
=> u=l·u => u is expressed as finite linear combination of the elements of S. => UE[S] Thus,
S~[S]
In order to show that [S] is the smallest subspace of V containing S, we assume that W is another subspace of V containing S. We shall be done if we can show that [S] ~ W. To do this, let v E [S]
=> v= UIVI +U 2V2 + ...... +u"vn , where Vj E Sand Since Vj ES and
S~W
=> Vj E W for all i
a/s
are scalars.
36
Introductory Linear Algebra
=> v =
CliVI
+ Cl 2V2 + ....... + Cln Vn E W as W is a subspace of V.
Hence [S]!;;;; W. This proves the theorem. Remark: Span of the set S, that is [S], is also called the subspace spanned by S or the subs subspace pace generated by S. Example 15. Let W = [u p u2 , ... ,un ] , where each uj is in a vector space V. Show that uk uk is in W for I <; k ::;; n. Solution. We may write uj as foHows: u l = l.u l +0,u2 +0,u3 + ... +O,un
u2 = O,u l + l.u 2 + 0,u3 + ... + O,un Un
This shows that Uk uk
E
= O,u l
+
+ O,un_ 1 + I,un
W for I::;; k ::;; n.
Example 16. Let S = {(x, 3x, 2x): x E~} . Find a vector v in J.j (= ~') such that S = [v]. Hence conclude that S is a subspace of V3 . S = {(x,3x,2x):XE~}
Solution: We have
=
{x(l, 3, 2) :XE~}
= [(1,3,2)] = [v], where v=(l,3,2). Note that by Theorem 2.4, S
=
[v] is always a subspace of V3 .
Example 17. Let S={(a+3b, a-b, 2a-b, 4b):a,bE~}. ShowthatSisa subspace of V4 . Solution: We have S
= {(a+3b,a-b,2a-b,4b):a,bE~} =
{(a, a, 2a, 0)+(3b, -b, -b, 4b): a, bE~}
=
{a(l, 1, 2, 0) + b (3, -1, -1, 4): a, b E~}
= [vI' vz ], where vI = (1, 1,2,0), v2 = (3, -1, -1, 4).
=> S = [vI' v2 ] is a subspace of V4 , by Theorem 2.4. Example 18. Determine if the vector (1, 1, 0) is in the span of the set S = {(t, 2, 1), (1, 1, -1), (4, 5, -2)}?
37
Vector Spaces
Solution. If (I, I, 0) is in [S], then (I, 1,0) = a (I, 2, 1) + b (I, I, -1) + c (4, 5, -2) =
(a+b+4c,2a+b+5c,a-b-2c)
~
a+b+4c=I, 2a+b+5c= I, a-b-2c=0. lfthe above linear systems is consistent, then (I, 1,0) would be in [S] and if the system is inconsistent, then (I, 1,0) will not be in [S]. And from chapter I, we know how to check the consistency or inconsistency of the system. To do this, the augmented matrix associated with the above linear system is
B
~ [~ - [~
4 5
-I -2 4
-3
-I
-l
R, --> R, -2R,
-2 -6 -I ,~~lS-RI 4
-
1]
0
3
0
3
,R2 ~(-I)R2
-1 2
,~~(-~)~
1 4
-
3
0 0
0
0
,Rl~Rl-R2
2
This show that row-rank of coefficient matrix :I: row-rank of augmented matrix. ~ The system is inconsistent, and hence (1, 1,0) does not lie in [S]. Example 19. Let S be a non-empty subset of a vector space V and u, v
E
V. Suppose that
UE
[S U {v}] and u E [S]. Then prove that v E [S U {u}] .
Solution: Since
UE
[S V {v}]
, 38
Introductory Linear Algebra
=> u = where ui
E
<XIU1I + <X 2U2+ ..... +<XnUn + <Xv <X1U
S and <x, <Xi are scalars, 1 $ i $ n .
Clearly a Therefore
'* 0, for if a = 0, then it would imply that
1 (-<X -U+ _ 1) U +
=>
V
=>
VE[SU{U}].
=
<X
<X
I
U E
[S]
+(_ <Xn)u <X
"
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ __ 1. Let S = {(l, 1,0), (2, 2, 1), (0, 0, 2)}. Determine which of the following vectors are in [S]: (a) (3,4,5) (b) (3,3, 5) (c) (0,0,0)
(d) (3,3, 1)
(e) (-11,7, -I)
if) (1,2,3) 2
2. Let S = {x\ x + x , 1+ x + x polynomials are in [S]: (a) 2y 3 + x + I 3
(c) x + 2x + I
2
} •
Determine which of the following (b) 3x3
(d) 2x3
+ x-I + x 2 + 2x +1
3. Let S = {(2a, 0, - a) : a E 9\}. Show that S is a subspace of V3" (Hints: Use the method discussedl,In, example 16).
4. Let S = {(5a {(Sa + 2b, a, b}: a, b E 9\}' Find vectors u and v such that S = [u, v]. Why does this show that S is a subspace of V3? (Hints: Use the method discussed in example 17) 5. Let S = {(a - 3b, b - a, a, b): a, b E 9\}. Show that S is a subspace of V4 . (Hints: Use the method discussed in example 17). - 6. Let S be a non-empty subset of a vector space V. Prove that (a) S is a subspace of V iff S = [S]. (b) [[S]] = [S]. 7. Let V be a vector space and u, v E V. Prove that [u, [It, v] = [u - v, u + v]. 8. For what value(s) of h will y be in the subspace of V3 spanned by Ill' u2' u lt33 if 1I1 = (1, -1, -2), u2 = (5, -4, -7), u3 = (-3,1,0), andy = (-4,3, h).
Vector Spaces
39
9. Let u 1l = (1, 4, -2), u2 = (-2, -3,7) and b = (4, 1, h). For what value(s) of h is b in the plane spanned by up u2? 10. Let Uu = (2, -1) and v = (2, 1). Show that the vector b = (h, k) is in [u, v] for all hand k. 11. Determine the subs subspace pace of V3 spanned by each of the following sets: (a) {(\, I, 1),(0, 1,2),(1,0,-1)} (b) {(2,1,0),(2,0,-I)} (c) {(I, 2,3), (1, 3, 5)}, (d) {(I, 2, 3), (1, 3, 5), (1,2, 4)}.
2.4
BASES AND DIMENSION
In this section we shall assign a dimension to certain vector spaces, and dimension is based on the concept of a basis for the space. But basis is based on the concept of a linearly independent set. Hence we first introduce to the readers with the linearly independent and linearly dependent sets. Definition. Let V be a vector space over the field F and S = {up u2 ' ••• , u,,} is a subset of V. Then S is said to be linearly dependent (or vectors up u2 , ••• ,u" are said to be linearly dependent) if there exists scalar a p a 2 , ••• , a" in F, not all of which are zero, such that CXIU1I +CX 2U2 + ... +cx"u" CX1U
= 0v.
A set which is not linearly dependent is called linearly independent, that is, the set S = {u {UI' 1' u2 , ••• ,u,,} is called linearly independent if and only if CXlli1l CX11l
+ CX 2 U 2 + ........cx"u"
= 0v
=> each cx; = 0.
Definition. An infinite subset S of a vector space V is said to be linearly independent if every finite subset of S is linearly independent. The set S is said to be linearly dependent if it is not linearly independent. Remark: Throughout this book we shall use the notation LD for linearly dependent and LI for linearly indendent.
How to check LD or LI of a given set S Let S = {u"u 2 ' ...... ,u,,} be a subset of a vector space V. To check whether the set S is LD or LI, conisder the equation
40
Introductory Linear Algebra
*°
lfit Ifit is possible to find atleast one a i that staisfy the above equation, LD. If the above equation has only the trivial solution, namely, then S is LO. a) = 0. = ... = an = 0, then S is LI. 2 Remarks:
(a) A 'lon-zero vector is always LI for if u
*
0V! then au = 0v => 0.=0. (b) Zero vector is always LO, LD, for 1I . 0v= 0V' 0v- and hence any set containing the zero vector is always LO. LD. (c) By convention, we take the empty set clcI>> as LI. Now we shall give some examples of LO LD and LI sets.
Example 20. Let e) = (1, 0, 0), e2 = (0, I, 0) and
e)
= (0, 0, 1). Then the set
S = {e {el' z', eJ is LI in V) over 9t as l , e2 aiel + a aze 2 e2 z + aJe) =
°
= (0, 0, 0)
=> (a" a 2z,, aJ = (0,0,0)
°
=> a ll = =a 2z =a JJ .• Exwmple 21. Let e) = (1, 0, 0, ... ,0), e2 = (0, 1,0, ... ,0), ... , en = (0,0, ... ,0, 1). Note that each ei has n entries. Then as in example I, it can easily be checked that the set S = {el' e2 , •.• , ell} is LI in Vn over 9t . Example 22. Suppose that we want to check whether the set S
={(I, 2, I,), (-I, I, 0), (5, -I, 2)}
is LD or LI in V) over 9t . For this, we let a" a 2z ,, a 3 be scalars (reals) such that a l (I, 2, 1) + 0. 2 (-1, I, 0) + 0. 3 (5, -1,2) = (0, 0, 0).
a ll -- a 2z + 5a3 3 = 0, 2a l + a 2z -- a 3 =0, a l + 2a3 =0. To solve the above system of linear equation, we write the augmented matrix associated with the above system as
=>
B =
[~ 1I
-:
-~ ~l
° ° 2
Vector Spaces
41
- [~
-I
5
3 -II -3 0
2
3 -3 0
0
2 11
--
0
3 2 3
0 0
-[ [ This shows that
(XI
°lR' -->
11 Rz+/3R, -- o ,Rz -'t R,
- [: -
~l ,R, --> R, -2R,
o ,~-'tR3-RI
0
0 0 ,~-'tR3-Rr
0
2
11 -3
1.-->mR,
0
I
0 0
0
I
0
o
0
I
= 0 = (X2 = (X3
•
Hence the given set S is LI. Example 23. Check whether the given set S = {I- x, x - x 2 , 1- x 2 } is LD or LI in P2 over 9t Solution: Let a, b, c be-scalars such that
a(l- x) + b(x - x 2 ) + c(l- x 2 ) ~
(a+c)+(b-a)x-(b+c)x 2 = 0
~
a+c = 0, b-a =0, b+c =0.
=0, a zero polynomial.
42
Introductory Linear Algebra
On solving, we get b = a, c =-a,
and a is arbitrary real number. Thus, the above system of equations has infinitely many solution and therefore a nontrivial solution which implies that the set S is LD.
Example 24. Show that the set S = {sinx,sin2x, ... ,sinnx} is a LI subset of C [-n, n] for every positive integer n. Solution: Let a" <x" a<x 2 , ••• , all <XII be scalar such that a<XIl sinx+a sinx+<X 2 sin2x+ ... +a +<X n sinnx = 0,
...(2.2)
where 0 E C[-1t, 1t] . Use the fact that x
Jsinmx. sinnxdx =
-x
{a, 1t
'
m::t-n . m=n
Multiplying equation (2.2) by sin x and integrating it between the limits -n to n, we obtain a l = 0. Next, we multiply equation (2.2) by sin 2x and integrate it between the limits -n to n, then we obtain a 2 = 0. Continuing this process n times, we get
This show that S is LI. Example 2S.lf {u, v, w} be LI subset ofa vector space Vover 9t ,determine whether the set {u+v,v+w, w-u} is LI or LD. Solution: Let a, band c be scalars such that
a(u + v) + b(v + w) + c(w-u) =0.
=> (a-c)u+(a+b)v+(b+c)w=O => a - c = 0, a + b = 0, b + c = 0 as {u, v, w} is LI. It is easy to check that the above system of linear equations has a nontrivial solution, in fact, infinitely many solutions given by a = c, b = -c, and c is arbitrary. Hence {u + v, v + w, w - u} is LD.
Vector Spaces
43
Example 26. If S = {up U2"'" un} is a subset ofa vector space V and U E V be such that
UE
Solution: Since
[S]. Then S u {u} is LO. LD. U
E
[S]
=> U =alul + (1,2U2 + ... + anun for some scalars =>
alu l + ... +anun +(-1)11
aj"
=Ov
=> S U {II} is LD. LO. LD and LI set, The following theorem follows from the definition of a LO and it will be very useful in deciding the LD LO or LI character of a given set. Theorem 2.6. Let V be a vector space. Then (a) The set {v} is LO LD iff v = Qv' 0v' (b) The set {vi'v 2 } is LO LD iffone of vI and v 2 is a scalar multiple of the
other. LD iff one of vI' v2 and v3 is a linear combination (c) The set {vI' V 2 ' v3 } is LO of the remaining two vectors. Proof: (a) Let {v} is LO. LD.
°
=> 3 a scalar a :1:- such that av = 0v => v = 0v' Qv' LO. Conversely, let v = Qv' 0v' then l.0 v = 0v implies that fO v} is LD. (b) Let {VpV2} is LD. LO.
=> 3 scalars a and 13 such that atleast one of a and 13 is not zero, say a:1:- 0, and aV I + I3v2 = 0v
=>
is a scalar multiple of v2 • Conversely, let one of VI and v2 is a scalar multiple of the other, say VI is a scalar multiple ofv2 . That is, there exists a scalar a such that VI
VI =
av2·
=> l,vl -av2 = Ov =>
{VI' v2 } is LO. LD.
44
Introductory Linear Algebra
=> 3 scalars
(Xi' (Xl' (X3
such that atleast one of
(Xi
is non-zero,
say, (XI
=>
*" 0, and
(XlVI
+ (XlV2 + <X3 V3 =0v·
VIE[Vl,Vl ]
Conversely, let one of vI' v 2 and v3 is a linear combination of the remaining two vectors, say VI E [vl , v3 ].
=(Xv2 + pv)
=>
VI
=>
t.vl -(Xvl -pv3
=>
{vi'
for some scalars n, Cl,
p.
=Ov
v2 ' v3 } is LO. LD.
Theorem 2.7. Let S = {ui' u2, ... ,un } is an ordered set in a vector space V. Then S is LO LD if! iff one of the vectors of the set S belongs to the span of remaining other vectors of S. Proof: Let S = {ui' ul' ... ,un} is LO. LD.
=> 3 scalars (Xk
(Xi' (X2' ••• , (Xn'
*" 0 (I :s: k :s: n),
(XIUI
not all of which are zero, say
such that
+ (XlU2 + ... + (XkUk + ... + (Xnun
=Ov Qv •.
Conversely, let one of the vectors of the set S, say uk uk is in the span of remaining other vectors of S, that is,
for some scalars
=>
(XIUI
(Xi'
i = I, 2, ... ,n and
j
*" k.
+ ... + (Xk_IUk_1 + (-I)u k + (Xk+IUk+1 + ... + (Xnun = Ov·
Vector Spaces
45
=> S ={up u2 ""'u,,} is LD. After knowing LD and LI set in a vector space, we can define the basis and dimension of a vector space. Definition. Let S be a subset of a vector space V. Then S is said to be a basis for V if (i) S is LI, and (ii) [8] = V. Remark: Since [8] is always a subspace of V and hence [S] ~ V. This shows that in order to have [8] = V, one needs only to show that V ~[S]. That is, S spans V if and only if every element of V can be expressed as the linear combination of elements of S. Definition. The dimension of a vector space V is the number of elements in a basis for V. If the number of elements in a basis for V is finite, the space V is said to be finite dimensional. Example 27. As in example 21,
e = (1, 0, 0, ... ,0), e2 over 9t. Let J
=
(0, 1, 0, ... ,0), ... , en
Then
=
(0, 0, ..... 0, 1) are LI in Vn (= 9t n )
x = xleJ +x2e2+ ... +xnen
= (xl' x 2'···, xn)' This shows that the set S = {el' e2, ... ,en} spans Vn. Hence S is a basis for Vn . Remark: This particular basis S = {ep e2, ... , en} is called the standard basis for Vn• Since the number of element in S is n, and hence dimension of Vn is
n, and we write dim Vn
=n.
Example 28. Consider the set S
=
{t, x, x\ ... ,xn} in Pn. Then S is LI in Pn, for if,
(lo.1 + (lIX + (l2X2
Then
+ ... + (l"X" = 0, a zero pOlynomial for some scalars (li'
(lo =O=(ll =(l2 = .. ·=(In'
46
Introductory Linear Algebra
Further, each polynomial of degree less than or equal to n is just the linear combination of the polynomials in S. Hence S spans Pw This implies that S is a basis for P n and dim p" = n + 1 = the number of elements in S. Remark: This particular basis S = {I,
X,
x 2 , ••• , xn} is called the standard
basis for Pn'
The Spanning Set Theorem Theorem 2.8. Let V be a vector space over the field F. Let S = {vI' v2 ' •.• , vn } be a subset of V and W = [s]. (a) If one of the vectors in S, say v k ' is a linear combination of the remaining vectors in S (i.e., the set S is linearly dependent), that is, if
(b) If W "# {Ov}, some subset of S is a basis for W, that is, we can find
a subset A of S such that A is LI and [A] Proof: Given that
= W = [S].
=> There exists scalars u" ... ,cxk_Pcxk+p ... ,cxn such that v k = (cx\v\ +CX 2V2 + ... +CXk_\Vk_1 +CXk+\Vk+\ + ... +unvn)
...(2.3)
To prove that
Let ue[vp v2 , ••• ,vn ] be any element.
for some scalars a p i = 1, 2, ... , n. Substituting the value ofvk from equation (2.3) into the above equation and re-arranging the terms, we get u = b\v\ +b2v2 + ... +bk_\Vk_\ +bk+1Vk+1+ ... +bnvn,
where b/s are some new scalars.
47
Vector Spaces
=> v = C,V, + C2V 2 + ... + Ck_,Vk _, + Ck+'Vk +' + ... + CnVn ' for some scalars ·c/s.
=>
v=c,v, +C2V 2 + ••• +Ck_'Vk _'
=>
VE[V"V 2
+0. Vk
+Ck+'Vk +' + ••• +cnvn
,···,vn ]
Thus, we have shown that
(b) If Ifthe the original set S is LI, then it is already a basis for Was lS] = W. If S is LD, then one of the vectors in S can be expressed as a linear combination of the remaining vectors of S. Then by part (a), the set S, SI S, is LI, we are done. formed from S by deleting that vector spans W. If SI Otherwise, one of the vectors of S, SI is a linear combination of the remaining vectors of S,. SI. Form the set S2 by deleting that vector of S,. SI. Then S2 spans S,. SI. If S2 is LI, we are done. Otherwise continue the above process until the spanning set is LI and hence is a basis for W = [S]. If the spanning set is eventually reduced to one vector, then this vector will be non-zero as W"* {Ov}, and hence linearly independent. This completes the proof of part (b).
Example 29. Let S = {(t, 0, 2), (0, - 2,5), (2, - 6, 19)} be a subset of V3. Determine whether the set S is LI or LD. If the set S is LD, find a LI subset A of S such that [A] = [S]. (Note that this LI set A is a basis for [Sj). Solution: It is standard to check (readers should verify) that the set S is LD. In fact (2, -6, 19) = 2 (1, 0, 2) + 3 (0, -2, 5). Then by the spanning :;et theorem, [(1,0,2), (0, -2,5)] = [(I, 0,2), (0, -2,5), (2, -6, 19)] One can check that the set A = {(I, 0, 2), (0, -2, 5)} is LI. Thus, A is a basis for [S]. Now in the following theorem, we shall establish an important result which is a basic tool in basis of a vector space. Theorem 2.9. Let S = {v" v2 ' ••• , VIII} spans a vector space V. Then any LI set of vectors in V contains no more than m elements.
Introductory Linear Algebra
48
Proof: In order to prove this theorem, it is enough to show that any set in V which contains more than m vectors is LD. Let W be such a set, that is,
W = {up U 2 ' ••• , un} be a set in V and n > m. Since S spans V,
=> Each u; (1 :;; j:;; n), which is in V, must be expressed as linear combination of elements of S. nI
=> uj
='L,Aijv; for
some scalars Aij; 1:;; j:;; n.
;=1
n
UIUI
+U 2U2 + .... +unun ='L,UjU j j=1
n ( =~UJ ~AijV; III
n
)
III
= 'L,'L,{AyUJ)v, J=I ,:1
=
~(~AijUj}..
