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Syllabus
S.Y.B.Sc. (C.S.) Mathematics Paper II
Linear Algebra
Unit 1 : Systems of linear equations and matrices
(a) Systems of homogeneous and non-homogeneous linear equations
(i) The solutions of systems of m homogeneous linear equations in n
unknowns by elimination and their geometric interpretation for
, 1, 2 1, 3 2, 2 2, 3 3, 3m n .
(ii) The existence of non-trival solution of such a system for m n .
The sum of two solutions and a scalar multiple of a solution of
such a system is again a solution of the system.
(b) (i) Matrices over , The matrix representation of systems of
homogeneous and non-homogeneous linear equations.
(ii) Addition, scalar multiplication and multiplication of matrices,
Transpose of a matrix
(iii)The types of matrices : zero matrix, identity matrix, symmetric and
skew symmetric matrices, upper and lower triangular matrix.
(iv) Transpose of product of matrices, Invertible matrices, Product of
invertible matrices.
(c) (i) Elementary row operations on matrices, row echelon from of a
matrix and Gaussian elimination method. Applications of Gauss
elimination method to solve system of linear equations.
(ii) The matrix units, Row operations and Elementary matrices,
Elementary matrices are invertible and an invertible matrix is a
product of elementary matrices.
Reference for Unit 1 : Chapter II, Sections 1, 2, 3, 4, 5 of Introduction to
Linear Algebra, SERGE LANG, Springer Verlag and Chapter 1, of Linear
Algebra A Geometric Approach. S. KUMARESAN, Prentice Hall of
India Private Limited, New Delhi.
Unit 2 : Vector spaces over
(a) Definition of a vector space over . Examples such as
(i) Euclidean space n .
(ii) The space of sequences over .
(iii) The space of m n matrices over .
(iv) The space of polynomials with real coefficients.
(v) The space of real valued functions on a non-empty set.
(b) Subspaces – definition and examples including
2
(i) Lines in 2 , Lines and planes in 3 .
(ii) The solutions of homogeneous system of linear equations,
hyperplane.
(iii) The space of convergent real sequences.
(iv) The spaces of symmetric, skew symmetric, upper triangular,
lower triangular, diagonal matrices.
(v) The space of polynomials with real coefficients of degree n.
(vi) The space of continuous real valued functions on ,a b .
(vii) The space of continuously differentiable real valued functions on
,a b .
(c) (i) The sum and intersection of subspaces, direct sum of a subset of a
vector space.
(ii) Linear combination of vectors, convex sets, linear span of subset
of a vector space.
(iii) Linear dependence and independence of a set.
(d) (The discussion of concepts mentioned below for finitely generated
vector spaces only) Basis of a vector space, basis as a maximal linearly
independent set and a minimal set of generators. Dimension of a
vector space.
(e) (i) Row space, Column space of an m n matrix over and row
rank, column rank of a matrix.
(ii) Equivalence of row rank and column rank, Computing rank of a
matrix by row reduction.
Reference for Unit 2 : Chapter III, Sections 1, 2, 3, 4, 5, 6 of
Introduction to Linear Algebra, SERGE LANG, Springer Verlag and
Chapter 2 of Linear Algebra A Geometric Approach S. KUMARESAN,
Prentice Hall of India Private Limited, New Delhi.
Unit 3 : Inner Product Spaces
(a) Dot product in n , Definition of general inner product on a vector
space over .
Examples of inner product including the inner product
.f g f t g t dt
on ,C , the space of continuous real
valued functions on , .
(b) (i) Norm of a vector in an inner product space.
3
Cauchy-Schwarz inequality, triangle inequality.
(ii) Orthogonality of vectors, Pythagorus theorem and geometric
applications in 2 .
(iii) Orthogonal complements of a subspace, Orthogonal
Complements in 2 and 3 .
(iv) Orthogonal sets and orthonormal sets in an inner product space.
Orthogonal and orthonormal bases.
Gram-Schmidt orthogonalization process, simple examples in
3 1, .
Reference for Unit 3 : Chapter VI, Sections 1, 2 of Introduction to Linear
Algebra, SERGE LANG, Springer Verlag and Chapter 5 of Linear
Algebra A Geometric Approach, S. KUMARESAN, Prentice Hall of India
Private Limited, New Delhi.
Unit 4 : Linear Transformations
(a) Linear transformations – definition and properties, examples including
(i) Natural projection from n to m . n m
(ii) The map : n mAL defined by AL X AX , where A is an
m n matrix over .
(iii) Rotations and reflections in 2 , Streching and Shearing in 2 .
(iv) Orthogonal projections in n .
(v) Functionals.
The linear transformation being completely determined by its values
on basis.
(b) (i) The sum and scalar multiple of linear transformations from U to
V where U, V are finite dimensional vector spaces over is
again a linear transformation.
(ii) The space ,L U V of linear transformations from U to V.
(iii) The dual space V where V is finite dimensional real vector space.