We know that if A is m x n matrix and m < n, then the homogeneous linear system Ax = has a non-trivial solution, that is, if the number of unknowns is more than number of equations (n > m), the equation Ax = has a solution x other than x = 0, (See Theorem 1.2). Using this fact, we note that the homogenous linear system of equations
°
°
'tAj;a; = 0, i = 1,2, ... ,m ,=1
has a non-trivial solution in aj . This is a system of m equation in n unknowns with n > m, given in the explicit form as Alla l + AI2 a 2 + ... + Alna n = 0, A21 a l + A27 a 2 + ... + A2n a n = 0,
Multiplying the above equations in right by vI' v2 ' ••• ' VIII' respectively, and adding we get
Vector Spaces
49
t(t.AiiClj} 0V' where atleast one a * O. =
~
j
alu l + a 2 u2 + .... + anun = Ov , where atleast one aj
* O.
~ W = {u l ,u2""'un} is LO. This prl)ves the theorem. The following corollaries are immediate consequences of the above theorem.
Corollary 1. If V is a finite-dimensional vector space, then any two bases of V have the same number of elements. Proof: Let S] and
=
{VI' v2 , ••• ,v.,}
is a basis for V
S2 = {ul , u2 , ••• ,un } is another basis for V.
Then to show that In = n. Since S] is a basis for V and S2 is LI in V, ~ Number of elements in S2 can not exceed number of elements in SI' that is, n ~ In. Alternative,ly, if S2 is a basis for V and S] is LI in V, then In ~ n. And thus~ In = n. Corollary 2. Let V be a finite-dimensional vector space V and dim V = n. Then (a) any subset of V which contains more than n vectors is LO, (b) no subset of V which contains less than n vectors can span V. Theorem 2.10. In an n-dimensional vector space V, any set of n linearly independent vectors is a basis for V. Proof: Let V be a vector space and dim V = n. Let S
=
{vI' v2 ,. .. , v,,} is a LI set in V.
To show that S is a basis for V, enough to show that S spans V. Let v E Vbe any element. lfv = vi for some i (I ~ i ~ n), then nothing is to prove. Hence assume that v Vi' I ~ i ~ n. Then the set
*
vn ' v} is LO in Vas the number of elements in SI is S] = {vI' V 2 '··., vn' greater than dim V. ~
3 scalars
Cl .. Cl 2 , ... ,Cl"
and a, not all of which are zero, such that ...(2.4)
50
Introductory Linear Algebra
We claim that ex::t:Cl::t:- O. If possible, let Cl ex = O. Then equation (2.4) reduces to alvli + a 2v2+ .... + anvn = aiv
°v'
=> a l = 0 = a 2 = ... = an as S is LI Thus, a ll = 0 = a 2 2 = ... = an = a => SI is LI, which is a contradiction. ex 0, and then equation (2.4) can be rewritten as Hence, Cl
*
v
=
(-~}I+(-C;:}2+ ... +(-:}n
=>
S spans V Hence the theorem follows.
Lemma 2.1. Let S = {vI' v2' ... , v,,} be LI in a vector space V. Suppose v E V be such that v is not in [S]. Then S u {v} is LI. Proof: Let al' a 2 ••• ,, a" and ex Cl be scalars such that 2 ,, ... alvli +a 2v2+ .... +a"v" +av =Ov. aiv
Then a
=0,
...(2.5)
for if a:,t: 0, then
And it would imply that v E [S], a contradiction. Hence ex I-Ience Cl
=
0, and equation (2.5) reduces to
alvli +a 2v2+.... +a"v" = 0 aiv
=> a ll = 0 = a 22 = ... = a" = 01' as S is LI, but ex Cl also equals to
o.
Thus, S u {v} is LI. Theorem 2.11. Let W be a subspace of a finite-dimensional vector space V. Then any LI set in W can be extended, if necessary, to a basis for W, and dimW :S;dimV. Proof: If W ={Ov}, {OJ'} , then certainly dimW = O:s; dim V. So assume W::t:- {Ov} and let So = {v" v2 ' •• "v,,} is any LI set in W. If So spans W, then So is a basis for Wand we are done.
Vector Spaces
51
If So does not span W, then by lemma 2.1, there is a vector uIl in W such that SI = So u{u,} v{u1} is LI in W.
If SI spans W, we are done; otherwise 3 a vector u2 in W such that S2 = S, SI U V {u 2 } is LI in W. If S2 spans W, we are done; otherwise continu'e the above process. Finally, we reach a set
Sm
=
Sov{u"U S Ou{u"U22,.··,u,.} as a basis for W.
Since Sm is LI and number of vectors in a LI set can not exceed the dimension of V, and hence dimW::;dimV The following corollary follows from the above theorem.
Corollary: In a finite-dimensional vector space, every non-empty LI set is either a basis or it can be extended to fonn a basis. Remark: Let W is a subspace of a finite-dimensional vector space V. Then dimW =dimV <=> W = V.
Proof: If W = V, then obviously dim W = dim V Suppose dimW = dimV = n <
00 •
Let SI= {u tu"~ 1, u2 , ••• ,u.} is a basis for W.
=> SI is LI in W (and hence in V) and [SI]
=
W
But SI has n elements that equal to the dimension of V. Hence SI is also a basis for V.
=> [Sd = V Thus W = [Sd = V. Before proving an important result of Linear Algebra, we define the slim of two subspaces of a vector space V. Definition. If W WII and W2 are two subspaces of V, then their sum W WII + W2 is defined as W WII + W2 = {X+Y:XEJt; and YEW2 }. }· Readers can verify that W WII + W2 is a subspace of V (see exercise 2). Theorem 2.12. Let W WII and W2 are finite-dimensional subspaces ofa vector space V. Then W WII + W2 is finite-dimensional and dim(Jt; + W2) = dimJt; +dimW2 -dim(Jt; flW2)
52
Introductory Linear Algebra
Propf: Let dim(W; nW2 )= k, dimW; = m, and dimW2 = n. =:>
k 5, m and k 5, n .
= {u
Let SI
p
u 2 , ••• ,Uk } is a basis for
W; nW2 •
=:> SI is Ll and SI!';;;; W;, SI!';;;; SI !';;;; W2
SI is Ll and SI!';;;;W; =:> SI can be extended to form a basis for W WI. I• • Let S2 = {u p u 2 , ••• ,Uk , VI' V 2 , ••• 'Vm _ k } is a basis for W WI. I
Similarly, let S3 = {up Uu2 ' Let us consider the set
u,'
••• , U,' WI' Wp W 2 '·.·, , ••• ,
w
wn _k } is a basis for W2 •
S4 = {up ... , Uk' uk ' vp Vp ... ••• ,, V vm_ k ' Wp wp ... ... , W,,-k} Our aim is to show that S4 is a basis for W WII + W2. First, we shall show that S4 is LI. To do this, let
•
a, (15, i 5, k), b; (15, i 5, m - k) and c; (15, i 5, n - k) be scalars such that
Q,
k
m-k
n-k
LQ;u;+ La;u;+ Lb;v;+ LC;w; =Ov· =:>
;=1
;=1
k
m-k
;=1
n-k
...(2.6)
LQ;u;+ La;u;+ Lb,v; =- Lc;w;. ;=1
;=1
;=1
Clearly It
",-k
LQ,U; La,u; + Lb;v; ;=)
E
W; as S2 is a basis for
II-k
k
n-k
- LC;w; = LO.u; + L(-c;)w;
and
;=1
E
W2 as S3 is a basis for W2 .
;=1
;=1
Hence from equation (2.6) it follows that n-k
- LC;w;
E
W; nW2
=
L d,u;
;=1 n-k
=:>
-
L
k
C; W;
;=1 n-k
=:>
for some scalars d;-
,=1 k
LC; W; + Ld,u, = Qv 0v ;=1
W WI' I,
;=1
;=1
Vector Spaces
=> eC j = 0 for i = 1, 2, ... , n - k and di = 0 for i = 1, 2, ... , k, as S3 is LI. Then equation (2.6) yields It
m-/{ III-/{
La;u; + Lb,v; = 0v 1=1
;=1
=>
a j = 0 for i = 1, 2, ... , k; and b i = 0 for i = 1, 2, ... , m - k as S2 is LI. Hence S4 is LI. Clearly S4 spans the subspace WJ + W2 . Thus, S4 is a basis for WJ + W2 , and dim(~
+W2 ) = k+(m-k)+(n-k) =m+n-k = dimf~ +dimW2 -dim(~ nWJ
It may be noted that
~
n W2 is a subspace of V and
~nW2~~,~nW2~W2'
It follows that dim(~
n WJ::; W2)::; dim f~ < 00 and dim(~ n W2)::; dim W2 < 00.
Thus, dim (~ +
WJ <
00.
Now we shall give some examples based on basis and dimension. Example 30. Is the set S
={(I, 2, I), (-I, 1, 0), (5, -I, 2)}
a basis for V3 over
9\?
Solution. To show that S is a basis for V3 , one needs to show that (i) S is LI and (ii) S spans V3. (i) Let a, b, eC be scalars (rea Is) such that a (1,2, I) + b (-I, 1,0) + eC (5, -1,2) = (0, 0, 0).
5c = 0, a - b + 5e 2a+b-c= 0, 2a+b-e= a+2c=0. a+2e=0. On solving we obtain a = 0 = b = e. c. => Sis S is LI. (ii) Since S S is LI and number of elements in S S= 3 => S spans V3, by Theorem 2.10. Thus S is a basis for V3 .
=>
=
dim V3
54
Introductory Linear Algebra
Example 31. Is the set
S = {x-I, x 2 + x -I, x 2 -x+I} a basis for P2 over 9\? In not, find a basis for [S] and hence dim [S]. Solution: Let a, b, c be scalars such that a(x-l) + b(x 2 +x-I)+ C(X2 -x+I)=O, a zero polynomial in P 2
=> -a-b+c = 0, a+b-c = 0, b+c = 0. On solving we obtain, a =2c,
b =--c, and c is arbitrary. This shows that the linear system has infinitely many solutions and thus a nontrivial solution. Hence S is LD. One can take c = I, so, that a = 2, b = - 1. Hence 2(x-I)= 1.(x 2 +x-I)-1.(x2 -x+ I) Now consider the set
B
2
2
= {x +x-I, x -x+I}.
Then clearly B spans [S], (why?) and it is standard to check that B is Ll. Thus B is a basis for JS] and dim [S] =' 2. Example 32. Let S = {(Xl'X2,X3) E ~ : XI
+ x 2 + X3 =O} be a subspace of V3
over 9\. Determine a basis for S and hence find dim S. Solution: We have
S ={(Xl' x 2, x 3 )
V3 :x1 =-Xl -X3 } = {(-X 2 - X3'X 2,X3): Xl ,X3 E 9l} = {(-xl,XpO) +(-X3,0,X3): Xl ,X3 E 9l} = {X2( -1,1, 0)+X3( -1,0,1): Xl ,X3 E 9l}
=> S
E
= [{(-I, 1, 0), (-1, 0, I)}]
Clearly the set B ={(-I, 1,0), (-1,0,1)} spans S and it is easy that B is LI. Hence B is a basis for S and dim S = 2.
;'0
check
Vector Spaces
55
Examplc 33. Let S = {p(x) E ~ : p' (x) = o} be a subspace of P4 over 9\ . Find a basis for S and hence dim S. Solution: We have
= a+bx+cx 2 + d~3 +ex4 : 2c+ 6dx+ 12ex2 = o} 4 2 1 = {p(x) = a+bx+cx +dx +ex :c = 0 = d =e}
S = {p(x)
= {p(x) = a.1 +b.x: a,b E 9\} = [{I, xl].
Clearly the set B = {I, x} spans Sand B is LI. => B is a basis for S and dim S = 2. Exa mplc 34. Let W\ WI and W W2 be two distinct subspaces of a finite-dimensional vector space V such that dimV=n,dimlf'; =dimW2 =n-1. Prove that dim dim(W, (WI n W2 ) = n - 2.
Solution: Given that If'; i:- W2 •
=> :3 u (;to 0v) E Wj WI such that u Let U = [{ u}] + W2 =>
~
W2 .
dimU = dim [{u}] + dimW2 - dim ([{II}]n W2) = 1+ n - 1= n, since [{Ill] [{II}] n W2 = {Dv}' {Ov}'
Since [{u}] is a subspace of WI' Wj, it follows that U=[{u}]+W, is a subspace of Wj WI + W2, and Wj WI + W2 is a subspace subs pace of V.
=>
n= dimU::;; dim (If'; +W 2)::;; dimV = n
=> n= dim(W, +W 2 ) =
dim If'; +dimW2 -dim(1f';
n W2 ) = 2(n-I)-dim(1f'; n WJ.
=> dim (If'; n W2 ) = n - 2.
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ __ I. Show that (a) Every subset of a LI set in a vector space V is also LI in V. (b) Every superset of a LD set in a vector space is also LD in V. 2. If Wj WI and W2 be two subspaces of a vector space V, then show that WI + W W2 is also a subspace of V. their sum W\
Introductory Linear Algebra
56
3. Determine which of the following sets span V3? (0) {(-I, 2, 3), (0, I, 2), (3,2, I)} , (a)
(b) {(O, 0, 2), (2, 2, 0), (0, 2, 2)} , (c) {(3, 3, I), (I, 1,0), (0, 0, l)} .
Hints: One may use here the fact that if the set is L1, it spans V3; otherwise not. 4. Determine which of the following subsets of V3 are LI or LD? (a) S
={(I, 2, 3), (2, 3, 1), (3, 4,2)},
(b) Cb) S = { (3, 2, 5), (I, 4, 5), (2, -1, I)} ,
(c) S
={(I, 1,0), (0, 1,2), (2, 1,3), (I, 2, 4)},
(d) S
= {(O, 0, 0), (I, 0, 0), (0, I, O)}.
5. Determine which of the following subsets of C (0, 00) are LI or LD? (a) {x 3e",x Ze',(x 3 +x z +2)e'}, (b) {lnxZ,l nx 3,l nx 5},
(c) {cos z x, cos2x, I}. (d
6. Show that (I, i) is LD in Cover C, but (1, i) is LI in Cover 917. 1Iff {II, {II, V, w} be a LI set in a vector space, then check the linearly dependence and independence of the sets
(a) {u -v, v- w, w- u}, (b) {u+v,v+w, w+u}.
8. If {II" {lI p 112""'U,,} ll2""'U,,} is LI in Vn and n is odd number, prove that {Ill {III
+ Uz , liz II z + 113 , U3 + U4 ,,·••• •• ,1I,,_1 + 11", II", lI linn + lIj } is also LI in Vn .
9. If none of the elements appearing along the principal diagonal of a lower triangular matrix is zero, prove that row (or column) vectors are
LI. 10. Find three vectors in V3 which are LD, and are such that any two of them are LI. 11. Let {UI' UZ, ... ,un } be a basis of Vn . Show that {up aUI + u2' aUI + u3, ... ,aul + un} is also a basis of Vn for every ex in {u" 9\.
Vector Spaces
57
12. Let x = (x I' x 2) and Y = (y" Y2) be vectors in V2 such that z Z XIYI + xzYz = 0, xl + xi = YI + Y: = 1. Prove that S = {x, y} is a basis for V2 . 13. Determine which ofthe subsets S of V3 given in Problem 4 form a basis for V3? 14. Determine the dimension of the subspace [S] of V3 for each S in Problem 4. 15. Are the vectors III
u3 1I3
= (1,1,2,4), Ulizz = (2, -I, -5, 2), = (I, -1, - 4, 0), 1I4 u4 =(2, 1, 1, 6),
Ll in V4 over 9\? Find a basis for the subspace of V4 spanned by the vectors
lip liZ' up IIII]J
and 1Iu4•
16. Find a basis for the subspace U = {(xI' XZ' x 3' xX44 )) E V44 :: XI + Xz = X3 + X44 }}
over 91. Which of the following sets also form a basis for U: (a) S ={(-1, I, -I, I), (-1,0,-1,0), (0, 1, 0, I)}, (b) S = {(l, -I, -I, 1),(-1, 0, -I, 0), (0, I, 0, I)}.
17. Let ~ = {(x" x Z' x J ) E ~ : 2x, + Xz + 4X3 = O}, W2 = {(XI' x 2' x 3) E V3 : XI - x 2 + X3 = O}. (a) Show that W WI1 and W2 are subspaces of V3 over 91.
(b) Determine bases of (i) W/ WI (ii) W2 (iii) W WI1 n W2 and (iv) W WI1 + W W2 (c) Find dim Wl' WI' dim W W2, dim (Wl (WI n W W2) and dim (Wl (WI + W W2 ).
18. Consider the vector space P 4 over 9\ of all polynomials with real coefficients of degree atmost 4. Let W={p(X)E~ :p(l)=O}. (a) Show that W is a subspace of P4 over 91. (b) Determine a basis for W, and hence find dim W. 19. If every vector of a vector space V has a unique linear combination up U u2Z' ••• ' un; then show representation in terms of the non-zero vectors Ill'
that the set S = {up U u2 ' ••• ' un} is a basis for V. 20. Determine if S = {x + 1, X2 + x -l, X2 - X + I} is a basis for P 2 over 9\ ? 11. Let S = {vI' v vZ' p ... ,' VII} be a Ll set in a vector space V over
~R
58
Introductory Linear Algebra
°
k
Write
wI Wj
VI' w k = ~>ikVi; 2 ~ k ~ n, where cik > = vI' for all i(l ~ i ~ k) 1=1
Show that the set T = {wl' {wI' w2 ' ••• , wn }
22. Let
ul'
u2'
... ,u"
IS LI in V over 9t.
be vectors in a vectol\ space V such that
(t) V = [UI'U 2 ,U3""u.l, (ii) II IIIj + u2 and u Ij - 2u2 both are in [u 3 , u 4' Prove that V =
[113'
... ,un ].
u4 , ... ,u,,].
23. Let S = {(I, I, 0), (0, I, - I), (I, 2, -I)} be a subset of a real vector space V)' Determine whether the set S is LI or LD. IfS If Sis is LD, find a TJ subset A of S such that [AJ = [S], [5J,
24. Let S = {(I, 1, 0), (0, 1, -1), (0, 1, 0), (1, 2, 3)} be a subset of a real vector space V)' Determine whether the set S is LI or LO. LD. If S is LO, LD, find a LI subset A of S such that [A] = [s]. 25. Let S = {(J, 0, 0), (0, I, 0), (I, I, 0), (2, 2, o)} be a subset of a real vector space V)' Determine whether the set S is LI or LO. LD. If S Sis is LD, find a Ll subset A of S such that [A] = CS]. [S]. 26. Let Wj WI and W2 be subspaces of V V8 and dim Wj WI = 6, dim W2 = 5. What are possible dimensions of Wj WI n W2? 27. Does there exist subspaces Wj WI and W2 of V7 such that dim Wj WI = 4 = dim W2 and IV WIj n W2 = {o}? {OJ? 28. Let V be the vector space of all 2 x 2 real matrices. Find the dimension and basis for V. Let Wj WI be the set of all matrices of the form [;
W
and W2 be the set of all matrices of the form [ a
-a
~x]
b]. Prove that W WI C
1
and ~2 are subs paces of V. Find the dimensions of W WI' WI + W2 I , W2 , W) and w) WI n W2 . 29. Let V be an n-dimensional vector space over the field F. Suppose that S is LD in V. Prove or disprove: S contains more than n vectors. 30. Let W) WI be a subspace of a vector space V. Let U (:;t:0v) E V be such that U ~ W). WI' Let W2 = [(u}]. [{u}]. If dim W1= 5, dim V = 10; find dim (W) (WI + W2 ). 31. Let W) WI he an m-dimensional scbspace of an n-dimensional vector space V. Prove that there exists an (n - m) dimensional subspace W2 of V such th,,' W] WJ + W2 = Vand Wj WI n W2 = {Ov}.
~3
I] LINEAR TRANSFORMATIONS
In many physical systemS- vector spaces are transformed into another vector spaces by means of an app'ropriate transformation. In this chapter, readers will be introduced with linear transformation that sends each element of a vector spaces into another vector space preserving some basic properties.
3.1
LINEAR TRANSFORMATIONS
Definition. Let U and V be 'two vector spaces over the same field F. A linear transformation from U into-V is a function T such that (I) T(u+v)=T(u)+T(v) for all £I, v in U and
(ii) T(au) = aT(u) for all u in U and for all a in F.