(c) (i) Kernel and image of a linear transformation
(ii) Rank-Nullity Theorem
(iii) The linear isomorphisms, inverse of a linear isomorphism
(iv) Composite of linear transformations
(d) (i) Representation of a linear transformation from U to V, where U
and V are finite dimensional real vector spaces by matrices with
4
respect to the given ordered bases of U and V. The relation
between the matrices of linear transformation form U to V with
respect to different bases of U and V.
(ii) Matrix of sum of linear transformations and scalar multiple of a
linear transformation.
(iii) Matrices of composite linear transformation and inverse of a
linear transformation.
(e) Equivalence of rank of an m n matrix A and rank of the linear
transformation : n mA AL L X AX . The dimension of
solution space of the system of linear equations 0AX equals n –
rank A.
(f) The solutions of non-homogeneous systems of linear equations
represented by AX B .
(i) Existence of a solution when rank A = rank ,A B .
(ii) The general solution of the system is the sum of a particular
solution of the system and the solution of the associated
homogeneous system.
Reference for Unit 4 : Chapter VIII, Sections 1, 2 of Introduction to
Linear Algebra, SERGE LANG, Springer Verlag and Chapter 4, of
Linear Algebra A Geometric Approach, S. KUMARESAN, Prentice
Hall of India Private Limited, New Delhi.
Unit 5 : Determinants
(a) Definition of determinant as an n-linear skew-symmetric function from
...n n n n such that determinant of 1 2, , ..., nE E E is
1, where jE denotes the jth
column of the n n identity matrix nI .
Determinant of a matrix as determinant of its column vectors (or row
vectors)
(b) (i) Existence and uniqueness of determinant function via
permutations.
(ii) Computation of determinant of 2 2, 3 3 matrices, diagonal
matrices.
(iii) Basic results on determinants such as
det det , det det dettA A AB A B .
(iv) Laplace expansion of a determinant, Vandermonde determinant,
determinant of upper triangular and lower triangular matrices.
5
(c) (i) Linear dependence and independence of vectors in n using
determinants.
(ii) The existence and uniqueness of the system AX B , where A is
an n n matrix with det 0A .
(iii) Cofactors and minors, Adjoint of an n n matrix A.
Basic results such as . det . nA adj A A I . An n n real
matrix A is invertible if and only if
1 1det 0;
detA A adj A
A
for an invertible matrix A.
(iv) Cramer‟s rule
(d) Determinant as area and volume
Reference for Unit 5 : Chapter VI of Linear Algebra A geometric
approach, S. KUMARSEAN, Prentice Hall of India Private Limited,
2001 and Chapter VII Introduction to Linear Algebra, SERGE LANG,
Springer Verlag.
Unit 6 : Eigenvalues and eigenvectors
(a) (i) Eigenvalues and eigenvectors of a linear transformation
:T V V , where V is a finite dimensional real vector space.
(ii) Eigenvalues and eigenvectors of n n real matrices and
eigenspaces.
(iii) The linear independence of eigenvectors corresponding to distinct
eigenvalues of a matrix (linear transformation).
(b) (i) The characteristic polynomial of an n n real matrix,
characteristic roots.
(ii) Similar matrices, characteristic polynomials of similar matrices.
(c) The characteristic polynomial of a linear transformation :T V V ,
where V is a finite dimensional real vector space.
Reference for Unit 6 : Chapter VIII, Section 1, 2 of Introduction to
Linear Algebra, SERGE LANG, Springer Verlag and Chapter 7, of Linear
Algebra A Geometric Approach, S. KUMARESAN, Prentice-Hall of India
Private Limited, New Delhi.
The proofs of the results mentioned in the syllabus to be covered
unless indicated otherwise.
Recommended Books :
1. SERGE LANG : Introduction to Linear Algebra, Springer Verlag.
2. S. KUMARESAN : Linear Algebra A Geometric approach, Prentice
Hall of India Private Limited.
6
Additional Reference Books :
1. M. ARTIN : Algebra, Prentice Hall of India Private Limited.
2. K. HOFFMAN and R. KUNZE : Linear Algebra, Tata McGraw Hill,
New Delhi.
3. GILBERT STRANG : Linear Algebra and its applications,
International Student Edition.
4. L. SMITH : Linear Algebra, Springer Verlag.
5. A. RAMACHANDRA RAO and P. BHIMA SANKARAN : Linear
Algebra, Tata McGraw Hill, New Delhi.
6. T. BANCHOFF and J. WERMER : Linear Algebra through Geometry,
Springer Verlag New York, 1984.
7. SHELDON AXLER : Linear Algebra done right, Springer Verlag,
New York.
8. KLAUS JANICH : Linear Algebra.
9. OTTO BRETCHER : Linear Algebra with Applications, Pearson
Education.
10. GARETH WILLIAMS : Linear Algebra with Applications, Narosa
Publication.
Suggested topics for Tutorials / Assignments :
(1) Solving homogeneous system of m equations in n unknowns by
elimination for , 1, 2 1, 3 2, 2 2, 3 3, 3m n .
(2) Row echelon from, Solving system AX B by Gauss elimination.
(3) Subspaces : Determining whether a given subset of a vector space is a
subspace.
(4) Linear dependence and independence of subsets of a subsets of a
vector space.
(5) Finding bases of vector spaces.
(6) Rank of a matrix.
(7) Gram-Schmidt method.
(8) Orthogonal complements of subspaces of 3 (lines and planes).
(9) Linear transformations.
(10) Determining kernel and image of linear transformations.
7
(11) Matrices of linear transformations.
(12) Solutions of system of linear equations.
(13) Determinants : Computing determinants by Laplace‟s expansion.
(14) Applications of determinants : Cramer‟s rule.
(15) Finding inverses of 2 2, 3 3 invertible matrices using adjoint.
(16) Finding characteristic polynomial, eigenvalues and eigenvectors of
2 2 and 3 3 matrices.
(17) Finding characteristic polynomial, eigenvalues and eigenvectors of
linear transformations.
8
1
SYSTEMS OF LINEAR EQUATIONS AND
MATRICES
Unit structure
1.0 Objectives
1.1 Introduction
1.2 Systems of Linear equations and matrices
1.3 Equivalent Systems
1.4 Exercise
1.5 Unit End Exercise
1.0 OBJECTIVES
Our aim in this section is to study the rectangular arrays of numbers
corresponding linear equation and some related concept to develop an
elementary theory of the same. Object of this section is to solve the
unknowns providing different and simplest methods for understanding of
the students in a very simple manner.
1.1 INTRODUCTION
In this section, we shall be concerned with sets of numbers
arranged systematically in the columns and rows, called rectangular
arrays. To take an example consider the system of linear equations:
11 1 12 2 13 3
21 1 22 2 23 3
31 1 32 2 33 3
0;
0;
0.
a x a x a x
a x a x a x
a x a x a x
We shall see that while solving this system of equations, we in
fact, find ourselves working only with the rectangular array of constants :
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
9
The unknowns 1 2 3, ,x x x act merely as marks of position in the
equations.
Also in this section we introduce basic terminology and discuss a
method for solving system of linear equations.
A line in the xy plane can be represented algebraically by an
equation of the form.
11 1 12 2a x a x b ,
An equation of this kind is called a linear equation in the variables
1x and 2x . More generally, we define a linear equation in the n variables
1 2, , ..., nx x x to be one that can be expressed in the form.
11 1 12 2 1... n na x a x a x b , where 11 12 1, , ..., na a a and b are
real numbers.
1.2 SYSTEMS OF LINEAR EQUATIONS AND
MATRICES :
We shall deal with the problems of solving systems of linear
equations having n unknowns.
There are two aspects to finding the solutions of linear equations.
Firstly, the formal manipulative aspect of computations with matrices and
secondly, the geometric interpretation of the system as well as the
solutions. We shall deal with both the aspects and integrate algebraic
concepts with geometric notions.
System of linear equations :
The collection of linear equations
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
...
...
...
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
… (*)
where ija , ib , 1 i m ,
1 j n .
is called a system of m linear equations in n unknown.
If 1 2 ... 0mb b b
10
i.e.
11 1 12 2 1
21 1 22 2 2
... 0
... 0
n n
n n
a x a x a x
a x a x a x
1 1 2 2 ... 0m m mn na x a x a x
It is called a homogeneous system of m linear equations in n unknowns.
If atleast one 0bi , where 1 i m , it is called a non-
homogeneous system of m linear equations in n unknowns.
i.e. If
11 1 12 2 1
21 1 22 2 2
... 3
... 0
n n
n n
a x a x a x
a x a x a x
1 1 2 2 ... 0m m mn na x a x a x
is called a non-homogeneous system of m linear equations.
In short,
Simple example
2 3 0
2 3 4 0
5 0
x y z
x y z
x y z
(homogeneous system)
is called a homogeneous systems in three variables x, y and z.
If
2 3 5
2 3 3
5 0
x y z
x y z
x y z
(non-homogeneous system)
is called non-homogeneous system in three variables x, y and z.
We may represent (*) in a short form as
1
, 1
n
ij j ij
a x b i m .
Any n-typle 1,..., nx x x which satisfies all the equations in the
system (*) is called a solution of the system. , 1ix i n . The
set of solutions is called the solution set or sometimes, the general
solutions of the system. If the system has no solution, we say the system
is inconsistent. We give few examples :
1. 1 + 2 = 4 x y E
2 + 2 = 6 – Ex y (inconsistent)
11
2. 12 4x y E
22 3 7x y E (unique solution)
3. 12 4x y E
22 4 8x y E (infinite many solutions)
We try to solve the above systems by eliminating one variable. In
system (1). 1 2E E gives O = 2 , thus the system is inconsistent. In
system (2), 1 22E E gives = 1y , substituting 1y = in 1E , we get 2x = .
Thus we have a unique solution set 2,1 for system (2).
In system (3), the equation (E2) is obtained by multiplying the first
equation by 2.
Therefore (s, t) is a solution of this system,
NOTE : If m n , the system 0AX has a non-trivial solution.