Example 1. Let U be any vector space over the field F. Then the identity transformation T = I from U into U defined by I (u) = u for all u in U is a linear transformation, because I (£I + v) = U + v = I (£I) + I (v) for all u, v in U and I (a u) = au = a I (u) for all u in U and a in F. Example 2. Let U be any vector space over F. Then the zero transformation T= 0 from U into U defined by 0 (u) = 0u for all u in U is obviously a linear transformation, where 0 u is the zero vector in U. Example 3. Let Pn be a vector spa~e of all polynomials with real coefficients of degree atmost n over m. Let the map D : Pn ~ Pn is defined by D (p (x» = p'(x), called differential map. Then D is a linear transformation. To see this, let
and
2 n p(x) = ao + alx+ a2 x + ... + anx E Pn 2 q(x) = bo +b1x+b2x + ... +bllx" E Pn ·
59
60
Introductory Linear Algebra
Then D(p(x) + q (x)) = D«ao +bo ) + (al +bl)x + ... + (an +bn)x n ) n = (al +q) + 2(a2 +b2 )x + ... + n(an +bn)x - I = (al + 2a2x+ ... + nanx n- I ) + (b l +2b2x+ ... + nbnx n- l ) = D(p(x)) + D (q(x)).
If a is any scalar, then D(a p(x)) = D«aao) + (aal)x + ... + (aan)xn) n I =(aal) + 2(a a2)x + ... + n (aan)x n I = a (al + 2a2x + '" + nanx - ) = aD(p(x)).
Thus D is a linear transformation. Example 4. Let T: V2
~
V2 be a map defined by T(ul' u2) = (u2 +4, ul)'
Then T is not a linear transformation, because for
a
E ~H,
U
= (ul' u2) E V2
and
we note that T(all) = T(aul,a1l2)=(au2+4,aul)' and aT(u) = a(u2 +4,lIl)= +4,1I1)= (au2 +4a,aul)
This show that T(au):t:-aT(u) if a :t:- 1. Hence T is not a linear transformation. Definition. If V be a vector space over the field F, then a linear transformation T from V into itself is called linear operator. In the following theorems, we shall give some basic properties of a linear transformation. Theorem 3.1. Let U and V be vector spaces over the same field F, and T: U ~ V be a linear transformation. Then (a) T(Ou) = Ov, where 0u and 0v are zero elements of U and V,
respectively. (b) T(-u) = - T(u) for all u in U.
Proof: Since T is linear transformation and hence for all u in U and for all scalars a,
linear Transformations
61 T(au)=aT(u)
Substituting a = 0, we obtain (a) and for a = - 1, we obtain (b). Theorem 3.2. Let U and V be vector spaces over the same field F. Then the map T: U ~ V is a linear transformation iff T(auj T(aul + u2) = a T(uj) T(lll) + T(u2) for all up u2 in U and a in F. Proof: First, suppose that T is a linear transformation. Then for all up ll2 u2
in U and a in F, we have
T (a Uj III + u2) ll2)
= T (a Uj) lll) + T(u2) T(1l2) = aT(uj) aT(uI) + T(u2).
Conversely, suppose that T (a Uj III + u2) ll2) = aT (Uj (ul ) + T (u2) for allll allu l , u2 in U and a in F. Then for a = I, we obtain T (lll (u j + ull2) (u j ) + T (u (1l2); 2) = T (lll) 2); and for U ll22 = v' u' we obtain
°
T(alll T(auj
+ 0v) 0u) = a T(lll) T(uj) + T(Ou)
T(auj) T(auI) = aT(uj)+Ov aT(lll)+Ov =aT(uj). =aT(lll).
Hence T is a linear transformation. The proof offollowing theorem is similar to Theorem 3.2 and hence left as an exercise for readers (see exercise 1). Theorem 3.3. Let U and V be vector spaces over the same field F. Then T is a linear transformation iff· inF. inFo
T(a1 Uj ul + a2 u2) ll2) = aj al T(uj) T(uI) + a2 T(u2) for all up u ll22 in U and aj' ai' a 2
In fact. by induction on n one can show that for a linear transformation T, we always have T(allil T(ajuj + a2u2 a2ll2 + ... + allll,,) = aj al T(llj) T(lll) + a2 T(1l2) + ... + an T(u lI ) ... , all in F.
for all lll' u j ' 112' ... , u" ll" in U and a p a 2 ,
Remark: Theorem 3.2 or 3.3 can directly be used to show that the given map is a linear transformation. In the following theorem we shall show that a linear transformation is completely determined by its values on the elements of a basis. Theorem 3.4. Let U be a finite-dimensional vector space over the field F and let {u {lll' ll2' j, u 2 , ... , un} be an ordered basis for U. Let V be a vector space
62
Introductory Linear Algebra
over the same field F and let vI' v22' ... , vn be n vectors (not necessarily distinct) in V. Then there exists a unique linear transformation T: U ~ V such that T(u;) = Vj
for
i
=
1,2, ... , n.
Proof: Let U E Ube any element, and given that B = {u l , u2, •.. ,un } is a basis for U.
=> :3 scalars ai' a 2 , .•• ,an such that U U = al ul + a2 u2 + ... + an un . For this vector u in U, we define T(u) = a l vI + a 2 v2 + ... + an un·
Clearly T (u (1I 1 I) = T (l.uI
+ O.u2 + ... + O.u n ) =
l.vI
+ O.v2 + ... + O.vn = vI
T (u (1I 2) = T (O.uI + l.u2 + O.u3 + ... + O.u O.lIn ) = O.vI + l.v2 + O.v3 + ... + O.vn = v2 T{lI n ) = T(O.ul + ... +O.un_l T{u
+ l.u n ) = O.vI + ... +O.vn_1 +l.vn =vn .
=> T(u) = vi for all i = 1, 2, ... n. Now we shall show that T is linear. To do this, let v element. => There exists scalars 13 1, f3 2, ... f3 n such that
E
U be any
v = 131 UI + 132 U2 + ... + f3 n Un . Now for any scalar a, we h!lVe T(au + v) = T«aal + f3l)ul + (aa2 + (32)U2 + ... + (aa n + f3n)U n )
+ (3 1)VI + (aa2 + (32)v2 (32)V2 + ... + (aa n + f3n)V f3n)vn Vn) (a\v\ (alvl + a2 v 2 + ... + anVn) anvn ) + (f31 vl + f3 2v2 + ... + f3n f3nVn)
= (aal =a
=aT(u)+T(v).
This implies that T is a linear transformation. Finally, we shall show that T is unique. To do this, let S : U a linear transformation such that S(ui) = Vj'
Then,
~
V be
i = 1, 2, ... ,n.
S (u) = Seal uI ul
+ a2 U2 + ... + an un)
= a I S(uI)+a2 S (u2)+···+a n S(un )
= al VI + a2 v2 + ... + an vn = T (u). The above relation is true for all u in U and hence S = T.
Linear Transformations
63
Example 5. Suppose that T: V2 ~ V2 be a map such that T (1, 2) = (1, 0) and T (2, 1) = (2, 3). We wish to know whether there is any linear transformation following the above rule or not. If yes, then what is the general formula for T. To see this, we shall apply Theorem 3.4. If B = {(l, 2), (2, I)} is a basis for V2 , it will imply that T is a linear transformation. It can easily be checked that the set B is LI and since number of elements in B = 2 = dim V2 , it follows that B is a basis for V2 . Thus T is a linear transformation.
To find a general rule that defines T, let (x, y) E V2 be any element. ~
(x, y) = a(1, 2) + b(2, 1) for some scalars a, b.
~
a
~
a=-3"x+3"y,b=3"x-3"Y'
+ 2b = x, 2a + b = Y 1
2
2
1
Since J' is a linear, therefore T(x, y) = T(a(1, 2) + b(2, 1» = aT(1, 2) + bT(2, 1) = a(1, 0) + b(2, 3) =(a+2b,3b) = (x, 2x - y).
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ __ 1. Prove Theorem 3.3.
2. Which of the following functions are linear transformation? (a)
T:V2 ~V2 defined by T(xt>x2)=(l+xl,x2)'
(b)
T: V2 ~ V2 defined by T (xl' X2) = (XI
(c)
T:V3~V2
(d)
T:P2~V2 defined by T(aO+alx+a2x2)=(aO-al,a2)'
(e)
T: V3 ~ P2 defined by
+ X2' 0) .
defined by T(xl,x2,x3)=(xl-x2'0).
T(x\> x2' x3) = Xl T(xb
+ (Xl
-
x2)x + (Xl - X3)x2.
3. In the following problems, determine ifthere exist a linear transformation. If it does exist, find a general formula for it.
Introductory Linear Algebra
64
(a) T: V2 ~ V2 such that T (2, 3) = (1, 0), T (1, - 1) = (1, 1). (b) T: V:; ~ V3 such that T( -1,1,0) = (1,1,1) and T(I, 2, 3) =(2, 2, 0). (c) T:V2 ~V2 such that T(3, 4) = (1, 2), T(I, 3) = (1, 0), and
T (2. 1) = (1,- 1). (d) T: V3
~ V2
such that T (1, -1, 1) = (1, 0) and T (1,1,1) = (0, 1).
(e) T: V3 ~ V3 such that T (1, - 1. 2) = (1,2,4), T(2, 3, 1) = (2,1,3) and T(3, 4, -2)=(3, 0,2).
3.2
RANGE AND NULL SPACES
In this section we shall study two interesting subspaces associated with a linear transformation and their properties. Definition. Let U and V be vector spaces over the field F and let T: U V be a linear transformation. Then R(T)= {v E V: T(u)=vfor some u E U} is called range space of T, and N(T) = {u
E
~
U: T(u) =Ov}
is called null space of T. Remark: Null space of T is also called kernel of T and is denoted by ker(7). However, we shall use in this text only N (1) to denote the null space of T. The following result in basic in nature about R (1) and N (7). Theorem 3.5. Let U and V be vector spaces over the field F, and T: U ~ V is a linear transformation. Then (a) Ca) R (1) is a subspace of V, and (b) N (1) is a subspace of U. (a) Since 0v= T(Ou) Proof: Ca) ~
0v EER R (T)
~
(T):;t:~. R CT):;t:~. Let vI' v 2 E R (1) and a
~ :3
Ut' !I" u2
E
E
F.
U such that T(ll,)
= v" T(ll?) = v2.
Now, av, + v2 = aT (u,) + T (u 2) = T (au Il + u2), as T is linear
Since aau: u: + 112
E
U as U is a vector space, it follow that a VI
+ v2
E
R (n (T) .
Linear Transformations ~
65
R (7) is subspace of V.
(b) Since T(Ou) = 0v~ 0u
Let lit, 112 EN (T)
(U t ) ~ T (Ut)
E
N(7) ~ N(7)"* <1>.
Op T (U 2 ) = 0v. = 0V' QV·
Then for any scalar a, we have T(aut +1I2)=aT(ut)+T(u2)' as Tis linear = a.Ov +O~ =Ov. ~
alit a lit + U2 u2 E E N(T)
~
N (7) a subspace of U.
Theorem 3.6. Let U and V be two vector spaces over the field F and T : U ~ V is a linear transformation. Then (a) T is one-one iff N (7) = {O u}. (b) If [Ut,U2, ... 'u n ]=U, then R(T)=[T(ut),T(u2), ... ,T(un )].
Proof: (a) Suppose T is one-one. Then to show that N(7) = {nu}· {au}· Since N(7) is a subs subspace pace of U, hence 0u
E
N (7)
~ {Ou} ~ N (T) .
To show that N(T)
~
{Ou} , we let
U E
N (T) be any element.
~ T(u)=Ov =T(Ou) ~
u = 0u as T in one-one u E {nu}. {au}. Hence N(T) ~ {Ou} .
~
Thus, N(T) = {Ou}. Conversely, let N (7) = {Ou}. Then to show that T is one-one. To do this, let Ut' u t ' 112 E U be any elements such that T(ut) = T(u2)
=> => => =>
T(Ut- u2)=T(ut)-T(u2)=OV
N(T) = {Ou}
lit -u2
E
Ut -- 1I 2
=0u
Ut
=u2·
This show that T is one-one. This completes the proof of part (a). (b) Given that [up u2' ... un] = U. Then to show that R(T) = [T(ut), T(u2), ... , T(un )]·
66
introductory Linear Algebra
Let v Since
E
U E
R (1) be any element => :1 U = [Ut' u2' '" un]
U E
U such that T (u) = v.
=> U = T(u) = v=at v=Ut T (ut)+a2 (Ut)+u2 T (u2)+···+ v E [T(ut), T(u2)"'" T(u n )] Hence R(T) <;;;; [T(l't), T(u2), ... , T(u n )]. Conversely, let
=>
W=
WE
[T(Ut), T(U2)' ... , T(u n )] be any element.
PtT(Ut) +P 2T(u2) + ... + PnT(un) for some scalars
Pi'
=> w=T(PtUt +P2U2 + ... +PnUn), as Tis linear => WE R(I) as PtU + '" + PnUn E U Hence [T(ut), T(u2)' ... , T(u n )] <;;;; R(T). Thus, R(T) = [T(ut), T(u2) , ... , T(un )] . This completes the proof of part (b). Example 6. Let T: V2 ~ V3 be a lineaNransformation defined by T(xt, x2) = (Xt, Xt -x2' x2) Suppose that we wish to know whether T is one-one or not. For this purpose, one may use part (a) of Theorem 3.6. We have N(T) = {x = (Xt, x2) E V2 : T(x)
= T(xt, x2) = ov)} = {(Xt, x2)E V2 : (Xt, Xt -x2, x2) = (0, 0, o)} = {(x\> x2) E V2 : Xt = O,Xt -x2 = 0,x2 = O} = {CO, O)} = {Ovz }'
=> T is one-one. Example 7. Let T: V3
~
V3 be a linear transformation defined by
T(l, 0, 0)
and
= (1,0,1),
T(O, 1, 0)
= (0, 0, 1)
T(O, 0, 1) = (1, 0, 0).
Suppose that we wish to know R (1). For this purpose, one may use pm1 (b) of Theorem 3.6.
Linear Transformations
67
We observe that the set
B ={(1, 0, 0),0, I, 0), (0, 0, I)} is a basis for V3 , and hence [B] = V3 . Therefore R (T) = [T(1, 0, 0), T(O, 1, 0), T(O, 0,1)] = [(1, 0, I), (0, 0, 1), (1, 0, 0)] = [(0,0,1), (1, 0, 0)], since (I, 0, 1) = (0,0,1)
+ (1, 0, 0)
= {x(O, 0, 1)+ y(l, 0, 0): x,y E 9\} = {(y, 0, x): x.y
E
9\}
Now we shall define rank and nullity of a linear transformation. Definition. Let U and V be vector spaces over the field F and T: U be a linear transformation. If U is finite dimensional, then
~
V
rank of T = dim R (7). and nullity of T = dim N (7). Remark: dim R (7) is sometimes denoted by r (7) and dim N (7) by n (7) .. The following theorem is a very important result in linear algebra, known as Rank-Nullity Theorem. Theorem 3.7. Let U and V be vector spaces over the field F and let T be a linear transtormation from U into V. Suppose that U is finite dimensional. Then, dimR(T) + dimN(T) = dimU. Proof: Let dim U = n and dim N (7) = k. ~ k ~ n as N (7) is a subs subspace pace of U. Let BI B,
= {u, ,u2, ... ,ud
is a basis for N (7).
Let B, is extended so that B2 = {11.u2.···. {1,.u2.···. uk'"k+I'···.u k'"k+"···.un } is a basis for U. We wish to show that B={T(Uk+l).T(Uk+2)' ...• T(un )} is a basis for R(7).
To do this, we shall show that (i) B is LI, and (ii) B spans R (7). To show (i), let <Xk+l' <Xk+2' .... , <Xn- be scalars such that
f
I=k+\
ui T (II / )=OV
~
T(
f
I=k+l
U/lliJ=OV U/IliJ=OV
as Tis linear
Introductory Linear Algebra
68 n
=>
L
Uilli
N(T)
E
,=k+1
n
k
i=k+1
i=1
=> L a,llj = L~,Uj, La/IIi L~/Ui' as BI is a basis for N(7) k
h
11
=> Lp,lIi+ L ,=1
=>
(-U,)Ui=OU
i=k+1
Pi = 0 for i = 1,2, ... , k;
~a i
and
= 0 for i = k + 1, k + 2, ... , n as B2 is LI.
This show that the set B is LI. Now to prove (ii), we note that any element in R (7) is of the form T (u) for some U in U. Since B2 is a basis for U and U is in U, therefore II il
U = LCiUi for some scalars ccj"r i=1
= T(±CiUi + i=1
±
CiUi)
i=k+1
(3.1)
=T(±CiUi)+T(.± CiUi) 1=1
=>
t
CiUi
E
l=k+1
N(T) => T(t CiUi ) = 0v.
Therefore, equation (3.1) reduces to T(U)=T(
f
CiUi)=
i=k+1
=> B spans R (7). Thus, we have shown that B is a basis for R (7). Hence dim R (7) = n - k. Thus, dim R(T)+dimN(T) = n-k +k = n = dimU.
f i=k+1
ciT(ui)'
Linear Transformations
69
Remark 1. Suppose T: U -+ V is a linear transformation and dim U <
00.
Then, dim R(T) :s; dim U, because dim R(T) + dim N (T) = dim U < 00. Remark 2. If T: U -+ V is a linear transformation and dim U <
00.
Then,
dimR(T) :S;min{dimU, dim V}.
Proof: Since R (1) is a subspace of V => dim R (1)~dim V. From Remark 1, dim R (1):S; dim U Hence dimR(T):S; min{dimU, min {dimU, dim V}. Remark 3. If T: U -+ V is a linear transformation and dimU < 00 • Then dim R(T) = dimU ~ T is one-one. Proof: .: dim R(T) + dimN(T) = dimU .. dimR(T)=dimU
~
dimN(T)=O
¢::> (:::>
N (T) ={Ov }
¢::> (:::>
T is one-one.
Remark 4. If dimR(T)
~
R(T) = V
~
dimR(T)=dimV
Remark 6. dim R (T) = dim U = dim V ~ T is one-one and onto.
Example 8. Let
T: V3
-+ V4 be a linear transformation defined by
T(xl> x2' x3)
=(Xl' Xl -
X2' Xl - X2 - X3' X3)
Determine N(T),R(T),dimN(T),dimR(T) and verify the RankNullity Theorem. Solution: N (T) ={(xl' X2' X3) E V3
= {(Xl' X2' X3) E V3 : (Xl' Xl - X2' Xl - X2 = {(Xl' X2' X3) E V3 : Xl =0, Xl - X2 =0, Xl ={(Xl,X2,X3):Xl =0=X2 =X3}
= {CO, 0, 0) = 0v Of/. }} 3
=0v4 } X3' X3) =(0,0,0, O)} - X2 - X3 = 0, X3 = O}
: T(Xl. T(XI. X2. X3)
70
Introductory Linear Algebra
=> dimN(T)=O One may note here that T is one-one. Now we have R(T)={T(x" x2, x3): (x" x2, x3) E V3}
= {{(x"x, -x2,x, -x2 -x3,x3):x"x2,x3 are any real number"} = {(x" x" x" 0) + (0, -x2, -x2, 0)+ (0, 0, -x3' x3) :x" x2' x3
are any real numbers} ={x, (1,1,1,0) + x2 (0, -1, -1, 0) + x3 (0,0, -1, 1): x" x2' x3
are any real numbers}
_=[{(l, I, I, 0), (0, -1, -
1, 0), (0, 0, - 1, I)}]
Clearly, the set
S = {(l, {(1, 1, I, 0), (0, -I, -I, 0), (0, 0, -I, -1, I)} spans R (T), and it is standard to verify that S is LI.
=> S is a basis for R (T), and hence dimR(T) = 3.
Since dim N (T) + dim R (T) = 0 + 3 =3 = dim V3, and hence it verifies the rank - nullity theorem. Remark for Example 8. Suppose that we do not wish to verify the ranknullity theorem. In such case, one need not check the linear independence of the set S. After knowing dim N (T), one may directly use the rank-nullity theorem to find dim R(T) . For example, in the above case dim N (T) = 0 . Hence. dim R (T) Example 9. Let T: P3 -+
~
=dim V3 -
dim N (T) =3
be a linear transformation defined by
T (ao + a,x + a2x2 + a3x3) = ao
+ (a, + a2 + a3)x + (a, - a2 + a3)x3 .
Find N (T), R (T), dim N (T) and dim R (T) . Solution: We have N (1') = {p(x)
=Qo + Q,X + Q2X2 + a)x 3 : T(p(x» =0, a zero polynomial}
Linear Transformations
=
71
{p(x) = ao + alx + a2x2 + a3x3 : aO + (a) + a2 + a3)x 3 +(a)-a2 +a3)X =o}
= {aO + alx + a2x2 + a3 x3 : aO = 0, al + a2 + a3 = 0, al - a2 + a3 = o} ={aO +aIX+a2 X2 +a3 x3 :aO =0,a2 =0,a3 =-a!l 3 = {a)x - alX : al e ill} =
[{x - x 3 }]
Clearly, {x - x 3 } is a basis for N (T) (7)
=> dim N (T) = I . Now we have, R(1) = {T(p(x»:p(x)=ao +alx+a2x2 +a3 x3 eP3}
°
3 = {ao + (al + a2 + a3)x + (al - a2 + a3)x : aj e 9t for all i = to 3} 3 = {ao +blx+b2x :ao,b) =al +a2 +a3 and b2 = al -a2 +a3
all are in ill} =
[{t, x, x 3 }]
And dimR(T)=dimP3 -dimN(T)=4-1=3. Remark for example 9. It may be noted here that the set S = {I, x, x 3 } spans R(T) and S is LI. Hence dimR(T) = 3.
EXERCISES _______________________________ 1. 1fT: V2
~ V2
be a linear transformation such that T (I, 2) = (I, I) and
T(2, I) = (0, I); find. R (7). (T).
2. If T: P2 T(x)
~
=(1,1)
V2 be a linear transformation .such that T(I) = (1,0),
and T(x 2 ) = (2,2) ; find R (T). (7).
3. Let T: Vm ~ Vn be a linear transformation with m > n. Is Tone-one? Justify your answer. 4. Let T: Vm ~ Vn be a linear transformation with m < n. Is Tonto? Justify your answer.
Introductory Linear Algebra
72
5. Find a linear transformation T: V3 N (T)
V3 such that
= {(x), x2' x3) E V3 : x)
6. Find a linear transformation T: V3 N (T)
~
= {(x), x2' x3) E V3 : x)
~
- x2 + x3
V3 sllch that
- x2 + x3
7. Find a linear transformation T: V3
~
= O}.
=0, 3x)
- x2 + x3
= O} .
V3 sllch that
R(T) = {(x), x2, x3) E V3: x) - x2 + x3 = O}. OJ.
8. Find a linear transformation T: V3
~
V3 such that
R(T) = {(x),x2' x3) E V3 :3x) - x2 + x3 = OJ. O}. 9. In the following problems, for the linear transformation T as defined in
the problem~, find N (T), R(T), dim N (T), dim R(T), a basis for N (1) and a basis for R (7). (a) T:V2 ~V2 defined by T(X),X2)=(x) -x2,x) +x2)' (b) T:V2 ~V3 defined by T(x),x2)=(x),x2'x) -x2)' (c) T: V4 ~ V3 defined by T(x), x2' x3, x4) = (x), x) - x3' x3 - x)). (d) (cl) T: V4 ~ V3 defined by T(x), x2' x3, x4) = (x) - x2' x2 - x3, x3 - x4)' (e) T: P2 ~ V2 defined by T (ao + a)x + a2x2) = (ao + a), a) + a2) .
(j) T:V3 ~V3 defined by T(x),x2,x3)=(x),x) -x2'x) -x2 -x3)' (g) T: P3
~ ~
defined by
T(ao + a)x + a2x2 + a3x3) = a)
+ (ao -a))x + (a) + a2)x2 + (ao + a) + a2 + a3)x3
2 (h) T: V2 ~ P2 defined by T(x), x2) = x) + (x) + x2)x + (x) - X2)X .
(I) T: V3
~
P2 defined by 2
T(X),x2,x3)=x) + (x) +x2 -X3)X + (x) +x2 +X3)X
(J) T: V3
~
V3 defined by T(xl' x2' x3) = (3x) , xI - x2' 2xI + x2 + x3)'
10. Let U and V be vector spaces over the field F. Let T: U
~
V be a
is oneTis linear transformation and suppose dim U = n < co . Prove that T one if and only if dim V = n .
Linear Transformations
3.3
73
THE ALGEBRA OF LINEAR TRANSFORMATIONS
First of all, we shall show that the set of linear transformation is a vector space. Theorem 3.8. Let U and V be vector spaces over the field F. Let T and S be linear transformations from U into V, then (a) T
+ S:U ~ V defined by
(T + S)(u) = T(u) + S(u) for all u in U
is a linear transformation. (b) If ce is any element in F, then cT: eT: U ~ V defined by (cl) (u) = eT(u) cT(u) for all U in U is a linear transformation. Proof: Let R = T + S. Then for any Ut' u t ' u2 in U and a in F, we have R(allt + u2) = (T + S)(a Ut + u2) =
T(aul + u2) + S(aul + u2)
=
lI l) + S(u2)' aT(ul) + T(u2) + as( 1I1)
as T and S are linear transformations. =
a(T(ul) + S (Ut» + T(u2) + S (u2)
= a «T
+ S)(Ut» + (T + S)(u2)
= aR(ut) + R(u2)' => R = T + S is a linear transformation. cT. Then Further, Let R* = eT.
R* (aUt + u2) = (cT)(aut (eT)(aut + u2) = eT(aut cT(aut
+ u2)
= e(aT(ut) c(aT(ut) + T(u2», as T is linear = eaT(u\)+eT(u2) caT(u\)+cT(u2) =
a (eT)(ut) (cT)(ut) + (eT)(u2) (cT)(u2)
= a R*(ut)
=> R*
= cT eT
+ R* (u2) for all up u2' in U and a in F.
is a linear transformation.
Remark: Theorem 3.8 shows that the set of all linear transformations from
U into V is a vector space over F under the operations as defined in this theorem, denoted by L( U, V). It should be noted that L( U, V) is defined only when U and V are vector spaces over the same field F.
Introductory Linear Algebra
74
Theorem 3.9. Let U and V be vector spaces over the field F and dim U = n < 00, dim V = m < 00. Then dim L (U, V) = mn < 00 • Proof. Let B2 = {vI' v2, ... , vm }
be ordered bases for U and V, respectively. For each pair of integers (p, q) with I $ P $ m, 1 $ q
$
EM:U
n, we define a linear transformation ~
V by
EP,q (u .) = { 0, j.* q Vp,j=q } =
ojqvp,l$j$n.
Here 0 jq is the Kronocker delta and it may be noted that such linear
.
transformation exists uniquely by Theorem 3.4. We claim that the set B= {EM:I$p$m,l$q$n}
forms a basis for L (U, V). It may be noted that B has mn elements. First of all, we show that B- spans L (U, V). Let TeL T e L (U, V) be any eleI:l1ent. m
=> T(uj) =
L
Apj Vp vp , for sQme sQrne scalars Apj' Api' 1 $j $ n,
p=1
(sinceT(uj)eV and B2 is a basis for V). m
Let
S
=
n
L L
Apq EM
p=1 q=1
m n
=>
S(Uj)
=
L L Apq EM (Uj), 1$ j $ n p=lq=1
m
=
n
L L
Apq Ojq Vp vp
p=1 q=1
m
=
L
Apj Vp vp
p=1
= T (u (U j),
1$ j
$ n
Linear Transformations
75
S =T. Hence B spans L (V, V). Finally, we show that B is Ll. n
1/1 III
Let S
=
LL
ApqEP,q = 0, a zero transformation.
p=1 q=\ 1/1 III
=> S(Uj)
=
n
L L Apq Ep,q (Uj) = O(Uj) = Qv, 0v, 1 S; j S;n p=1 q=\
III
=>
n
L L
Apqojqvp =Ov
p=\ '1=\ 1/1 III
=>
L
Apj vp =Ov
p=\
=> Apj = 0, I S; j
S;
n, 1 S; P pS; S; m as B2 is LI => B is LI
Thus B is a basis for L(V, V) and dim L(V, V)
= mn.
Now we shall define the composition of two linear transformations. Definition. Let V, V and W be vector spaces over the field F. Let T: V ~ V and S: V ~ W be linear transformations. Then composition of Twith S, defined by SoT or simply ST, is a map from V to W, that is, ST:V ~W defined by (ST)(u)=S(T(u)) for all U in U.
In the following theorem we show that the romposition of two linear transformations is again a linear transformation. Theorem 3.10. Let V, V and W be vector spaces over the field F. Let T: V ~ V and S : V ~ W be linear transformations. Then composition map ST: V~ W defined by (S1) (u) = S (T (u)) for all u in V is a linear transformation. Proof: We have (ST)(au\
+ u2)
~ S(T(au\
+ U2)) , by definition.
= S(aT(u\)+T(U2)), = as(T(u\)) =
as Tis linear
+ S(T(u2)) ' as S is linear
a (ST)(u\ ) + (ST) (u2) , for all u\' Ut, u2 in V and a in F.
=> ST is a linear transformation.
Introductory Linear Algebra
76
Example 10. Let T and S be linear operators on V2 defined by T(xl' x2) = (x2, xI),
S (xI' x2) = (xI' 0) .
Find a general formula for S + T, ST, and T. Solution : We have
f2 and S2 like one defines S
(I) (S+T)(xl,X2) = S(xl,x2)+T(xl,x2) =
(xI> 0) + (x2' xI)
= (XI + x2, xI) . (ii)
(ST)(xl' x2) = S(T(xl,x2»
= S(x2' xI) = (x2' 0). (iii)
(TS)(xl' x2) = T(S(xl> x2»
= T(xl'O) = (0, xI). (iv)
T2 (Xl' X2) = T(T(XI' X2»
= T(X2,XI) = (XI' X2)· (v)
S2 (XI' X2) = S(S(XI' X2)) = S(XI'O) = (XI' 0).
__ __ __ __ __ __ __ __ __ __ __ _ __ EXERCISES _ _ _ _ I. Let T and S be linear operators defined on V2 by T(xl' x2)
= (0, x2),
S(xl' x2)
= (-xI, -
x2)
Find a general rule likes the one defining Sand T for the linear transformations S + T, ST, TS, S2, T2. 2. Let T: V3
~
V2 be a linear transformation defined by T(xl,x2,x3)=(xl -x2,x2 -X3)
77
Linear Transformations and S : V2 -+ V2 be another linear transformation defined by S(xI' x2) = (X2' Xl)· Xl). S(xl'
Find a general formula for ST like one defines for Sand T. 3. Let T: V2 -+ P 2 be a linear transformation defined by T(xi' x 2) = XI Xl
+ (Xl (XI
- x 2) X + (Xl (XI
+ x 2)x2,
and S : P2 -+ P2 be another linear transformation defined by x » = p' 2(x). S Find a general formula for ST like one defines for Sand T.
(Pix» (Pi
INVERSE OF A LINEAR TRANSFORMATION
3.4
Definition. Let U and V be vector spaces over the field F, and T: U -+ V be a linear transformation. Then T is said to be invertible if there exists a function S from V into U such that ST is the identity function on U and TS is the identity function on V. If T is invertible, then the function S is unique and is called inverse of T; and we write S = rl. Further more, a linear transformation T: U -+ V is invertible if and only if (a) T is one-one, that is, T(ul) T(uI)
= T(u2) => ul = u2;
and
(b) T is onto, that is, R(T) = V.
An invertible linear transformation is also known as non-singular or isomorphism. That is, a linear transformation T is called non-singular if and only if it is both one-one and onto. Remark: If T is not one-one or if T is not onto or both fail, then T is not non-singular. Such a map is called singular. Theorem 3.11. Let U and V be vector spaces over the field F and let T : U -+ V be a linear transformation. If T is non-singular, then rl : V -+ U is also linear, one-one and onto (i.e. rl is also non-singular). Proof: Step I: To show that Let vi' v2
E
rl : V -+
U is linear.
V and a. E F be any elements.
Introductory Linear Algebra
78
Now,
aVI
+V2 = aT(ul)+T(u2)
T (a ul + U2) as T is linear.
=
T-I(avl +v2) = aUI +u2 =
=>
rl
a T- I (vI) + T- I (v2)
is linear.
II: To show that T- I : V ~ U is one-one. Step 11: Let VI' V2
E
V be such that
Enough to show that vI Since vI' v2
E
=
V and T: U
rl (vI) = T- I (v2)
v2 . ~
V is one-one and onto
=> 3ul,u2 EU such that T(ul)=vl,T(u2)=v2 => ul
=T- I (vI)
=>
= u2 as by assumption
Uj
and u2
=T- I (v2) rl (vI) = T- I (v2)
=> T(u) = T(u2) => vI = v2' Step Ill: III: To show that T- I : V ~ U is onto Let
U E
U any element.
=> 3v E V such that T (u) = vasT: U
~
V is one-one and onto
=> u=r) (v) Thus, for each u in U there exists v E V such that T- I (v) = U
•
Hence rl is onto. In the following theorem, we shall show that non-singular linear transformations preserve linear independence. Theorem 3.12. Let U and V be vector spaces over the field F and let T : U ~ V be a linear transformation. Then T is one-one if and only if T maps linearly independent subsets of U onto a linear independent subset ofV.
Proof: Let T is one-one. Let S
=
{u), u2, ... , un} is L/ in U.
Linear Transformations
79
Then to show that S· = {T(uI)' T(u2)' ... ' T(u n )} is LI in V. Let ai' a2' ... , all be scalars such that n
L ajT(u/) Qv a;T(u/) = 0v j=1 ;=1
=> T ( [ t ajuI)=Qv a;uI)=ov as T is linear /=1
n
=>
L aja; Uju; = Qv 0u j=1 ;=1
as T is one-one
°
=> a; aj = Q for all i = I, 2, ... , n; as S is LI => S· is LI in V. Conversely, let T maps every LI sets in U onto a LI sets in V. Let Uu E U such that Uu
*" Qv 0u
{u} is LI in U => tu}
=> {T(u)} is LI in V => T(u) *" 0v Qv Thus,
1I
*" Qv Qv 0u => T(u) *" 0v
i.e T(u)=Qv=>u=Qv T(u)=Ov=>u=Ou
=> T is one-one. Theorem 3.13. Let U and V be vector spaces over the field F, and let
T : U ~ V be a linear transformation. Suppose that dim U = dim V < 00. Then T is one-one iff T is onto. Proof: We have T is one-one <=> N (T) = {Qv} {Ou}
°
<=> dim N(T) = Q <=> dimR(T) = dimU, (":dimN(T) + dimR(T) = dimU) dim V, as given
<=> R(T) = V <=> T is onto.
Introductory Linear Algebra
80
Now we shall give some examples based on the theories discussed in this section. Example 11. Let T be a linear operator on P2 defined by
2 T(ao + a,x alx + a2x2) = ao + (a, (al - a2)x + (ao + a, al + a2)x . [s T non-singular. If yes, find a rule for 11 l ' like the one which defines T. Solution: To show that T is non-singular, it is enough to show that T is onc-one as T : P 2
~
P2 is linear and dim P2
= 3 < 00 •
We have, N(T) = {(p(x) E P2 : T(p(x)) = 0, a zero polynomial in P2} 2 = {ao + a,x alx + a2x2 : ao + (a, (al - a2)x + (ao + al a, + a2)x = O} =
{ao +a,x+a2x2 +alx+a2x2 :ao =O,a, =O,al -a2 =O,ao +a( +a2 =O}
=
aO=0=al=a2} {ao+a,x+a2 {aO+alx+a2 x2 ::aO=0=a,=a2}
=> N (T) = to} , where 0 is a zero polynomial in P 2 => T is one-one. In order to find 1', 1 1, let alx + a2x2) = bhoo + htx + b2x2; ai' bi T- I (ao + a,x
=>
ao +alx+a2x2
=
E
91 9l
T(bo +htx+b2x2)
2 (bl - b2 )x + (bo + ht + b2)X = bo + (b,
=> bo == ao, bb,l - b2 = = ai' a" bo + b b,l + b2 = = a2 1 2
1 2
=> bo == ao, ht == -- (a, (al + a2 - ao), bz = = -- (a2 - al - Go) Thus,
2 rl (ao + a,x alx + a2x2) = ao +.!.. (a, (al + a2 - ao)x +.!.. (a2 - al - ao)x .
2
2
Example 12. Let T be a linear operator defined on V3 such that
T(el)
=
el + e2'
T(e2)
=
e, el - e2 + e3'
T(e3) =3e( +4e3, where {el' e2' e3} is a standard basis for V3·
81
Linear Transformations
(a) Find a general formula for T. (b) ]s T non-singular? If so, find So]ution:
rl.
(a) Let (xI' x2, x3) E V3 be any element ~
(xI' x2, x3) = xlel
+ x2 e2 + x3 e3
~ T(xl,x2,x3) = T(xlel +x2 e2 +x3 e3)
= xI T (el) + x2T (e2) + x3T (e3) as T is linear = xI (e l + e2) + x 2 (e l - e2 + e3) + x3 (3e l + 4e3) = xI
(I, 1, 0) + x2 (I, -1, 1) + x3(3, 0, 4)
~ T(xl,x2,x3) = (xI +x2 + 3x3,xl -x2,x2 + 4x3)
(b) We have N (T)
= {(xI' x2, x3) E V3 : T(xl> x2, x3) =(0, 0, O)} = {(xI' x2, x3) E V3 : XI + x2 + 3x3 =0, 4 - x2 = 0, x2 + 4x3 =O} = {(xI' x2, x3) E V3 : XI XI = = x2 = x3}
°
= {O, 0, O} :..~ T is one-one and hence non-singular. Note: Readers should solve the linear system of equations appearing above and see that it has only trivial solution. To find rI, rl, let
I T- (xI> x2, X2, x3) = (YI' Y2, Y3)
where YI, Y2, Y3 ~
E
9l and are to be determined. (Xl' X2' X3) = T(y" Y2, Y3) = (YI
YI + Y2 + 3Y3
+ Y2 + 3Y3' Y) - Y2, Y2 + 4Y3)
= xI'
Y) - Y2 =x2, Y2 + 4Y3 =x3·
On solving, we get
Y)
=5"1 (4xl
Y2 =
+x2 - 3X3),
5"1 (4x) -
4x2 - 3x3),
Introductory Linear Algebra
82
Thus, l T- (xl' X2' X3)
= (.!.(4Xl +X2 -3X3),.!.(4xl-4x2 - 3X3), .!.(-Xl +X2 +2X3»). 5 5 5
Example 13. Let Tbe a linear operator on a finite-dimensional vector space
n N (T):::; {Ov} ~ T2 (x):::; 0v => T (x):::; 0v Solution: Let R (T) n N (T) :::; {Qv {Ov } . V. Then R(T)
for all x in V.
Suppose that fl(x) :::; 0v for all x in V. Then to show thatT(x) = Qv. 0v. Since T2(x) = T(T(x» = 0v
=> T(X)EN(T) But for x E V, T(x) E R(T) Thus, T(x) E R (7) n N (7) = {Qv} {Ov} => T(x) Conversely, assume that T2 (x)
=0v => T(x) =0v
Then to show that R (T) Let y
=> y
E E
R (T)
= Qv· 0v.
for alI x in V.
nN (T) = {Ov} .
nN (T)
R (T) and YEN (T)
=> T(z)=y forsomezin V and T(y)=Ov => T2 (z) =T(T(z»:::; T(y) =0v => T (z) = Qv, 0v, by hypothesis => y=Ov, as T(z)=y => yE {Qv} {Ov} Hence R(T) n N (T) ~ {Ov} But {Ov}
~ R (T)
Thus, R(T)
n N (T);
as R (7) and N (T) are subspaces.
n N (T) = {Ov} .
Linear Transformations
83
EXERCISES ______________________________ EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ I. In all problems of Exercise 9 in Sec 3.2, check whether Tis non-singular (that is, one-one and onto) or not. If T is non-singular, find rl. 2. Let T: V3 ~ V3 be a linear transformation such that T (1, 0, 0) =(-1, 0, 0), T (0, 1, 0) = (0, 0, -1) and T (0, 0, 1) = (0, 1, - 1).