2 4
2 4 8
s t
s t
4 2 ,t t is a solution of (3) for any t . The solution set is
4 2 , :t t t . Thus system (3) has infinitely many solutions.
Geometrically, the first system has two parallel lines which do not
intersect. The second system has two intersecting lines intersecting in one
point. The third system has one line and every point on the line is a
solution. Thus, the system has infinitely many solutions.
“Thus, a system of non – homogeneous equations may have no
solutions, a unique solution or infinitely many solutions.”
However, a system of homogeneous equations in n unknowns,
where all ci = 0 1 i n This is called a trivial solution. Any solution
1, ..., nc c where at least one 0ic , 1 i n is called a non-trivial
solution.
We note that in order to solve the given system of equations, we
eliminate one variable. The most fundamental technique for finding the
solutions of a system of linear equations is the technique of elimination.
We multiply an equation by a non-zero scalar and add it to another
equations to produce an equation where one of the unknowns may be
absent. We assume that the new system obtained has same solutions as
the original system.
12
1.3 EQUIVALENT SYSTEMS :
Consider the system of linear equations.
1
n
ij j ij
a x b
; for 1 i m
A system of equations obtained by
i) Multiplying an equation in the system by a non-zero scalar.
ii) Adding scalar multiple of an equation in the system to another
equation is called asystem equivalent to the given system.
Problem: We now find solution of m homogeneous linear equations
in n unknowns when 1 2 1 3 2 2 2 3 3 3m,n , , , , , .
i) m = 1, n = 2
11 12 0a x a y , atleast on of 11 12 0a ,a .
If 11 0a , and s,t is a solution of the system.
11 12 0a s a t
12 11/ ,S a a t t
The solution set is
12
11
,a
t t ta
i.e. 12
11
,1a
t ta
If 11 0a , then 12 0a and multiplying the equations by 112a we
get equation 0y .
The solution set is 0t , t or 1 0, t t . Thus the
system has infinitely many solutions.
Geometrically, the system represents a straight line through origin
and every point on the line is a solution.
13
a x a y 011 12
m = n = 2
ii) 11 12 0a x a y E1
21 22 0a x a y E2
If 11 12
21 22
a aλ
a a , 0λ λ , then multiplying equation (i) by λ , we
get that two equations are same and the system is 11 12 0a x a y , which
we discussed in (i) as 11 12
21 22
a a
a a , then 11 22 12 21 0a a a a .
If s,t is a solution then 21a times 1 11E a times 2E gives
21 12 11 22 0a a a a y
i.e. 11 22 21 12 0a a a a t
t 0 11 22 21 12 0a a a a
Substituting in ( E1) and (E2 ) we get
11 0a s
21 0a s
s = 0
021 22
a x a y
011 12
a x a y
14
The system has unique solution, namely trivial solution
(Geometrically, the system represents two lines passing through origin.
iii) m = 1, n = 3
11 12 13 0a x a y a z
If 11 0a and (r, s, t) is a solution of the above system
11 12 13 0a r a s a t .
1312
11 11
,a ta
r sa a
s,t .
The solution set is
1312
11 11
, , ,aa
s t s t s ta a
i.e. 12 13
1111
,1, 0 , 0,1 ,a a
s t s taa
Thus, the system has infinitely many solutions. Geometrically the system
represents a plane passing through origin and any point on the plane is a
solution of the system.
iv) m = 2, n = 3
11 12 13 0a x a y a z ( 1E )
21 22 23 0a x a y a z ( 2E )
Let 1311 12
21 22 23
λaa a
a a a , then ( 1E ) and ( 2E ) are same equations.
(multiplying 2E by λ ) and we have already discussed this in (iii).
If 11 12
21 22
a a
a a or 1311
21 23
aa
a a or 1312
22 23
aa
a a
We eliminate x by 21 1 13 2a E a E . Similarly, we eliminate y by
22 1 12 2a E a E , and z by 23 1 13 2a E a E . Thus we get
12 23 22 13 21 13 11 23 11 22 21 12
x y z
a a a a a a a a a a a a
If (r, s, t) is a solution of the system.
If 12 23 22 13 21 13 11 23 11 22 21 22
r s t
a a a a a a a a a a a a
λ , for λ
The solution set is 11 22 21 22λ λa a a a
= 12 23 22 13 21 13 11 23 11 22 21 22λ , , λa a a a a a a a a a a a
At least one of the x, y, z co-ordinate is non-zero and we have infinitely
many solutions. Geometrically, the system represents two planes passing
through origin and the planes intersect in a line through origin.
15
v) 3, 3m n
11 12 13 0a x a y a z ( 1E )
21 22 23 0a x a y a z ( 2E )
31 32 33 0a x a y a z ( 3E )
If any equation in the above system is a linear combination of the other
two equations the system is equivalent to a system of one equation in three
unknowns or two equations in three unknowns and these cases have been
discussed. Otherwise
11 12 13
21 22 23
31 32 33
0
a a a
a a a
a a a
If r, s, t is a solution of the above system, we consider equation (E2)
and (E3) and get
22 33 23 32 31 23 21 33 21 32 22 31
r s t
a a a a a a a a a a a a
λ , for λ
Substituting in 1E
11 22 33 23 32 12 31 23 21 33 13 21 32 31a λ a a a a a λ a a a a a λ a a a
22 0a
11 12 13
21 22 23
31 32 33
0
a a a
λ a a a
a a a
and so λ 0 .