Is T non-singular? If so, find rl. 3. Let T: V3 ~ V3 be a linear transformation such that T (1, 1, 0) =(1, 0, 0), T (0, 1, 0) =(0, 0, 1) and T (0, 1, - 1) = (0, 1, 0).
[s T non-singular? If so, find rl. 4. Let Tbe a linear operator on a finite-dimensional vector space V. Suppose that rank of
3.5
f2
=
rank of T. Prove that R (T) fl N (T) = {Ov }.
IDEMPOTENT AND NILPOTENT OPERATORS
[n this section we shall mention a linear operator having some special properties. Definition. Let V be a vector space. Then a linear operator T on V is said to be idem idempotent potent if f2 = T.
Example 14. The 7.ero operator and the identity operator are always idempotent as 0 2 = and P = I.
°
Example 15. Let T be a linear operator on a vector space V3 defined by T(xl' x2, x3) = (0, x2' x3)'
Then T is idempotent, because T2 (xI' x2' x3)
=T(T(xl ,x2,x3» =T(O, x2, x3) = (0, x2' x3)
=T(xl> x2, x3) for all (xI> x2, x3) in V3 =>
f2 =
T.
Example 16. Let T be an idempotent operator on a vector space V. Then N(1) =R (/-1).
Solution. Let u e N(1) => T(u) = 0v. Now (J-T)(u)=I(u)-T(u)=u => ueR(I-T) Hence N (1)
~
R (/ - 1).
Let v e R (l - 1)
Introductory Linear Algebra
84
=>
3 U E V such that (1- 1')(u) = v
=> =>
u-1'(u)=v T(u-T(u»= T(v)
=> 1'(/1) - T2(u) = T(v) => 1'(1/) - T(u) = T(v), as T2 = l' => 0v = 1'(v) =>
VE
N(T)
Hence R(I- 1') <;;, N(1') . Thus, N(T) = R(I- 1') .
Definition. Let T be a linear operator on a vector space V. Then 1'is Tis said to nilpotent if there is some positive integer n > I such that T Tnn = 0. The smallest integer n > I for which 'P' = is called the degree of nilpotence.
°
nilpotent. potent. Example 17. The differential operator D defined on P n is nil Solution: Let D: P"
~
Pn be defined by
D(PII(x» = p~(x) = PII_I (x),
where Pk (x) denotes the polynomial of degree atmost k. 2
•
Then D (Pn(x» = D(PII_I(x» = Pn-2(x), 3 D (Pn(x» = Pn-3(x),
Continuing this process we note that Dn(Pn(x» = pn-n(x);:::: Po, a constant polynomial.
=> Dn+I(Pn(x»
= 0.
Hence D is nilpotent and the degree of nilpotence of D is n + 1. Example 18. Let Sand T be nilpotent linear operators on a vector space V such that ST = 1'S . Then show that ST is nilpotent. Find the degree of nilpotence of ST. Solution: Since Sand T Tare are nilpotent => 3 n > 1, m > 1 such that Sn =0, T m =
°.
Given ST= TS
85
Linear Transformations
Then (ST)2
=(ST) (ST) = S(TS)T = S(ST) T = (SS)(TT) =(SS) (TT) =S2T2.
In fact, by induction it can be shown that ( ST)k
Then (ST)n
= SkTk
for all k
=SnTn =O.Tn =0,
= 1,2, ... ,r
and (ST)m
=SmTm =Sm.O =0 .
~ ST is nilpotent. If nnand and m be the degree of nilpotence of Sand T, respectively, then the degree of nilpotence of ST = min (n, m).
EXERCISES ______________________________ I. Let T be an idempotent operator on a vector space V. Prove that (a) ve R(T) (b)
<=> T(v) =v .
V = R(T) + N(T).
(c) R(T)
n N(T) = {Ov} .
-Hints for (b): Note that any u e Vcan be written as u = T(u) + (u-T(u», where T(u)eR(T) and u-T(u)eN(T). 2. Let T be an idem idempotent potent operator on a vector space V. Show that J + Tis T is one-one. 3. Let T j and T2 be linear operators on a vector space V such that
1\ + T2 = J .
Prove that 1\ T2 =0 <=> 1\2 =1\, Tl =T2 . 4. Let T be an idem idempotent potent operator on a vector space V. Prove that J - Tis T is also idempotent on V. 5. If T be an idempotent operator on a vector space V, prove that R (1) = N(l-n. 6. Let Tbe a linear operator on a vector space V3 defined by T(xj, x2' x3) =
(0, Xj, xI + x2). Is Tnilpotent? Ifso, what is the degree of nil potence ,
ofT?
7. Let Tbe a linear operator on a vector space P3 defined by T(ao + alx + a2x2 + Q3x3) = (ao - Q) )x2 + (Qo + Q) )x3 .
Introductory Linear Algebra
86
Prove that (a) T is nilpotent of degree 2. (b) For a non~"Zero scalar A, I + AT is non-singular.
3.6
LINEAR FUNCTIONALS
Definition. Let V is a vector space over the field F. Then a map f: V ~ F
is called linear functional iff iff is a linear transformation, i.e., if f(a vI +v2) =af(vl)+ f(v2) for all vI' v 2 in V and a in F.
Example 19. Le! Lei f:
m2 ~ mbe defined by
f(XI,X2) = alxl +a2 x 2; al,a2 E9\.
Then f is a lillear functional. Solution. Let X=(XI,X2)E~H2'Y=(YI'Y2)E~H2 and aE~H. Then ax+ y =(axl =(axi + YI, ax2 + Y2) Now, f(ax+ y) = al (a xI + y\)+a2(ax2 + Y2); a\ ,a2 E
m
= a (a\x\ + a2x2) + (a\y\ +a2Y2) = af(x) + fey).
Hence f is a linear functional. Example 20. Let [a, b] be the closed interval on the real line m and C [a, b] bJ be the space of real-valued continuous functions defined on [a, b]. Let b
L: C[a, qa, b] ~ 9t be defined by L(g) = fg(t) fg(/) dt. dl. a
Then L is a linear functional on C[a, qa, b]. Solution. Let g, hE C [a, b] be any elements.
=>
L(g) =
b
b
a
a
fget) dt,dl, L(h) = fhet) h(t) dt.
For any scalar a, we have b
b
L(ag+h) = f(ag+h)(t)dt
= f(ag(t) + h(l»
dl dt
a
a b
= a fg(t) dt a
=> L is a linear functional.
b
+ fh(t) dt = aL(g) + L(h). a
Linear Transformations
87
Definition. Let V be a vector space over the field F. Then L(V, F) , the set of linear functionals on V, is called the dual space of V, and Lev, LCV, F) is clenoted by V*, that is, L(V, F) = V* = dual space of V.
= if: / is a linear functional on V}. and dim V*= dim L(V, F) = dim V. dim F = n.l, = n = dim V. Theorem 3.14. Let V be an n-dimensional vector space over the field F. Let B={vl'v2' .... ,v"} is an ordered basis for V. Then (a) There is a unique basis B* = {fi,/2"",f,,} for V* such that fi (v)=8lj. (The basis B* is called the dual basis of B). (b) For each linear functional/on V,
" /= L/(vJfi. i=1
(c) For each vector v in V,
" v = Lfiev) LfiCv) Vi' i=1
Proof. (a) Given that B = {vI' v2,"" VII} is an ordered basis for V. Then from Theorem 3.4, for each i = 1,2, ... , n; there exists a unique linear functional 1; on V such that
o, {
v j)=8lj= I, fiCvj)=8lj= fie
i"l= j i=j'
To show that B* = {fi, /Z'''''/II} is a basis for V*, it is enough to show that B* is LI. Let
(Xl'
a2' ... , all be scalars such that
L" a
i
f, =
°, a zero functiorral.
i=l
=>
(fai 1=1
f,)evj) =OCv)= f,)CVj) =Oev)= 0, j == 1,2, ... , n
Introductory Linear Algebra
88
~
n
2,(X;Jj(Vj )=O ;=1
~
n
2,(X;Oij=O ;=1
~
(X j
=0
for all) = 1, 2, ... ,n
Thus B* is LI and hence a basis for V*. (b) Given thatf E V* by any element ~
n
f
= 2, (X;!; for some scalars (X;, since B* is a basis for V·. ;=1
Now, n
=
2,(X;!;(Vj) ;=1
n = 2,(x;oij ;=1
= (Xj for all) = 1,2, ... , n. n
Hencef= 2,f(v~)Jj . ;=1
(c) Given that v E V be any element. n
~
V =2,~;Vi for some scalars ~i" ;=1
Now,
fj (v)
=
Jj (t~; Vi) /=1
n
=
L~;fj(Vi) i=1
n
=
L~;Oij
=
~j
;=1
for all} = 1,2, ... , n.
89
Linear Transformations n
Hence, v
Iii Lii (v) v; .
=
;=1
Remark. Letfbe a non-zero functional on V. Then R (f), the range space ofJ,isasubspaceofscalarfield 9t ore. Hence dimR(!) =I. IfVisan n-dimension~l
vector space, then dimN(f) = n -1.
Definition. Let V is a v,ector space over the field F and S is a subset of V. Then the set i
SJ
=
{f E V· : f ( u ) =0 for all u in S}
is called ann·ihilator of S. Theorem 3.15. Let S be a subset (need not necessarily be a subspace) of a vector space of V. Then SO is a subspace of V*. Proof.' We have SO = {fe
v· :f(u) =ofor all u inS}
Since 0 (u) = 0 all u in S, therefore 0 e SO. (Here 0 is the zero functional in V). Thus So ¢#; cp. Let J, g e So and (X is any scalar. Then for all u in S, (a.f + g) (u) = (a.!) (u)+ g(u) =
a.f(u) + g(u)
= a..0+0 =0
=> CI/+geSo. Thus, SO is
(X
subspace of V*.
Theorem 3.16. Let Vbe a finite-dimensional vector space over the field F and S be a subspace of V. Then dimS + dim So =dimV. Proof. Let dim V = n and dim S = k . Let Let B 1 is extended so that B2 = {ul' .... ,uk'uk+l'· .. ,un } is a basis for V.
Let B~ B2 for V.
=
{ft,h, .... ,jn}
is a basis for V*, which is dual to the basis
Introductory Linear Algebra
90
Hence 1; (u) tu) = Oij' We claim ~hat B = {h+IJk+2, ... Jn} is a basis for S". We prove our claim in the following three steps. Step I: To show-that that~
/i E So
for all) = k + I, k + 2, ... ,n. That is, to show
(1I) = 0 for-all U in S and for all) = k + 1, k + 2, ... , n.
Let
U
E S be any element.
~
U
=
k
I, a; ui i=1
~
Jj(u)
fj Ij
=
(~ai Ui),) = k + I, k+2, ... ,n /=1
i=1 k
= I,aioiJ = 0 as i"#j. i=1 ~
JjESO for all )=k+l,k+2, ... ,n.
Step 11: II: B is obviously LI as Br Bi is LI (since every subset of a LI set is LI). Step Ill: To show that B spans S". Let
If
E So k: V * n
f= I,f(Ui)/i, ~ 1= I,/(Ui)/i, by part (b) of Theorem 3.14. i=1
k
n
= I,f(Ui)/i I,/(Ui)/i + i=1
I,
f(ui)!; I(ui)!;
i=k+1
n
I,
=0+
J(ui)/i, sinceJES"
~/(u)=Ofori=1,2, ... ,k ~f(u)=Ofori=1,2,
i=k+1 n
I,
f(ui)/i I(ui)/i
i=k+1
Hence B spans S", and dim S" = n - k Therefore, dim S + dim S" = k + n - k = n = dim V.
Linear Transformations
91
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ __ l. Let V be a vector space over the field F. Then prove that (a) S
= {O} =>
(b) S = V =>
so =
so =
V*, (Here 0 is the zero vector of V).
{O}, (Here 0 is the zero functional in V*).
2. Let Vbe an n-dimensional vector space over the field F.If 1I('" 0v) Dv) E V, then show that there exists 1 lEE V* such that f(lI) *- O. 0. 3. Let V be an n-dimensional vector space over the field F. If u E V be such that f(lI) = for all f E V*, then show that u = 0v' Dv, 4. Let St SI and S2 be two subspace of a finite-dimensional vector space V. Prove the following:
°
(a) St SI =S2 ~Sp =sg (b) (SI (St +S2)O =SP nSg
(c) (SlnS2)o=sP+sf. (StnS2)o=sP+sf.
5. If 1 I is a linear functional on V3 such that 1(1,0,0) l(l, 0, 0) = I, f(O, I, 0) =-1,1(0, 0, 1)= 1, and ifu = (x,y, z), findf(u). Iff 6. If f is a linear functional on V3 such that
10,0, l(l, 0, I) = -I, f(O, 1, -1)= 1,/(2, -1, 0) = 2 feu).
and if u = (x, y, z), find
MATRIX REPRESENATION OF A LINEAR TRANSFORMATION
In this chapter, we first explain the co-ordinates of a ve~tor in a vector space relative to an ordered basis. It may be noted here that if Vis a finite-dimensional vector space, then an ordered basis is a finite sequence of vectors which is linearly independent and spans V.
4.1
COORDINATES
Let V be a finite-dimensional vector space, and B ordered basis for V. Let =:)
U E
= {ul, U2, ... , un}
is an
V be any element.
There exists scalars 0.1' 0.2' ... , an such that U = alul
+ a2 u 2 + ... + anun ·
Then a; is called ithcoordinate of U relative to the ordered basis B. The coordinate matrix of u relative to the ordered basis B, denoted by [u]B,
;,
defin~ ~ ~ [~J lulB
Example 1. Let B = {(I, 0, 0), (0,1,0), (0,0, I)} be an ordered standard basis for V3 • Suppose that we wish to find the coordinate matrix of the vector U=
(1,4,5) relative to the basis B. Then there exists scalars
0.1' 0.2'
0.3 such
that 1I
=
(1,4,5)
=
0.1 (1,0,0) + 0.2(0, I, 0) + 0.3(0, 0, 1)
= (0.1,0.2,0.3) =:)
0. 1 = 1,
0.2 = 4,
0. 3 = 5.
93
Matrix Represenation of a Linear Transformation
Hence In general, it can easily be seen that the coordinate matrix ofthe vector U = (UI,U2'·'"un)
relative to the ordered standard basis B
=
(el' e2' ... ' en) for Vn is
Example 2. Let B = {(2, I, 0), (2, I, I), (2,2, I)} be an ordered basis for V3 • Then find [u]8 for U = (I, 2, 0). Solution: Let u" u 2' u 3 be scalars. Then (1,2,0) = u l (2, 1,0) + u 2 (2, I, I) + u 3 (2, 2, I) = (2u l + 2u2 + 2u3 , u l + u 2 + 2u3, u 2 + u 3) ~ 2u I + 2u2 + 2u3 = I, u l + u 2 + 2u3 = 2, u 2 + u 3 =0. On solving, we get 1
Hence
I
1 I 2'
=-
[u]8 =
u
2
3 2'
=--
u3
3 =-. 2
I/2]
-3/2 . 3/2
[
Example 3. Suppose that we wish to find an ordered basis for V4 relative which the vector u = (-I, 3, 2, I) has the coordinate matrix as [4, I, -2, 7] T. Solution: Let B = (u l ' u 2' u 3, u 4 ) be an ordered basis for V4 . Given that [u]8 = [A, 1, -2, 7f. ~ U = (-1,3,2, I) = 4u + 1 . u - 2u + 7u 2 3 4 I = 4 ( 0, I, 0, 0) + 1 I (-I, -I, 0, 0) - 2 ( 0, 0, -1, 3) + 7 ( 0, 0, 0, I).
Introductory Linear Algebra
94
Consider the set B = {( 0, 1,0,0), (-I, -I, 0,0), (0,0, -1,3), (0,0,0, I)}.
"4
It can easily be checked that the set B is LI in and hence it also spans as B has four elements which is equal to dim Thus, B is an ordered basis for
"4
"4'
"4'
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ l. Determine the coordinates of (i) (1,2, 1), (ii) (3, 1,2) and (iii) (4, -2,2) with respect to the ordered basis B = {(2, 1, 0), (2, 1, 1), (2, 2, I)} for over ~n. a vector space 2. Show that B = {x + 1, x 2 + x-I, x 2 - x + I} is a basis for P2 over
"3
~H. Hence determine the coordinates of (i) 2x - 1, (ii) 1 + x 2 , and (iii) x 2 + 5x - 1 with respect to the ordered basis B. 3. Show that B = {( 1,0, -I ), ( 1, I, I ), ( 1,0, o)} is a basis for over ~H . What are the coordinates of the vector (a, h, c) with respect to the ordered basis B.
"3
4. Let" be the real vector space of all polynomial functions from m minto into ~H of degree atmost 2, that is, the space of all polynomial functions fof the form f(x) = a o + a1x + a~.
Let t be a fixed real number and define gl{x) = 1, g2(x)=x+t,
g3(x)=(x+t)2.
B = {gl' g2' g3} is a basis for V. If 2, = a o + a1x + a find the coordinates off with respect to this ordered basis B.
Prove that
r
f{x)
5. Show that the vectors U1
"4
= (I, 1,0,0),
u 3 = CO, 0, 1, I),
1I2
= (0, I, 1,0)
u 4 = CO, 0, 0, 1)
form a basis for over 91 . Find the coordinates of each ofthe standard basis vectors with respect to the ordered basis B = {u I' 11 2, U3 , u4 }. 6. Find an ordered basis for relative to which the vector u = ( 1, - 1, 1 ) has the coordinate matrix [ 1,2, 3f. 7. Find an ordered basis for P2 relative to which the polynomial 1 + x 2 has the coordinate matrix [ 1, -1/2, 112
"3
f.
95
Matrix Represenation of a Linear Transformation
4.2
REPRESENTATION OF A LINEAR TRANSFORMATION IN TERMS OF MATRIX
Let U and V be vector spaces over the field F and let dim U = n, dim V = m. Let 81 Let T: U
u2' ... , un) be an ordered basis for U, and (v\, v2' ... , vm) an ordered basic for V.
= ( up
82 = ~ V
be a linear transformation defined by
T( u) = aljv l + a 2j v2 + ... + amj vm,j = 1,2, ... , n. Then the coordinate matrix of T ( uj ) relative to the ordered basic B) is given by
,j = 1,2, ... , n ;
any
and matrix of Trelative to BI and B 2 , denoted by ( T: BI' B2 ), is defined as
(T:BpB2) = [::.:
:::
amI
a m2
...
:::].
a mn
Note that first column of (T: B I , B 2) is the coordinate of T(u l ) relative to 8 2 ( i.e., [T(ul)]B ), second column of (T: BI' B2) is [T(u 2)]B ' and so 2 on, finally nth colun~l of(T: B I , B 2) is [T(un)]B . 2
Example 4. Let D : Pn ~ Pn be a differential linear operator and B = {I, X, x 2, ... , x"} be the standard ordered basis for Pn • Suppose that we are interested to know (D : B, B), the coordinate matrix of D relative to B. To see this, we note that D(I)
=0 =0.1+0.x+0.x 2 + ... +O.xn
D(x) = 1 = 1.1+0.x+0.x 2 + ... +O.xn D(x 2 ) =2x =0.1 +2.x+0.x2 + ... +O.xn
D( X fl) =nxn-I
-IO n =.0 l+O .x+n ... +n.x +.x
Introductory Linear Algebra
96
This shows that
(D: B, B) =
°2 ...... °
0
1
0
0
0
...° ...
0
... 0 ... ...
000
n
000
o
...
0
(n+l) x(n+l)
Example S. Let T be the linear operator on V2 defined by T (x l' x 2 ) = (0, x2 ). Let B, = {( 1, 0), (0, 1 )} ~e the standard ordered basis for V2 and B2 = {(I, 1), (-1, I)} be an ordered basis for V2. Suppose we wish
to determine
(a) (T:B"B2) (b) (T:B2,B,) (c) (T:BI,B,) (d) (T:B2 ,B2 ).
We shall calculate each of them one-by-one. (a) To compute ( T: B I , B2) We have B, = {(I, 0), (0, 1), B2 = {( 1, 1 ), (-1, 1 )}. Also, T(x"x2) ~
=
(0,x2).
T(I,O)=(O,O)=<X 1 (1, 1)+~(-1, 1). P, (1, 1) + P2 (-1, 1).