0 0 0r, s, t , ,
Geometrically, the three equations represent three planes passing through
origin and they intersect in a unique point, name the origin.
We observe that, in the above systems, the system has infinitely many
solutions of m < n . But, for m = n , the system has only trival solution
provided certain conditions were satisfied by the coefficients ija .
NOTE : The system
1
0n
ij jj
a x
, 1 i m of m homogeneous linear
equations in n-unknowns has a non-trivial solution if m < n .
Solve examples :
1) Find atleast one non-trivial solution of the system 3 0x y z .
16
Solution : We take 1 1y , z and get 1 2
23 3
x
then
21 1
3, ,
is a non-trivial solution of the system.
2) Write the general solution of the system and give geometrical
interpretation.
2 3 4 0x y z
3 0x y z
Solution : The given system is
2 3 4 0x y z ( 1E )
3 0x y z ( 2E )
To eliminate y , by multiplying 2E by 3 and adding to 1E , we get
11 7 0x z ( 1E )
2 3 4 0x y z ( 2E )
If r, s, t is a solution of the system, then 11 7 0r t , i.e. 7
11r t
Substituting in 1E , we get 2 3 4 0r s t
i.e. 14
3 4 011
ts t
30
311
ts ;
10
11
ts
Thus, the solution is 7 10
11 11
tt , ,t ,t
.
The set of solution is 7 10
111 11
, , t t
This represents a line passing through origin and having direction
7 101
11 11, ,
in 3 . The given system is representing two planes in
3 passing through (0, 0, 0) . They interesting line passing through
0 0 0, , .
3) Show that the only solution of the following system is the trivial
solution.
4 7 3 0x y z
0x y
6 0y z
Solution : The given system is
4 7 3 0x y z ( 1E )
0x y ( 2E )
6 0y z ( 3E )
17
To eliminate y by performing 2 3E E we get
4 7 3 0x y z ( 11E )
0x y ( 12E )
6 0x z ( 13E )
If r, s, t is a solution of the system,
12E gives 0r + s = i.e. r s .
13E gives 6 0r + t = i.e.
6
rt .
11E gives 4 7 3 0r s t
i.e. 4 7 3 06
rr r
0r
0s r , 06
rt . Thus the only solution of the system is
0 0 0, , .
Tutorial / Assignment :
Topic :- Solving homogeneous system of m linear equations in n
unknowns by elimination for 1 2 1 3 2 2 2 3 3 3m,n , , , , , , , , , .
Theorem 1 :- Show that the system of linear equations.
1
0 1n
ij jj
a x , i , m
has a non-trivial solution for
1 2 1 3m,n , , , and 2 3, .
Theorem 2 :- Show that the system of linear equations
1
0 1n
ij jj
a x , i m
has unique solution for 2 2m,n , and 3 3, .
If 11 12
21 22
0a a
a a (for 2m n ) and
11 12 13
21 22 23
31 32 33
0
a a a
a a a
a a a
(for 3m n )
Exercises 1 : Show that the following systems of linear equations have
infinitely many solutions. Write the solution set and the geometrical
interpretations of the system as well as solution.
i) 2 0x y z
ii) 2 0x y z
3 2 4 0x y z
18
iii) 3 0x y z
0x y z
Exercises 2 : Show that the following systems of equations have only
trivial solution.
i) 2 3 0x y
0x y
ii) 4 5 0x y
6 7 0x y
iii) 3 4 2 0x y z
0x y z
3 5 0x y z
We observe that :
1. Proposition : For a homogeneous system of m linear equations in n
unknowns, the sum of two solutions and a scalar multiple of a solution
is again a solution of the same system.
2. Proposition : A necessary and sufficient condition for the sum of two
solutions or any scalar multiple of a solution to be the solution of the
same system of linear equations
1
, 1n
ij j ij
a x b i m
is that 0ib
for 1 i m .
System of Linear Equations :
General form of the system of linear equation is
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
...
...
...
n n
n n
m m mn n m
a x a x a x b
a x a x a x b
a x a x a x b
(A)
This is called as a system of m‟ equation in n-unknown variables.
Here ija ‟s and ib ‟s are real numbers and 1 2, , ... , nx x x are unknowns 'ija s
are called constants.
Examples
2 5 7 8 7
3 4 8
3 5 9 0
x y z w
y w
x z w
is a system of 4 variables. The above system (A) can be written as
19
1
, 1n
ij j ij
a x b i m
The system (A) can be written in the matrix form as
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m m mn n mm n
a a a x b
a a a x b
a a a x b
Again we can write this as AX B ; where A is a matrix of order m n
also it is called as matrix of coefficient. X is a column matrix of
variables and B is a column matrix of constant.