T(O, 1) = (0, 1) =
Clearly,
°
<x, = = <X2 and
P, - Pf"",O, P, "
T)ms,~
= [::
[T:B"B 2 ]
+ P2 = 1 give
~:J = [~ ~~~l
(b) To compute ( T: B2 , B,) Since T(x"x2) = ( 0, x 2 ). ~
~
°
T(1, 1) = (0, 1) = (1, 0) + 1.( 0, 1 ), T(-I,1) =(0,1 )=0.( 1,0)+ 1.(0,1). (T: B2 , B,) =
[~ ~
l
(c) To compute (T: B, , B,)
Since T(x" ~
X2) =
T(l, 0)
=
P, = P2 = 1-.
(0, x2) . CO, 0) = 0.(\, 0) + 0.(0, I) ,
2
Matrix Represenation of a Linear Transformation
T(O, 1)
97
= (0, 1) = 0.(1,0) + 1.(0, 1) .
l
~ (T: BI , BI) = [~ ~ J(d) To compute ( T : B2 , B 2 )
T (xI' x 2 ) = ( 0, x 2 ).
Since
1
1
1
1
T(1,I) = (0,1) = "2(1,1) + "2(-1,1),
= (0,1) = "2(1,1) + "2(-1,1).
T(-I,I)
( T: B,. B,)
Example 6. Let T: V3
~
~ [i
n
V3 be a linear operator such that
T(1, 0, 0) = (-1,0,0), T(O, 1, 0)
Let BI for V3 and
=
= (0, 0, -1) and
T(O, 0, 1)
= (0, 1, -1) .
{(l, 0, 0), (0, 1,0), (0, 0, I)} be the standard ordered basis
B2 = {(-I, 0, 0), (0, 0, -1), (0, 1, - I)} is an ordered basis for V3• Then determine (T:BI ,B2), (T:B2 ,BI ), (T:BI ,~), and (T:~,B2).
Solution: First of all, we shall find a general formula that defines TexpIicitly. Let (XI' x 2' x 3 ) be any element in V3 . Then (XI' x2' x3) ~
=
XI (1, 0, 0) + x2(0, 1, 0) + x3(0,0,I)
T(xl' x2' x3) = XI T(1, 0, 0) + x2 T(O, 1, 0) + x3T(0, 0, 1) =
XI (-1, 0, 0) + x2 (0, 0, -1) + x3 (0, 1, -1)
=
(-XI' x3' -x2 - x3)·
(a) To compute (T: B I, B2 ).
We have
= (-1,0, 0) = 1.(-1, 0, 0) + 0.(0, 0, -1) + 0.(0, 1, -1), T(O, 1,0) = (0, 0, -1) = 0.(-1,0,0) + 1.(0, 0, -1) + 0.(0, 1, -1),
T(l, 0, 0)
T(O,O,I)
=
(0,1,-1)
=
0.(-1,0,0)+0.(0,0,-1)+1.(0,1,-1)
98
Introductory Linear Algebra
:B,,~)
Hence Hen'e
n
I 0 0] ~ [[~
0 : O. 001
(T:B (T l ,B2 ) =
Remark: To compute (T: BIl , B2 ), the general formula for T obtained is not
•
required, and hence it is not used. (b) To compute (T: B 2 , B Il ) We have T(xl' x2, x3) = (-xI (-xl 'X3' ' x3, - x2 X2 - x3)· X3)·
=>
T(-I,O,O) = (\,0,0) (1,0,0) = 1-(1,0,0)+0·(0,1,0)+0·(0,0,1),
T(O, 0, -1) == (0, -1, -I, 1) = =
O· (1,0,0) - 1-(0, 0· 1· (0, 1, 0) + 1· (0, 0, 1), and
T(O,I,-I) = (0,-1,0) = 0·(1,0,0)-1·(0,1,0)+0·(0,0,1)
0] (T:B2'~) ~ [[~1 -~0 -!l
Hence (T : B2 , Bl ) Hen,.
~
=
-1
-~ .
(e) Bl , BIl ) (c) To compute (T: RI'
We have T(1,O,O) = (-1,0,0) = -1-(1,0,0)+0·(0,1,0)+0·(0,0,1), -J.(1,0,0)+0·(0,1,0)+0·(0,0,1), T(O, 1, I, 0) = (0,0, -1)
=
0·(1,0,0) + 0·(0, 1,I, 0) -1·(0, 0, 1), I),
T(O, 0, 1) I) = (0,1,-1) = 0·(1,0,0)+1·(0,1,0)-1·(0,0,1).
Hence
(T : BIl , B,) Bl ) = (T:
-1 0~ 0]~l. [-~ [ °o 0
0
1.
-1 -I -1 -I
(d) To compute (T : B 8 2, B 2 ) T(xl,x2,x3) = (-x"x3,-x2 (-Xl,x3,-x2 -x3). We have T(x"x2,x3)
=> T(-I,O,O) == (1,0,0)= (\,0,0)= -1.(-1,0,0)+0.(0,0,-1)+0.(0,1,-1), -1.(-I,O,O)+O.(O,O,-I)+O.(O,},-I), T(O, 0, -1) = (0, -1, 1) = 0·(-1,0,0) + 0·(0, 0, -1) -1·(0, 1, -1), T(O,I,-I) = (0,-1,0) = 0·(-1,0,0)+1·(0,0,-1)-1·(0,1,-1).
l
Matrix Represenation of a Linear Transformation
99
Hence
° °
-1
~l·
-1
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ 1. Let D be the linear differential operator defined on P3' the space of all polynomial functions from 9\ into 9\ of degree atmost 3. Let
B = {I, x, x2 ,x3} be an ordered basis for P3 • Determine (D: B, B). 2. Let T be the linear operator on V2 defined by T(xI' x2) = (x2' - xI). (a) Determine (T: B, B), where B is the standard ordered basis for V2 over 9\ . (b) Determine (T: B, B), where B = {( 1,2 ), (-1, I)} is an ordered basis for V2 over 9t (c) Determine (T: BI , B2 ), where BI = ({l, -I), (1, 2)}, B2 = ({l, 0), (3, 4)} are two ordered bases for V2 over 9\. (d) If B is any ordered basis for V2 over 9\ and (T: B, B) = A = (A i)' then prove that A I 021 ::F 0. 3. Let T be the linear operator on V3 defined by T( x I,x2,X3) = (3 x I +x3' -2xI +x2' -xI +2x2 + 4x3)· (a) Determine (T : B, B), where B is the standard ordered basis for V3
over 91(b) Determine (T: B, B), where B = {(l, 0, 1), (-1, 2, 1), (2, 1, I)} is an ordered basis for V3 over 9\.
(c) Determine (T: Bp Bi' B2 ), where
BI = {(l, 0, 1), (-1, 2, 1), (2, 1, I,)},
B2 = {(1, {(l, 0, 0), «0, 1, 1) (1, 0, I)} are ordered bases for V.? over 9\. 4. Let T: V3 ~ V3 be a linear operator such that T( 1,0,0) =( 1,0,1), T(O, 1,0)=(0, 1, 1) and T ( 0, 0, 1 ) = ( 1, 0, and
°).
If B = {(I, 0, 1), (0, 1, 1), (1, 0, O)} be an ordered basis for V3 over 9\, determine ( T: B, B ).
Introductory Linear Algebra
100
5. Let V be an n-dimensional vector space over the field m, and let B = { u)' u2' ••• , un} be an ordered basis for V. Define a linear operator on V such that T(u) = uj + I'j = 1,2, ... , n - 1; and T(un ) = 0 (Note that such linear operator exists uniquely by Theorem 3.4). (a) Determine (T : B, B). (b) Prove that 1" = 0 but 1" -) 0
*
6. Let T: P2 = p(x -1).
~ P2
be a linear transformation defined by T(P(x»
Let B) = {I, x, x 2 } and 2 B2 = {x - 1, x .:.. x
+ 1, x 2 + x + I}
be ordered bases for P 2. Determine (i) (T: B), B), (ii) (T: B2, B 2 ) and (iil) (iii) (T: B), B 2 ). 1. Let T: V2 ~ P 2 be a linear transformation defined by T(x),x 2) = x)
B,
Let
=
+ (x) - x 2) x + (x) + x 2) xl.
{(l, - 1), (0, 4)} and
B2 = {x + 1, xl + x + 1, xl - x} be ordered bases for V2 and P2 respectively. Determine ( T : B), B 2 ).
8. Let T: P2 Let
~
P3 be a linear transformation defined by T(p(x» =xp(x).
B) = {I, x, x 2 } B2 = {I, x + 1, xl + x, xl- I} be ordered bases for P 2 and P 3 respectively. Determine (T: B), B 2 ).
4.3
REPRESENTATION OF A MATRIX IN TERMS OF LINEAR TRANSFORMATION
In the previous section, we have seen that if a linear transformation T: U ~ V is given, then we can find a matrix (T: BI' B2 ) associated with T, where B J and Blare ordered bases for U and V respectively. Now one can ask a reverse question. Suppose that a matrix B is given. Then can we fmd a linear transformation T: U ~ V such that B = (T: B» B 2 ), where U and V are suitable vector spaces with dimU = n, dim V = m; and B" B2 are ordered bases for U, V. respectively? Yes, we can because the process of getting the matrix (T: B" B2 ) from the linear transformation T is reversible. To understand this process, let U and V be vector spaces over F and
101
Matrix Represenation of a Linear Transformation BI = (up u2' ... , un) be an ordered basis for U, and B2 = (vI' v2' ... , vm) an ordered basis for V.
Suppose that the matrix B = ( T : B I' B2 ) = (a r ) is given, where j = I. 2.... , m and j = 1. 2..... n; and we are interested in knowing a linear transformation T: U ~ V. One must note here that [alj' a 2j' .... a mj f is the ill column of the matrix B and it is the coordinate matrix of T(uj ) in the ordered basis B2' That is.
Hence, we have T(u l ) = all VI T(u 2) = al2 VI
+ a21 v2 + ... + amI vm ' + a22 v2 + ... + a m2 vm'
+ a2n v2 + ... + a nm vm • From Theorem 3.4, it follows that T (obtained from the above process) is a unique linear transformation such that B = ( T : BI' B2 ). Now to find a general formula that defines T, we extend T on the whole space U. To do this, let u be any element in U. T(un) = al n VI
=>
U
=
al "l
+ a2 u2 + ... + an un' for some scalars
as BI is a basis for U => T(u) = aIT(u l ) + a 2T(u 2) + ... + anT(un), as Tis linear. After substituting the values ofT(u l ), T(u 2), ... , T(un) in above equation, a little simplification yields T(u) = Plv l + P2v2 + ... + Pmvm, where PI'''''Pm are some scalars. This is the required linear transformation.
E.ampl. 7. Let A
=
[~
I :
aI' a2'"'' an
~l
Then find the linear transformation T: V4
~
V3 such that
A = ( T: BI' B2 ), where BI and B2 are standard ordered bases for V4 and VJ •
respectively.
Introductory Linear Algebra
102
Solution: We have BI = {el,e2,e3,e4}' B2 = {el,e2,e3}'
[t may be noted that ei appearing in BI and B2 have different meaning, that is, e l = ( I, 0, 0, 0) in BI and e l = ( I, 0, 0) in B2, and so on. Further, it should be noted that first column of A is the coordinate matrix of T (e l ) in the ordered basis B2 , second column of A is the coordinate matrix of T (e2 ) in B2' and so on. Therefore, we have T(e l ) = I·el +0'e2 T(e2) = I·el
+ l·e2 + l'e3
= (
I, I, I),
O·el
+ l·e2 + 0'e3 = (0, 1,0),
T(eJ = O·el
+ 0'e2 + l·e3 = (0,0, I ).
T(e3)
=
+ 0'e3= ( 1,0,0),
Now let (xl' x 2' x 3' x 4) is any element in V4 •
=> (x\>x2,x3,x4)
= xJ el +x2 e2 +x3 e3 +x4 e4
=> T(xl, x2' x3' x4)
= Xf, T(el) =
+ x2 T (e2) + x3 T (e3) + x4 T (e4)
x1C1, 0,0) + x2 (I, I, 1) + x3 (0,1,0) + x4 (0,0, I)
= (x~ + x2' x2
This is required linear
+ x3' x2 + x4)'
tran~formation.
Example 8. Consider the matrix:A of example 7. Let BI = {(I,I, 0, 0), (0, I, 1, 0), (0, 0, I, 1), (0, 0, 0, I)} B2 = {(I~ 1, 1), (1, 2, 3), (I, 0, O)}
are ordered bases for V4 and Y3' respectively. Find the linear transformation T: V4 ~ V3 such that A = (T: B 1I , B 2 ). Solution: As explained in Example 7, one can write directly, T(I, 1, 0, 0) = 1'(1, I, I) + 0·(1,2,3) + 0'(1, 0, 0)
=(1,1,1),
T(O, 1, 1, 0) = 1'(1, 1, 1) + 1·(1, 2, 3) + 1'(1, 0, 0) = (3, 3, 4),
+ 1'(1, 2, 3) + 0'(1, 0, 0) = (I, 2, 3), T(O, 0, 0, 1) = 0'(1, 1, 1) + 0 '(1,2,3) + HI, 0, 0) = (I, 0, 0). T(O, 0, 1,1) = 0,(1,1, I)
Now let (xI' x2' x3' x4) be any element in V.I" Then (xI' x2' x3' x4) = a(1, I, 0, 0) + b(O, 1, 1, 0) + c(O, 0, 1, 1)
+d(O, 0, 0, 1)
...(4.1)
for some scalars a, b, c and d.
Matrix Represenation of a Linear Transformation
103
:::::> T(xl' x2' x3' x4)
= aT(1, 1, 0, 0) + bT(O, 1, 1, 0) + cT(O, 0, 1, 1) + dT(O, 0, 0, 1) =
a (1, 1, 1) + b(3, 3, 4) + c(1, 2, 3) + d(1, 0, 0)
= (a + 3b + c + d, a + 3b + 2c,a + 4b + 3c).
... (4.2)
Now from equation (4.1), we have (xl,x2,x3,x4) = (a,a+b,b+c,c+d). a = xI' b = x 2 - xI' C = xI - x 2 + x 3' d = - xI + x 2 - x3 + x 4. Substituting the values of a, b, c, d in equation (4.2), we obtain
:::::>
T(XI,X2,X3,X4)=(-2xl + 3x2 +x4,x2 + 2x3,x2 + 3x3). This is required linear transformation.
EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
t. Consider the matrix
(a) If B is the standard ordered basis for V3 over m, determine a linear operator T on V3 such that A = ( T: B, B). (b) If BI = {(l, 0, 1), (1, 1, 0), (0, 1, I)}
and B2 = {(1, 1, 1), (1, 2, 3), (1, 0, O)} be ordered bases for V3, determine a linear operator Ton V3 such that A = ( T : B I' B2 )· 2. For the matrix
°1 °1 0, 0] 1 1 I find the linear transformation T: V4 where
~
V3 such that A
=
(T: B I , 8 2),
8 1 = {(l, 1, 1,0), (0,1,1,1), (0, 0, 1, I), (1, -1, 0, O)},
B2
=
{(t, 1, 1), (0, 1, 1), (1, 0, I)}
be ordered bases for V4 and V3 , respectively.
104
Introductory Linear Algebra
3. Consider the matrix
-l ~l
A { Let B I
=
{I, X, x 2 } and B2
=
{I, x + 1, x 2 + x, x 3
-
I} be ordered bases
for P2 and P3' respectively. Determine a linear transformation T: P2 ~ P 3 such that A = (T: BI' Bp B2 ). 4. Consider the matrix
o I
11
I .
1 OJ (I) If B = {x + I, x 2 + X + I, x 2 - x} be an ordered basis for P 2' determine a linear operator Ton P2 such that A = (T: B, B). 2
is an standard ordered basis for P2 and B2 = 2 {x + I, x + X + I, x - x} is another ordered basis for P2' determine Bp B2 ). a linear operator Ton P 2 such that A = (T: BI' 5. Consider the matrix (ii) If RI B) = {I, x, x
}
2
A =
[~ ~ ~l. -I
If RI B) =
=
{(l, {(1, 0, -1), (0, 0, I), (-I, 1, 1) is an ordered basis for V3 , and B2
{I, x, x 2
- x}
is an ordered basis for P 2; determine a linear
transformation T:
4.4
1
V3 ~ P 2
such that A
= (T:
Bp B 2 ). BI'
ROW AND COLUMN SPACES OF A MATRIX
Consider an m
x
n matrix
A=
all
al2
a21
a22
ln
...
[ a",1
alll 2
~~~ a
... a mn
]
.
Matrix Represenation of a Linear Transformation
105 n
Letrj = (ail' ai2.""ajn), i=I,2, ... ,m be the ithrowofA in Vn (=91 ).
Let
c; C;
ali az'
j
.'.' i=
= [
1,2, ... ,n
anll
be the ith column of A in VII1 (= 9\nl). 1. The row-space of A, denoted by R(A), is the subspace in Vn spanned by the row vectors r" r 2, ... , rll1 , that is, [rl, r2, .. , rll1 ] • R (A) = [ri' 2. The column space of A, denoted by C(A) , is the subspace in VIII spanned by the column vectors cl' cI, c2, ... , c n , that is, C(A) = [CI,c2, .... ,cn ].
3. The solution set ofAx of Ax = 0 is called the null space of A, that is, N (A) = {x: Ax = O}.
4. dim R (A) is called row-rank of A. 5. dim C (A) is called column rank of A. 6. dim N (A) is called nullity of A. 7. Since the row vectors of A are just the column vectors of its transpose, AT, and the column vectors of A are the row vectors of AT, it follows
that R(A)
= C(AT)
and C(A) = R(A T ).
Remark: It is unfortunate that range space and row space both start with the letter' R' , and we are using the same notation for both of them. However, it may be noted that R(A) stands for row space of A only in this section of this chapter, elsewhere R(1) denoted the range space of T. Lemma 4.1. Let A = (a r ) be a non-zero m x n matrix over the field F. Suppose that A is in roJ'-reduced echelon form. Then the set of DOD-zero rows of A is a basis for the row space of A. Proof: Let
,p,}, B = {PI,P2, ..... ·····,P r },
where PI' P2, ..... , P, Pr be the DCD-zero rows of A and other m - r rows of
106
Introductory Linear Algebra
A are the zero rows. Then certainly row space of A is spanned by the vectors PI' P2' ..... , Pr' Pr · Hence it is enough to show that B is LI. Let the leading 1 of Pi be in the column kr Then we have (I) kl < k2 < ..... < kr • (ii) a'k; = 1. (iii) ajk; = 0, j"# k.j.
Let aI' a2' ..... , a r be scalars such that r
...(4.1)
'LaJPj =0. j=1
The lth component in equation (4.1) is given by r
'L a
j
aF
=
0,
1 5./ 5. n.
...(4.2)
j=1
When /
=
ki' equation (4.2) yields r
'Lajajk, =0 ./=1
=> =>
ai =
0, 1 5. i 5. n as aik, = 1.
This shows that the set B is LI, and hence it is a basis for the row space of A. With the help of following theorem, one can find a basis for row-space of A and hence R(A). Theol
11
4.1. Let U be the row-reduced echelon form of a matrix A. Then R(A)
= R(U)
and N(A)
= N(U) .
Moreover, if U has r non-zero rows containing leading 1's, then they form a basis for the row space R(A), so that dim R(A) is r. Proof: Since A and U are row-equivalent, hence by Theorem 1.1, Ax = 0 and Ux = 0 have same solution set. Thus, N(A) = N(U). Let r i be the ith row of an m x n matrix A. Then R(A) =
Ilj,r2, ... ,rml IIj,r2,
If we apply three elementary row operations on A, then A is transformed into the matrices Ai of the following form:
107
Matrix Represenation of a Linear Transformation
rl Ij
Ij
rj Al
=
for i <j, A3 = r;+krj
kIj ,A2 = rj rill
rill
rill Here we note that row vectors of Ai' A2 and A3 are the linear combination of row vectors of A. By applying inverse elementary row operations on A I' A2 and A 3 , these matrices can be changed into A. This shows that row vectors of A can also be written as linear combinations of rows of Ai 's. Hence if two matrices are row-equivalent, they must have the same row space. Since A and V are row-equivalent, therefore, RCA) = RCV). Since V has r non-zero rows, from Lemma 4.1 it follows that the set of r non-zero rows of V forms a basis for the row space of A. Thus, the numbers of non-zero rows in V = row-rank of A. In the following theorem, we show that row-rank of a matrix is same as the column rank of A. Theorem 4.2. For any matrix A, dim R(A) = dim CCA). Proof: Let
dim RCA) =r.