Theorem : If 1 2, , ..., nc c c c and 1 1 1 11 2, , ..., nc c c c
are solutions of
homogeneous system 0AX .
1
0, 1n
ij jj
a x i m
then
a) 1 2. , , ..., ,nc c c c is also a solution.
b) 1 1 1 1
1 1 2 2, , ..., n nc c c c c c c c
is also a solution.
Proof (a) : Since 1 2, , ..., nc c c c is the solution of the system therefore,
we have
1
0, 1n
ij jj
a c i m
(1)
Consider L.H.S. =
1
n
ij jj
a x
=
1
n
ij jj
a c
=
1
.n
ij jj
a c
= . 0 [from equation (1)]
= 0 = R.H.S.
1 2. , , ..., nc c c c is the solution of the given system.
b) Since 1 1 1 1
1 2 nc c ,c ,...,c
is the solution of the given system.
We have 1
1
0 1n
ij jj
c c , i m
(2)
Consider L.H.S. =
1
n
ij jj
a x
= 1
1
n
ij j jj
a c c
= 1
1 1
n n
ij j ij jj j
a c a c
= 0 [from (1) and (2)]
= R.H.S.
1 1 1 11 1 2 2, , ..., n nc c c c c c c c
is the solution of the system.
20
Theorem : Consider a system of m-equation in „ n ‟ unknowns i.e. a
system
1
0, 1n
ij jj
a x i m
. If m n , then the system has non-
trivial solution.
Proof : We prove the result by induction method on „ m ‟.
Let 1m , 1 n . The system of equation reduce to
11 1 12 2 1... 0n na x a x a x .
Let 1 0, 1pa p n .
Consider, 11 1 12 2 1 1... ... 0p p n na x a x a x a x (I)
Let 1 2 3 11, 0, 0, ..., 0px x x x
111
1
, 0, ..., 0p p np
ax x x
a
Substituting the above values of 1 2, , ..., nx x x in RHS of (I). We get,
L.H.S. of (I) which is equal to zero.
111 2
1
1, 0, ..., , .., 0p np
ax x x x
a is the non-trivial solution of (I).
for 1m the system has non-trivial solution.
result is true for 1m .
Let the result be true for m k .
i.e. the result is true for system of k number of equation. Then to prove
that result is true for 1m = k + and 1n > k + . Consider a system of
1k + equation.
11 1 12 2 1 0n na x a x ... a x 1E
21 1 22 2 2 0n na x a x ... a x 2E
21 1 1
1 20nk k k
na a x ... a x 1kE
In 1E , all the coefficients are not equal to zero.
Let 1 0pa for some p, 1 p r .
from 1E
11 1 12 2 1 1 1 1 1 1 1 1 0p p p p p p n na x a x ... a x a x a x ... a x
Since 1 0,pa we get
1 11311 121 2 3 1
1 1 1 1
pp p
p p p p
aaa ax x x x ... x ...
a a a a
1 1 11
1 1
p np n
p p
a ax ... x
a a
(II)
21
Substituting the values of px in remaining equation 2 3 1kE ,E ,..., E , we
get, the system k equations. In variables 1 2 1 1p p nx ,x ,...,x ,x ,...,x we
denote there equations by 1 1 12 3 kE , E ,.., E & 1
1kE .
This is the system of k–equations therefore, by induction it has non-trivial
solution.
Let 1 1 2 2 1 1 1 1p p p p n nx c , x c ,...,x c , x c ,...,x c be the non-
trivial solution of the above system.
Substituting the values of 1 2 1p nx ,x ,...,x ,..,x in (II), we get the values
px . Therefore, we get the values of 1 2 nx ,x ,...,x which are not all zeros.
Therefore the system of 1k equation has non-trivial solution.
Note :
i) If, m n , then system has non-trivial solution but the converse may
not be true. i.e. if the system has non-trivial then it does not mean
m n .
ii) If 1 2 nc c ,c ,...,c is the non-trivial solution of homogeneous system
than . ,c is also a solution.
If the system has non-trivial solution then it has infinitely many
solution.
Examples :
1) Find the solution set of the following problems.
a) 2 3 5 0x y z (i)
0x y z (ii)
Solution :
(ii) y x z
Substituting y, in (i), we get 2 3 5 0x x z z .