Let V be the row-reduced echelon form of A. Then r = the number of non-zero rows in U. If we can show that r columns of A corresponding to the leading 1's in V form a basis for C(A), then it would imply that dim C(A) = r = dim RCA). We shall prove it in two parts: (i) They are LI, and (ii) they span C(A). (i) To show that r columns of A corresponding to the leading 1's in V are Ll. Let A be the submatrix of A whose columns are those columns of A that correspond to the r leading I's in U. Let V be the submatrix of V corresponding to the r columns containing leading I's in V. Then clearly V is the row-reduced echelon from of
A
A.
Hence A x = 0 and Vx = 0 have the same solutions set as and [j are row-equivalent. But the columns of V containing the
108
Introductory Linear Algebra
leading l's are LI and hence columns of U are LI. Therefore, = 0 has only a trivial solutions and so has
Ax = O.
Ux
Thus, the column
of Ii are LI. This shows that the columns of A corresponding to the leading I's in U are LI. \ (ii) Now we prove that r columns of A corresponding to the leading l's in U span C(A). To do this, we note that columns of A span C(A). Further, the columns of A corresponding to the free variables are not included in Ii. These column vectors of A (which are not contained in Ii ) can be expressed as linear combination of the column vectors of A. Hence by Theorem 2.8 (The Spanning Set Theorem) it is clear that span of columns of Ii = span of columns of A = C(A) , and the result follows. Definition. For an m x n matrix A, dimension of row space of A (or the dimension of column space of A) is called rank of A, that is, rank (A) =dim R(A) = dim C(A)
= row rank of A = columns rank of A. Remark. For any m x n matrix A, rank (A) :s; min (m, n). It follows from the fact that dim R(A) :s; m and dim C(A) :s; n. Example 9. Consider the matrix
A = [-;
-~
-~l'
-;
3-6-68
Note that the matrix A is the coefficient matrix of example lOin Chapter I, and the row- reduced echelon form of A is
-2 0 10 3 U =
0
0
o
0 0
3 0
109
Matrix Represenation of a Linear Transformation
Then row space of A = R(A) = R(U) = [up u2 ], and dim R (A) = row-rank of A = 2 = the number of non-zero rows in U. One can verify that u l ' u2 are Ll and they span R(A), RCA), thus SI = {ul' u2} is a basis for R(A). Let be the columns of A corresponding to the leading l's in U. Then C(A) = [vI' v2]·
One can readily see that S2 = {vI' v2} is Ll and S2 spans C(A) and hence dim C(A) = 2 = column rank of A. It also verifies that rank (A) = dim R(A) RCA) = dim C(A) =2.
EXERCISES ______________________________ In the following problems, find bases of row space and column space of the given matrix A, and hence determine their dimensions.
1. A =
[~o ~ ! ~l o
3. A"
0 0 0 0 0 0
[1 ~~ ~l
:il
_ [00 01
2. A _
=
4. A
= [-
0 0
~
-1
6. A"
8. A
=
~l
-1 3 4
[~ ~ rl [~
0 0
-2 0 0
10. A =
[~ ~ ~l. 000
-:l 2
EIGENVALUES AND EIGENVECTORS
In chapter 4, we have seen that a linear transformation can be represented by a matrix. In this chapter we shall restrict ourselves to square matrices only (unless otherwise stated) as we shall see later that eigenvalues and eigenvectors are studied only for square matrices. Eigenvalue of a rectangular matrix does not make any sense. The basic concept of eigenvalues and eigenvectors are very much useful in physical sciences, medical sciences and engineering. Eigenvalues provide many critical information about a continuous dynamical system. For example, stability and instability of a dynamical system completely depends on eigenvalues of the matrix associated with the system. Before studying eigenvalues of a matrix, one should be aware of the null space of a matrix. Although null space of matrix is defined in Chapter 4, we restate the same in detail which is required in this chapter.
5.1
NULL SPACE OF A MATRIX
Definition. The null space of a matrix A is the set of all solutions to the linear homogeneous equation Ax = 0, and is denoted by N (A), that is, N(A) = {x:Ax=O}.
Remark 1. If A is III x n matrix, then solutions of Ax = 0 belongs to
r:,(= 9\").
Remark 2. For an m x n matrix A, null space of A, N (A) is a subspace of VII because (i) AO = 0 => OE N(A) => N(A) *<j>
(ii) For any
1I,
v in N (A), we have
A(u+v)= Au+Av=O+O=O
=> u + V E N (A). (iii) For any u in N (A) and for any scalar c, we have A(cu) = cAu
=0
=> cUEN(A). 110
Eigenvalues and Eigenvectors
Example 1. Suppose
th~t
A
111
we wish to find N (A) for the matrix
=[~ ~l
Then by definition N(A) ={x E V2 : Ax =O}. We note that
=> XI = 0, 2xI +Sx1 = 0 and XI +4X2 = 0 => XI =0=x1 • :. N(A)={(Q--D)}. Example 2. SupPQse that we are interested in knowing N (A), where
A
=
[~ -~ -~l. o
-1
1
To find N (A), CA), w, need to know solutions ofAx of Ax = matrix associated with ·the system Ax = 0 is given by
B=
[~
[~ - [~ - [~ -
3 -2
-1
3
-1 3 -2
-7 -1
7
3 -2
-1 -1
o.
The augmented
~l
n
R, --+ iI, R, - 211,
niHR,l,
I O} OJ
--+ R, iI, 17
0 R, --+ II, 11, +3R, -1 1 0 0 o 0 , R3 ~ RJ - R2
112
Introductory Linear Algebra
This shows that the system Ax = 0 has infinitely many solution given by XI +X3
-X2 +x] -x
X3
=0, 0,
=
is a free variable. xl=-xJ '
::::>
x 2 = Xl' and x3 is free variable. Hence N(A) = {x = (xi' x2 ' xJ E J-; : Ax =
o}
E9t} E 9t}
={C-X3,Xl'X3):X3
= {x 3(-I, I, I): X3 = [((-I, I, 1, I)}]
We note that the set S ={(-I, I, I)} spans N (A) and S is Ll. Thus, S is a basis for N (A).
EXERCISES ______________________________ EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ In the following problems, for the given matrix A determine the null space of A and hence find a basis for the nuIJ space of A.
[-) [-3 6-1 -7] -7] I
l. A=
I
-2
2 -4
[1[' -1 4]
3. A = 2 0 3 2 -2 8
-I 2 3 -1 5 8 -4
2. A=
4.
[-:
0
0 -1
1 I
I 1
0
0
0
I -1 1
j
A~[~ -1] 2 -I -1
2
0
Now we shall bring our attention to eigenvalues of a matrix A, which is always a square matrix from here onwards.
5.2
EIGENVALUES AND EIGENVECTORS
Definition. A scalar A. is called an eigenvalue of a matrix A if there exists x (~ 0) such that Ax = A.x, ... (5.1) Ax, and Xx is called an eigenvector corresponding to the eigenvalue A..
Eigenvalues and Eigenvectors
113
Remark 1. Eigenvector is always non-zero, however, eigenvalue may be zero. Remark 2. Eigenvalue is also known as charactcristic value or proper value and eigenvector as characteristic vector or proper vector. In this book, we shall use only the terms eigenvalue and eigenvector. Remark 3. Corresponding to one eigenvalue A of a matrix A, there are infinite number of eigenvectors. In fact, if x is an eigenvector corresponding to an eigenvalue A, then for every non-zero scalar c, Y =cx, is an eigenvector corresponding to A. Proof: Since x is an eigenvector corresponding to A, therefore Ax = Ax. We note that Ay=A(cx)=cAx=c('-x)=A(CX)=AY and hence the result follows. Remark 4. Equation (5.1) can be re-written as (A - IJ)x
=0,
... (5.2)
where I is the identity matrix whose order is same as that of A. (a) The set of all non-zero solutions of (5.2) is the set of all eigenvectors corresponding to the eigenvalue A. (b) The set of all solutions (including zero solutions) of (5.2) is called the eigenspace of A corresponding to the eigenvalue A. (c) Eigenspace of A corresponding to A = null space of (A - AI) =
N(A -AI). (d) (cl) Set of eigenvectors corresponding to A
=N (A -
AI) - {O} .
Now we prove a basic result regarding eigenvalue of a matrix A. Theorem 5.1. Let A be a square matrix. Then following statements are equivalent: (a) A is an eigenvalue of A. (b) A - A.J is singular. (c) det (A - AI) = O. Proof: The proof of the above theorem is based on the fact that the homogeneous linear system of equation Ax = 0 has a non-trivial solution if and only if A is singular (See Theorem 1.5).
Introductory Linear Algebra
114
We note that A is an eigenvalue of A<:=:>3 x (;t 0) such that Ax = Ax. <:=:> (A - AI) x = 0 has a non-trivial solution. <:=:> A - ')..[ is singular. <:=:> det (A - AI) = 0 is singular. Remark 1. For an n x n matrix A, det (A - AI) is a polynomial in A of degree n. This polynomial is known as characteristic polynomial of A, that is, characteristic polynomial of A = det (A - AI). Remark 2. From Theorem 5.1, it follows that A = 0 is an eigenvalue of A if and only if A is singular. That is, A is non-singular if and only if all eigenvalues of A are non-zero. Theorem 5.2. Let AI'
A2 , ... ,Ak be distinct eigenvalues of a square matrix
A of order n. Let Vi be an eigenvector corresponding to Ai' i = I, 2, ... , k. Then the set SK = {vi' v2 ' ... ··, vk } is LI. Proof: We prove this theorem by the method of induction.
Since ~
VI
is an eigenvector corresponding to AI
0 ~ SI = {vI} is LI Now assume that the set vl;t
Sr = {vI' v2 '···, vI'} is LI SI' Then to show that
Let ui' u 2 ' ••• , u r +1 be scalars such that ...(5.2)
UIVI +U 2V2 +",+urvr +ur+lvr +1= 0 ~
A(u , VI +U 2 V2 +",+u r VI' +Ur+IVr +l ) = AO = 0
~
u,Av, +u 2 Av2 + ... +urAvr +u"",Avr+' = 0 as A is linear
~
UI/'IVI +U 2 A2V 2 + ... +UrArVr +Ur+IA r + 1 vr+1 = O.
...(5.3)
Now multiply equation (5.2) by Ar+ I and then subtract equation (5.3) from this to get . u l (Ar+1 - AI )VI + ~
!:x 2(A,.+I - A2)V~ + ... + Ur(Ar+1-
A,.) V,. = 0
UI(A,.+I-AI)=O=U2(A.,+I-Az)=···=U,.(A,.+I-A,.),
as the set SI' S,. is LI.
Eigenvalues and Eigenvectors
115
=> u U Il = 0 = u U 12 = ... = u, U, as A,
:t A,+I
for i = 1, 2, ... , r.
Hence equation (5.2) reduces to U,+IV,+I
0
=
=> a r + 1 = o'as Vr + 1 :t 0 Thus, the set SI' Sf' + 1 is LI. Hence by induction, it follows that Sk is LI Theorem 5.3. Let 1..1'1..12 "'" A" be n distinct eigenvalues of a square matrix A of order n. Let VI be an eigenvector corresponding to Ai' i = 1, 2, ... , n. Then the set B ={vI' v1 ' " ' ' v,,} is a basis for the domain space of A, and matrix of the linear transformation A relative to the ordered basis B is
0
,0
.~. ~~
o
o
o
AI
(A:B,B)= [
0
Proof: From Theorem 5.2, it follows that B is LI. Since A is of order n, hence the dimension of the domain space of A is n. Therefore, B is a basis for the domain space of A. Since VI is an eigenvector of A corresponding to Ai
=>
Av, = A., v"
=>
AVI
i = 1, 2, ... , n.
=AIVI =AIVI +0,v2 +0,V3 + ... +O.v" 1
AVl = A A1V A1,V 2V 1 = O,VI + A 2 ,V1 + O,vJ + ... + O.v", Av" = A."v" = O,vl + ... +O.V,,_I +A"v". Hence AI
(A:B,B)= [
0
0
.~. ~~ .~. o
0
0
This completes the proof of the theorem.
·~l·
A"
Introductory Linear Algebra
116
Theorem 5.4. Let A." A. 2, ... ,A." be n eigenvalues (need not necessarily be distinct) of a matrix A of order n x n. Then (a) det A = A.I A.2 ...A. n = product of the n eigenvalues
(b) tr(A)=A. 1 +A.2+···+A.n = sum of the n eigenvalues,
where tr (A) = trace of A = sum of the diagonal elements of A. Proof: (a) Since eigenvalues A." A.2' ... ,A." are the roots of characteristic polynomial, hence we can write For A. =
det(A - A.I) = (A.I - A.)(A. 2- A.) ... (A. n - A.) 0, we get det (A) = A.I A. 2.. ·A.n •
This proves part (a). (b) We have
(A.I - A.) (A. 2- A.) ... (A. n - A.) = det (A - A.I) al2 all - A. al2 -A. a21
aln a2n
...{5A)
a -A. all2 ani The left hand side and the right hand side of the above equation are polynomials in '). . of degree n of the form p(A.) =(-I)"A.n +an_lA. n- 1+ ... +alA.+ao By expanding both sides of equation (5.4), we can compute the coefficient all _ 1 of A.n - 1 to get llll
A.I + A.2 + ... + A.n = all + azz + ... +anll • This proves part (b). Proposition 5.1. Let A. be an eigenvalue of A. Iff is any polynomial, then
f(A)x =f(A.) x for some x
=I:-
O.
Proof: Since')... is an eigenvalue of A ~ 3 X (=I:- 0) such that Ax = Ax. We have,
A 2x = A(Ax)
=A (Ax) =A. (Ax) =A. (Ax) =A. 2x.
In fact, by mathematical induction, it is easy to show that Anx = ')...nx.
...(5.5)
Eigenvalues and Eigenvectors
117
Since f is any polynomial we may write
f(A)
=
aoAn +aJAn- J+ ... +an_JA+aJ.
f(A)x = aoAnx+aJAn-Jx+ ... +an_JAx+anIx ~
n
~
=aoll. x+aJII.
=
n-J
~_.
x+ ... +an_J~+anx
f(A)x.
...(5.6)
From the above proposition we note the following obvious remarks. Remark 1. If A A. is an eigenvalue of A, then An A.n is an eigenvalue of An. It follows from equation (5.5). Remark 2. Let A A. be an eigenvalue of A. If x is an eigenvector of A corresponding to A, A., then x is an eigenvector of An corresponding to the eigenvalue An. It also follows from equation (5.5). Remark 3. Let A be an eigenvalue of A andfbe any polynomial. Thenf(A.) ThenfCA) is an eigenvalue of f(A). It follows from equation (5.6). Remark 4. Let x be an eigenvector of A corresponding to the eigenvalue A.. Then x is also an eigenvector off{A) corresponding to the eigenvaluef(A.), A. eigenvaluef(A), where f is any polynomial. It also follows from equation (5.6). Theorem 5.5. (CayIey-Hamilton (Cayley-Hamilton Theorem) Every square matrix satisfies its own characteristic polynomial. Proof: Let A be a square matrix of order n x n. Let p(A) =An +an_JAn- J+ ... +aJA+ao peA) be the characteristic polynomial of A. Then to show that p (A) = O. Since p (A) (A.) is the characteristic polynomial of A
=>
peA) p(A) = det{Al- A)
From elementary matrix theory we know that
{det{Al- A»l =(A.!- A) adj(Aladj(A.l- A)
...(5.7)
This shows that adj(Al - A), adjoint of AI - A, is a polynomial of degree (n - 1). Suppose that
adj(AI-A) = Bo+BJA+B2A2 + ... +Bn_JAn-J , where each Bj is an n x n matrix. Thus, equation (5.7) can be written as
118
Introductory Linear Algebra
=
(').J - A) (Bo + BIA + B2A2 + ... + B,,_IA"-I)
... (5.8)
Equating the coefficients of equal powers of A in equation (5.8), we obtain
-ABo = aof, Bo -ABI = a/, B /, BII - AB2 = a2I,
B"_I
=
I.
Multiplying the above equations by I, A, A2, ... , A", respectively, and then adding them we get
A" + a"_IA"-1 + a"_2A"-2 + ... + alA + aoI = O.
=>
p(A)=O.
Before giving some examples we emphasize the following important steps involved in calculating eigenvalues and eigenvectors. Step 1: Compute det (A - AI) . This is a polynomial of degree n, if A is of order n. Step 2: Find the roots of the polynomial obtained in step 1. These n roots are the eigenvalues of A. Step 3: For each eigenvalue A, find N (A -AI) . This is the eigenspace corresponding to A. Step 4: The set of eigenvectors = N (A - AI) - {O} . Remark: In step 3, N(A -AI) ={x: (A -AI)x = O}.
Since det (A - AI) = 0, there are solutions to (A - A/)x AI)x = 0 other than
x= O. In case, in solving (A - AI) x = 0 yields x = 0 only, then A is NOT an eigenvalue. Example 3. If A is a triangular matrix, then eigenvalues of A are exactly the diagonal elements of A.
Eigenvalues and Eigenvectors
1HI
Solution: Suppose that A is an upper triangular matrix, and A is any eigenvalue of A. Then
det(A - AI) =
~
all-A 0
al2 a 22 -A
a 2n
0
0
ann -'}..,
det(A -AJ) =(all -A)(a22 -'}..,)",(ann
- '}..,)
a ln
=0
=0
~
A = all' a 22, .. " ann' If A is lower triangular, the result follows in a similar way.
Example 4. Determine all eigenvalues and the corresponding eigenspaces for the matrix A =
[7 4]
-3 -1 .
Solution: Let A is an eigenvalues of A.
~
det(A-'}..,/)=
~
A2 -6A+5 =0
~
'}..,=1,5.
/
7-A -3
4 / -I-A =0
These are the eigenvalues of A. To find eigenspace corresponding to A = 1. We have, For A = 1,
N(A-AJ) = {x:(A-AJ)x=O}. N (A - I)
= {x
~
: (A - I) x
= 0].
6x, +4X2 =0, -3xl -2x2 =0.
~
3x, +2X2 =0
~
x I =--x 3 2'
2
and x 2 is a free variable.
120
Introductory Linear Algebra
Thus,
N(A-I)={(X p X2):XI
=-~X2'X2 E~H}
={( -~X2'X2 }X2 Em} ={X2(
-~, I}X2 Em}
aE = {a {Cl (-2, 3): Cl
m}
=[(-2,3)]. Hence, the eigenspace corresponding to A. = I is N (A -1) = [(-2, 3)], and set of eigenvectors corresponding to A. = I is [(-2,3)] - {(O, O)} . Now to find the eigenspace corresponding to A. = 5. We have,
N(A-51)
We note that
(A - 51) x
=
{x:(A-5I)x=0}
~ [_~ ~
_:]
[;J =[~]
2xI +4X2 =0,
- 3xI - 6x2
=0.
~ XI
=-2x2' and x 2 is a free variable.
Hence
N(A-5J)
=
{(XI' X2):XI =-2x2 and x2 E9t}
=
{(-2x2 ,X2 ):X2 e9t}
=
{x2 (-2, 1):X2 e9t}
=
[(-2, I)]
= eigenspace for A. = 5. ~
The set of ofeigenvectors eigenvectors corresponding to A. = 5 is [(-2,1)] - {CO, O)} .
Remark: In example 4, eigenvalues I and 5 are distinct, hence by Theorem 5.2, the set B ={(-2, 3), (-2, I)} is LI. In fact, by Theorem 5.3, B
=
{(-2, 3), (-2, I)}
is a basis for the domain space of A.
121
Eigenvalues and Eigenvectors
Example 5. Determine all eigenvalues and the corresponding eigenspaces for the matrix
A
=
[=: : :]. -16
8 7
Solution. Let A is an eigenvalue of A. Then det(A - AI) =
-9-A
4
-8
3-A.
4
8
7-A
-16
4 =0.
=> (A + Ii (A - 3) = 0 => A = -1, -1, 3 are eigenvalues of A. Now to find eigenspace corresponding to A = -1. We have, N(A + I) = {x: (A + I)x =.Q}. In order to solve (A + I) x = 0, we note that the augmented matrix associated with this system is
[ -s B = (A + J, 0) = -8 -16
-
=> -2xl +X2 +Xl =0. Hence
r
4
4
4
4
8 8
s
4 4
0 0
0 0 0 0
~] 0]
o ,R2~R2-RI o ,Rl ~ Rl -2RI
N(A +1) = {(XpX2,Xl):-2xl+X2+Xl=0} =
{(xl' X 2 ' Xl): X 2 = 2xI -Xl}
=
{(Xp2XI-Xl,Xl):XpXl E9\}
= {(Xl' 2xp O)+(O,-Xl , Xl): XI' Xl E9\} = {XI (1, 2, 0)+ Xl (0,
-1, 1): XI' Xl E 9\}
N(A + I) = [{(1, 2, 0), (0,-1, I)}]
= eigenspace corresponding to A = -1, and set of eigenvectors is [{(I, 2, 0),(0, -1, I)}] - {CO, 0, O)}.
122
Introductory Linear Algebra
Now to find the eigenspace corresponding to f... = 3. We have, N(A -31) = {x: (A - 31) x = O}. The augmented matrix associated with (A CA - 31) x = 0 is given by B = (A - 31, 0) =
: ~]
[-~! ~ -16
8 4 0
- > R,I4 111,-->11,/' [-3 1 O]'R' 1
- ~
[[-3-3 - -~ [[-3-3 - -~
[[-I-I [-I
-~
-
0 1 0 ,R2
-'t~/4
2 1 0 ,R) -'tR)/4
1 0]
1
1 0 0 0 -1 0 1 1
0
1
0 0 1 0
0 0 0 0
0 -2 0 0 O·
,R)-~
R) -2RI
~t -->R,+~ n~-->~+~
n~ -->~-R,
~r-->R'-~
n~ -2R, HR, - > ~R, -2R, -->
+ x 2 =0, -2x2 + x3 = 0, and -xI
x3
=> and
x3
is a free variable. xI
= x2 =
I
'2 x ),
is a free variable.
Thus,
N (A - 31) =
=
{(X" X2' Xl): XI =~Xl' X2== ~Xl' Xl E 9t}
{(i Xl ' iXl' Xl }x) E9t}
123
Eigenvalues and Eigenvectors
=
[(1,1,2)]
= eigenspace corresponding to A. = 3. Hence, set of eigenvectors
=
[(1, 1,2)] - {CO, 0, O)}.
Example 6. Determine eigenvalues and
ei~envectors
for the matrix
A=[~o ~ ~l. 1 1
Solution: Note that A. 0, 1 an,d 2 are eigenvalues of A. To find eigenspace for A. = o. We have, N(A -O.l) -OJ) = N (A) = {x : Ax = O}. OJ. The augmented matri~ associated with Ax = 0 is given by
B
~
~ G)~ e)~ [~ 0 ~ ~l (A,
0
-[~ ::~l xI
=0,
+ x3
0, and x3 is a free variable. Thus, x2
=
N(A) = {(XPX2,X3)~Xl'=0,X2=-X3,X3 is a free variable} = {(O,-Xl,Xl):Xl
e9t}
== {Xl (0, -1, 1):Xl e 9t} = [(0, -1, 1)] = eigenspace corresponding to A. = And set of eigenvectors corresponding to~A. = 0 is [(0, -1, I)] - {CO, 0, O)}. To find eigenspaces corresponding t~ A. = 1.
We have, N(A-I)={x:(A-I)x=ol:
o.
124/
Introductory Linear Algebra
The augmented matrix associated with (A - 1) x
B ~
=
0 is given by
~ [~ ~ r ~]
x 2 = 0 = x) and xI is a free variable.
N(A-1) =
{(X.,X2 ,X3 ):X2 =0=X3 ,XI
= {(X.,
e9\}
0, 0): XI e 9\}
= {XI (1,0,0): XI
e 9\}
=
[(1,0,0)]
= eigenspace corresponding to A. = 1. Set of eigenvectors corresponding to A. = 1 is [(1,0,0)] - {(O, 0, O)}. Similarly, readers can verify that N (A - 21) =
{(X.,
x 2 ' x3 ): XI
= {(O, xl' x3 ) : X3 = {X3 =
=0, x2 =xl' X3 e 9\} e 9\}
(0, 1, I) :X3 e 9\}
[(0, 1, 1)]
= eigenspace corresponding to A. = 2, and set of corresponding eigenvectors is [(0, 1, I)] - {(O, 0, O)}. Example 7. For the matrix A given in example 6 ofthis chapter, verifY CayleyHamiton Theorem. Solution: The characteristic polynomial of A is p (A.) = 1.3 -3A? +21. ~
3
2
P (A) = A -3A +2A
~ [~ :]-[~ ~ [~ ~] 0
0
4
6
4
6
0 0 0
P (A) = 0, a zero matrix.
:H~ ~] 0 2
2
Eigenvalues and Eigenvectors
125
EXERCISES ____________________________ EXERCISES _ _ _ _ _ _ _ _ _ _ _ _ _ ___ _ In problems 1-14, matrix A is given. Determine the eigenvalues and the corresponding eigenspaces of A. I.
A=G ~]
3. A = [;
5. A
~]
=[~ =~ ~l 6 -6 4
7. A
=[; ~ -~l 4 8 15
9. A{~ -~ ~l 11. A
=[~o =~ ~l 4-1
13. A=[i ;
~i il
15. Let A be an invertible matrix. Prove that if A is an eigenvalue of A, then A-I is an eigenvalue of A-I. (Note that A cannot be zero as A is invertible). 16. Prove that A and AT have the same eigenvalues. Do they necessarily have same eigenvectors? Justify your answer.
b]
.
17. For the matrix A =[ ac d ' prove the followmg:
Introductory Linear Algebra
126
(l) A has two distinct real eigenvalues if (a- d)2 + 4bc > o. (ii) A has one eigenvalue if (a- d)2
+ 4bc = o.
(iii) A has no real eigenvalue if (a - d)2 + 4bc < 0 .
(iv) A has only real eigenvalues if b = c. 18. Show that if A is an eigenvalue of an idempotent matrix A (i.e, A2 = A) , then A must be either 0 or 1.
DIAGONALIZATION OF MATRICES DIAGONALlZATION
5.3
In this section, we show that a square matrix with some specific features can be diagonalized. Before understanding diagonalization of matrices, readers should be aware of similar matrices. Hence, first of all, we define similar matrices and the associated properties. Definition. Let A and B be n x n matrices. Then we say that A is similar to B, written as A '" B, if there exists an n x n invertible matrix P such that A=p-IBP. Remark: A '" B => B '" A
Proof:
A", B
=> A = p-I BP for some invertible matrix P. => PAp-1 = P(P-IBP)p-1 P(P-1BP)p-1 = (pp-I) B (pp-I) = lBl= IBI= B
=> (p-1r (p-Ir ll AP-I = B Q-1AQ= B, where Q = p-I => Q-IAQ=
=> B '" A. Thus, we simply say that A and B are similar. Proposition 5.2. Similar matrices have same characteristic polynomial, but converse need not be true. Proof: Let A and B are similar matrices => 3 an invertible matrix P such that A =p-IBP
Now the characteristic polynomial of A is det(A - A1) = det(p-I BP - ')..J)
det(p-1BP-AP-1p) = det(p-IBP-AP-Ip)
Eigenvalues and Eigenvectors
127 = det(p-I(B-AI)P) =
det(p-I).det(B- A.!).det(P)
= (detprl.det(B-AI)detP
= det(B-A.!)
characteristic polynomial of B. To see that the converse is not necessarily true, consider the two matrices =
A =[
~ ~] and B = G ~]
Then it is easy to see that characteristic polynomial of A
A2 - 2A. + I = characteristic polynomial of B. But A is not similar to B, because if A ~ B, then B = rlAP P-lAP for some invertible matrix P. =
~ B = P-IIP = J, which is a contradiction. This completes the proof. Now we are in a position to define the diagnolization of square matrices. The basic aim of this section is to find an answer of a question: Is it p-l AP is a diagonal matrix possible to find an invertible matrix P such that rl for a given square matrix A? We shall see that this is sometimes possible, but not always. If it is possible, then the process of doing so is known as diagonalization of matrices. Now we give its formal defmition.
Definition. A square matrix A is said to be diagonalizable if there exists an invertible matrix P (order of A and P should be same) such that rlAP P-lAP is a diagonal matrix (that is, A is similar to a diagonal matrix). The following theorem gives a necessary and sufficient condition for the matrix A to be diagonalizable. Theorem 5.5. Let A be a square matrix of order n. Then A is diagonalizable if and only if A has n linearly independent eigenvectors. Proof: Let A is diagonalizable. ~ 3 an invertible matrix P of order n such that D = rl p-l AP is a diagonal matrix. Let AI' A2 , ... ,An be the diagonal entries ot D, i.e.,
Introductory Linear Algebra
128
Let vI' V 2 ' ••••• , vn be the columns of P. Then
AP
A[vl
V2
= [Avl
AV2
=
!
1..1 and
PD
~
P p
[
0
[A.IVI
Avn1, Avnl, 0
il
1..2 0 0
=
vn]
0
A. 2V2 ... A.nVn]·
...(5.9)
... (5.10)
But D = rlAP P-IAP ~ PD = AP. Hence from (5.9) and (5.10) we get
... Avn] Equating columns, we get [Avl
AV2
=
[A.IVI
A.2 V2
...
A.nvn]
AVj = A.;Vp i = 1, 2, ... ,n. Av; ...(5.11) Since P is invertible, its columns must be LI and therefore each column v; Vj must be non-zero. The equation (5.11) shows that 1..1' 1.. 2"'" A.n are v2' ... ,Vn vn are corresponding LI eigenvectors of A. eigenvalues and VI' V Conversely, let vI' v2' ... , vn be n LI eigenvectors of A corresponding to the eigenvalues 1..1' 1..2,,,,, A.n , respectively. Now we construct a matrix
P whose columns are vI' v2 ' ... , vn ; and we also construct a diagonal matrix D with diagonal entries as 1..1' 1..2'"'' A.nn .• Then from (5.9) - (5.11), we note that ...(5.12) AP =PD Since the columns vI' V V 22'' ... ... ,, Vn vn of P are LI, therefore P is invertible. Thus, equation (5.12) yields ~
A is diagonalizable.
Eigenvalues and Eigenvectors
129
Remark 1. The above theorem shows that A is diagonalizable if and only if there are enough eigenvectors to form a basis for Vn•
[~ ~].
Remark 2. AIJ matrices are not diagonalizable. For example, take A =
If A is diagonalizable, then for some invertible matrix P we note that rlAP =
[AIo
0]
A2 '
where A( = 0 and A2 = 0 are eigenvalues of A. => A must be zero matrix as P is invertible, which is a contradiction. Theorem 5.6. An n
x
n matrix with n distinct eigenvalues is diagonalizable.
Proof: Let v" v2' ... , vn be the eigenvector corresponding to the n distinct eigenvalues of an n x n matrix A.
=> The set {vI> v2 ' ••• , vn } is LI by Theorem 5.2. => A is diagonalizable by Theorem 5.5. Now we emphasize the important steps involved in diagonalization of a matrix. How to Diagonalize an n x n matrix A? Step 1: Find the eigenvalues
AI' A2' ... ,An
of A.
Step 2: Find n LI eigenvectors vI' v2 ' ••• , Vn of A corresponding to
A" A2"'" An' If it is possible, go to step 3; otherwise stop the process and conclude that A is not diagonalizable. Step 3: Construct the matrix P with vi as its ith column, that is, P = [VI v2
•••
vnl.
Step 4: Construct the diagonal matrix D with diagonal entries as AI' A2' ... ,A n. In fact, D = rIAP. Example 8. We consider the matrix
A=[_~ _~] as given in example 4 of this chapter. As we have seen in example 4 that ")... = I, 5 are the eigenvalues of A. Since these eigenvalues are distinct, therefore, by Theorem 5.6, A is diagonalizable.
130
Introductory Linear Algebra
From example 4, we also note that VI = (-2,3) and v2 = (-2,1) are eigenvectors corresponding to A = 1 and A = 5, respectively. One may note here that the set {vI' v 2 } is LI and hence it also suggests that A is diagonalizable. Then
-2 [-~
p= P = [ 3 In fact, one can verify that
-2]
[1 0]
1 and D= 0 5
P-IAP =D,
since
P-IAP =
~[_~
2] [7 4] [-2 -2]
2] [7 4] [-2 -2] -2 -3 -1 3 1
=[~ ~]=D. Remark 1. As in example 4, we have seen that S, SI = [(-2, 3)] - {CO, O)} = set of eigenvectors corresponding to A = 1. S2 = [(-2, 1)] - {CO, O)}= set of eigenvectors corresponding to A = 1.
One can ask a natural question here that can any eigenvectors from 8S,1 and S2 will diagonalize A? Answer is yes. Because any two vectors vI and v 22 (v, (vI from S, SI and v2 2 from S2) are LI and hence the set whose columns are vI and v22 will diagonalize A. For example, let us take v, vI = (--4, 6) and v2 = (6, -3).
v,
v,
Then
P
=
[-4 6]
[1° 0]5
6] and D = [1 6 -3
0]
It can be verified that p-I AP = D . Thus, P diagonalizes A. Remark 2. A change in the order of eigenvectors does not change the diagonalizability of A. But the eigenvalues appearing on the main diagonal. of the resulting diagonal matrix D would appear in accordance with the order of the eigenvectors in the matrix P.
For example if
P =
[-2 -2] 1
[5 0]
3' then D = 0
1 .
Eigenvalues and Eigenvectors
131
Example 9. Now wc we consider the .,atrix matrix A Exa.,ple
~ [_~: ~ ~l as given in ;]
example 5 of this chapter. Eigenvalues of A are A = -1, -1, 3. Since eigenvalues are not distinct, therefore, at this stage we can not conclude whether A is diagonalizable or not. For A = -1, the set of eigenvectors is
Sl SI
=
[{(l,2,0),(0,-1,1)}]-{(0,0,0)}.
For A = 3, the set of eigenvectors is S2 = [(1, 1,2)] - {CO, 0, O)}. One can check that the vectors vi Vi
= (1,2,0), v2 = (0, -1, 1) and v3 = (1, 1,2)
are Lt. Ll. Hence A is diagonalizable and
p~[! p~[~ -; .... Example 10. Consider the matrix A =
n D~[~l ~I~l n il D~[~' [~ ~ ~l~] 005
We wish to check whether A is diagonalizable or not. One can note that A = 4, 4, 5 are eigenvalues of A. Since eigenvalues are not distinct, therefore, no conclusion can be made at this stage. Now,
N(A-4J) = {x:(A-4J)x=0} = {(XI' Xl' Xl): XI
°
= =Xl' Xl E 9t}
= [(0, 1, 0)] . N(A-51) = {x:(A-5J)x=0} = =
{(x" Xl' Xl): XI [(0, 0, 1)].
°
= =Xl' Xl E 9t}
=
Hence every eigenvector of A is either a multiple of VI (0, 1, 0) or V l = (0, 0, 1). Thus it is not possible to have three linearly independent eigenvectors of A, i.e., one can not construct a basis for V3 with the eigenvectors of A. Hence A is not diagonalizable.
132
Introductory Linear Algebra
EXERCISES ______________________________ I. In problems 1-14 of Sec 5.2, determine whether the given matrix A is diagonalizable. Ifso, find an invertible matrix P over 9\ and a diagonal matrix Dover 91 such that P-IAP = D.
2. I, the matrix A
~ [~ =~
:]
diagonalizable? If so, find an invertible matrix P and a diagonal matrix D such that P-IAP = D.
3. Prove that the matrix
[~
A=
diagonalizable.
n
a
a i, any real number. i, not
0
l
~ [~ ~ 0
4. Prove that the matrix
A
a
a i, any real number. i, not
diagonalizable. 5. Without any calculation, justify why these matrices are diagonalizable.
[~ ~] 0
(a)
2
(b)
0
[~ ~]
[~
0
;]
2 0 0
0
0
2
0
0
2
(c)
(d)
2
3
6. Consider the matrix
A
~ [~
0 0 0 0 b
0
o
c
[-i
~]
~]
Find condition(s) on a, band c such that A is diagonalizable.
Bibliography I. Halmos, P., Finite-Dimensional Vector Spaces, D. Van Princeton, 1958.
N(\~trand
Co.,
2. Hoffman, K. and Kunze, R., Linear Algebra, Pearson Education, Second Indian Reprint, New Delhi, 2005.
3. Krishnamurthy, V., Mainra, V. P. and Arora, J. L., An Introduction to Linear Algebra, Affiliated East-West Press Pvt. Ltd., New Delhi, 2003. 4. Kolman, B. and Hill, D. R., Introductory Linear Algebra with Applications, Pearson Education, Seventh Edition, 200 I. 5. Kwak, J. H. and Hong, S., Linear Algebra, Springer International Edition, Birkhauser, Boston, Second Edition, 2004.
6. Lay, D. C., Linear Algebra and its Applications, Pearson Education, Second Indian Reprint, New Delhi, 2004.
7. Strang, G., Linear Algebra and its Applications, Thomson Learning, Third Edition, 1988.
8. Gilbert, J., and Gilbert, L., Linear Algebra and Matrix Theory, Academic Press, 1995.
"This Page is Intentionally Left Blank"
Index Adjoint, 117 Annihilator, 89 Associativity additive, 1 multiplicative, 2 Augmented matrix, 3 Basis, 39,45 ordered, 92 standard basis of 9\", standard basis of P",
Eigenvector, 1l0,112 Elementary row operations, 5 Equivalent system of equations, 4 Field, 1 Finite-dimensional vector space, 45 Idempotent, 83 operator, 83 matrix, 126 Identity additive, map, 59 matrix, 16 Inconsistent sys~~m, 5, 9 Independence (linear), 39 Invertible matrix, 15 linear transformation, 77 Largest subspace, 28 Leading entry, 5 Linear combination, 34 Linear equations (see system of linear equations), 3 Linear functional, 86 Linearly dependent, 39 Linearly independent, 39 Linear operator, 60 Linear transformation, 59 composition of, 75 invertible, 77 matrix or; 95
45 46
Cayley-Hamilton Theorem, 117 Characteristic polynomial, 114 value (see eigenvalue), 113 vector (see eigenvector), 113 Column rank, 105 Column space, 105 Commutative addition, multiplication, 2 'Consistent system, 5, 9 Coordinate matrix, 95 Degree of polynomial, 24 Dependence (linear), 39 Diagonalization, 127 Dimension, 39,45 Direct sum, 33 Eigenspace, 113 Eigenvalue, 110, 112 135
(r1cj t_ '{:\
136
r
\(
,.."..
\':,S:. \ ".,,1" ( ...
range of, 64 rankof, 67 singular, 77 Lower triangular matrix, 119 L(U,V) space, 73 Matrix augmented, 3 of Iinear transformation, 95 row rank of, 9, 105 row-reduced echelon form, 6 similar, 126 trace of, 116 triangular, 118, 119 n - tuple, 22 Nilpotent operator, 83,84 Non-singular, 77 linear transformation, 77
matrix, 15 Nullity, 67, 105 Null space, 64, 110 oflinear transformation, 64 ofmatrix, 110 One-one map, 77 Onto map, 77 Operator (linear), 60 Ordered basis, 92 Polynomial, 24 characteristic, 114 degree of, 24 Range space, 64 Rank of linear transformation, 67 of matrix, 108
Introductory Linear Algebra. Rank-nullity Theorem, 67 Row equivalent, 5 rank, 9,105 Row echelon form, 5 Row-reduced echelon matrix, Row space, 105 Scalar, 21 Similar matrices, 126 Singular matrix, 16 Solution set, 4 Span, 45 Standard basis, 45 ofPn , 46 of 9\",
6
45
Subspace, 26,27 spanned by, 36 sum of subs paces, 51 zero, 28 System of linear equations, 3 homogeneous, 4 inhomogeneous, 4 Trace of a matrix, 116 Transformation linear, 59 matrix of, 95· Upper triangular matrix, 119 Vector space, 21 basis of, 39, 45 dimension of, 39, 45 finite-dimensional, 45 Zero linear transformation, 59 Zero polynomial, 24 Zero subspace,
28