2x z
Again,
2 3
3
y x z z z z
y z
Solution set is x, y,z / x, y,z 2 3z, z, z z
2 3 1z , , z
22
EXERCISE:
i) 2 4 3 0x z y
3 0x z y
ii) 2 4 0x y z w
3 2 3 0x y z w
0x y z
iii) 2 3 0x y z w
2 4 4 0x z w
2 0x y z w
2 2 0x y z
iv) 7 2 5 0x y z w
0x y z
0x z w
2 0y z w
Example 2: 2 3 4 0x y z (1)
3 0x y z (2)
Solution : (2) 3 y x z
Substituting y in (1), we get 7 z x
Since 3 3 7 10y x z x x x
10 y x
Solution set is , , : , ,S x y z x y z ,10 ,7 :x x x x
1,10, 7 :x x
Example 3: 2 4 0x y z w … (1)
3 2 3 0x y z w … (2)
0x y z … (3)
Solution : (3) y x z
Substituting y x z in (2), we get 5 5 w x z
Substituting the values of y and w in (1), we get 4
3
x z
4 1 5, ,1,
3 3 3S z z
Example 4:
7 2 5 0x y z w (1)
23
0x y z (2)
0x z w (3)
2 0y z w (4)
(2) y x z (5)
By substituting on (1), we get
5 3 w x z (6)
(3) 2 z x 0 x
(5) 0y
(6) 0w
Solution set is , , , , , ,S x y z w x y z w
0 0 0 0S , , , .
1.4 EXERCISE :
Solve the following homogeneous systems by any method :
i) 1 2 32 3 0x x x
1 2
3 3
2 0
0
x x
x x
Ans : 1 2 3 1 2 3, , : , ,x x x x x x i.e. 0, 0, 0
ii) 1 2 3 43 0x x x x
1 2 3 45 0x x x x
Ans : 1 2 3 4, , 4 ,x s x t s x s x t
iii) 2 2 4 0x y z
3 0w y z
2 3 0w x y z
2 3 2 0w x y z
Ans : , , , 0w t x t y t z
iv) 2 3 0x y z
2 3 0x y z
4 0x y z
Ans : (Only the trivial solution)
v) 1 2 43 0x x x
1 2 3
2 3 4
1 2 3 4
4 2 0
2 2 0
2 0
x x x
x x x
x x x x
24
Ans : (Only trivial solution)
vi) 3 2 0w x
2 4 3 0
2 3 2 0
4 3 5 4 0
u w x
u w x
u w x
Ans : 7 5 , 6 4 , 2 , 2u s t s t w s x t
Systems of linear equations : ( points to be remembered )
1. Determine whether the equation 5 7 8 16x y yz is linear.
Solution : No, since the product yz of two unknowns is of second degree.
2. Is the equation log 5x y ez linear?
Solution : Yes, since , , log 5e are constants.
1. Property : Consider the degenerate linear equation:
1 20 0 0 nx x ... x b
i) If the equation‟s constant 0b , then the equation has no solution.
ii) If the constant 0b , every vector in n is a solution.
2. Property : Consider the linear equation ax b .
i) If 0a , then x b / a is the unique solution.
ii) If 0 0a , b , there is no solution.
iii) IF 0a , 0b , every scalar k is a solution.
Property – A system of linear equations, AX B has a solution of and
only of the rank of the coefficient matrix is equal to the rank of its
augmented matrix.
2
25
MATRICES AND MATRIX OPERATIONS
Unit structure
2.0 Objectives
2.1 Introduction
2.2 Multiplication of Matrices
2.3 Matrices and Linear Equations
2.4 Transpose of a Matrix
2.5 Diagonal
2.0 OBJECTIVES :
Rectangular arrays of real numbers arise in many contexts other
than as augmented matrix for systems of linear equations. In
this section we consider such arrays as objects in their own
right and develop some of their properties for use in our later
work.
2.1 INTRODUCTION :
Rectangular arrays of real numbers arise in many contexts other than as
augmented matrices for systems of linear equations. In this section we
consider such arrays as objects in their own right and develop some of
their properties for use in our later work.
In this section, we define elementary row and column operations
for a matrix A. They are also known by the common name of elementary
transformations of the matrix A.
Matrices : A matrix is a rectangular array of numbers, real or complex.
An mn matrix is a rectangular array of mn numbers arranged in m rows
and n columns. Thus,
11 12 1
21 22 2
1 2
n
n
m m mn
a a a
a a a
a a a
is an mn (real m by n) matrix, usually denoted by a single letter A. Each
of the mn numbers 11 12, , ...a a is called an element of the matrix. The
26
element ija occurs in the ith
row and jth
column. The matrix is denoted by
ijA a .
Transpose of a matrix : If ij m nA = a
, then transpose of A is the
n m matrix ji n mB b
where ji ijb a for 1 i m , 1 j n .
If
11 1
1
n
m mn
a a
A
a a
then
11 1
1
m
t
n mn
a a
A
a a
The element ija occurs in the ith
and jth
column. The matrix is denoted by
ijA a .
Symmetric Matrix :
An n n matrix A over is said to be symmetric if tA A .
ij jia a for 1 i n , 1 j n .
Skew – symmetric Matrix :
An nn matrix A over is said to be skew symmetric if tA A .
Exercise 1 : Let A be a nn matrix over . Show that
i) tA A is symmetric.
ii) tA A is skew – symmetric.
iii) A can be expressed as a sum of a symmetric and skew – symmetric
matrices.
Exercise 2 : If A and B are symmetric nn matrices over , show that
A B and 2A are symmetric matrices over .
2.2 MULTIPLICATION OF MATRICES :
Let ijA a be an mn matrix over and jkB b be an np
matrix. We define the product,
, 1 , 1ikAB c i m k p to be an mp matrix, where
1
n
ik ij jkj
c a b for 1 , 1i m k p .
27
Note 1 : The product AB may be defined but the product BA may not be
defined. However, if A, B are mn matrices over , AB and BA are both
defined. However they may not be equal.
For example, let 0 1
0 0A
, 0 0
1 0B
, then 1 0
0 0AB
,
0 0
0 1BA
, AB BA .
Note 2 : , m nA B M are said to be equal. If ij ija b for 1 i m ,
1 j n , where ,ij ijm n m nA a B b
. However, we have
associated law for multiplication of matrices.
If m n n p p lA M , B M , C M . Then
AB C A BC and we also have distributive law.
If m n m pA M , B,C M then A B C AB AC
proof are obvious.
We define nA inductively for nA M .
1.n nA A A , , 1n N n . Although, we introduce matrices as
a motivation for the study of linear equations. We are studying matrices
for their own sake. The matrices form is an important part of linear
algebra.
Identity Matrix :
1 0 0
0 1 0
0 0 1
nI
is called an identity nn matrix.
Kronecker’s delta :
δ , δ 1n ij ijI if i j
0 if i j
Upper triangular matrix :
ijA a is called an upper triangular matrix if 0ija for i j
11 12 1n
22 2n
nn
a a a
0 a aA
0 0 a
Lower triangular matrix :
ij n nA a
is called a lower triangular matrix if 0ija for j i .
28
11
21 22
1 2
0 0
0
n n nn
a
a aA
a a a
Diagonal matrix :
ij n nA a
is called a diagonal matrix if 0ija for i j
11
22
0 0
0 0
0 0 nn
a
aA
a
Scalar Matrix :
ij n nA a
is called a scalar matrix if diagonal is c for some c .
0 0
0 0
0 0
c
cA
c
Invertible Matrix :
ij n nA a
is said to be an invertible matrix if nB M R such that
nAB BA I .
An invertible matrix is also called a non-singular matrix. The
matrix B is called an inverse of A.
Theorem : If nA M is a invertible matrix, the inverse of A is
unique.
Proof : If nAB BA I (i)
nAC CA I (ii)
Then n nBA I ( BA)C I .C ( post multiplying by C )
C ( In =1 )
Thus B(AC) = C nBI C (from ii)
B C ( In =1 )
We shall denote the inverse of an invertible matrix A by 1A .
Proposition 1 : If m n n pA M , B M then t t tAB B A .
29
Proposition 2 : If A is an nn invertible matrix over , then TA is
invertible and 1 1AT A T
Proof : 1 1nAA I A A
1 1T TT
nAA I A A
1 1T T
T TnA .A I A A
TA is invertible and 1 1T
AT A
Proposition : If nA,B M are invertible matrices, then the product
AB is invertible and 1 1 1AB B A . More generally, if 1, ..., kA A are
invertible matrices then 1 1 11 2 1..., , ...,k kA A A A A .
Exercises 1.2 :
1) Let
1 1 1
0 1 1
0 0 1
A
find 2 3 4, ,A A A .
2) Let a,b and let 1
0 1
aA
, 1
0 1
bB
find AB , 2 3,A A .
Show that A is invertible and find 1A .
3) i) Find a 2 2 matrix A such that 2 1 0
0 1A I
can you find
all such matrices?
ii) Find a 2 2 non-zero matrix A such that 2 0A .
4) Let nA M of 3 0A show that I A is invertible.
5) Let nA M of 3 0A A I show that A is invertible.
6) Let nA, B, P M . If P is invertible and 1B P AP then show
that 1n nB P A P for n .
7) nA,B M of A, B are upper triangular matrices, find if AB is
upper triangular.
8) Let cosθ sinθ
θ , θsinθ cosθ
R
show that
1 2 1 2θ θ θ θR R R for 1 2θ , θ .
30
9) Let
1 2 3 4
0 2 3 4
0 0 3 4
0 0 0 4
A
then find 1A ?
10) Let
1 1
2 2
1 0
A
, 3 1
4 4B
Is there a matrix C such that
CA B ? Justify your answer.
2.3 MATRICES AND LINEAR EQUATIONS :
We now introduce an algorithm that can be used to solve a large system of
equation. This algorithm is called “Gaussian Elimination method.”
We write the system of equations as :
1
1n
ij j ij
a x b , i m
. (A)
i.e. 1
2ij m n
n
x
AX B, A a , X x
x
, 1
ij m n
m
b
B , A a
b
is called
the matrix of coefficients and the m n matrix.
11 1 1
1
,
n
m mn m
a a b
A B
a a b
the augmented matrix.
For example the system of linear equations:
2 3 3 1x y z
5 3x y z
is denoted by matrix notation as 2 3 3 1
1 5 1 3
x
y
z
. The
augmented matrix of the system is 2 3 3 : 1
1 5 1 : 3
.
Examples
2 5 7 8 7
3 4 8
3 5 9 0
x y z w
y w
x z w
is a system of 4 variables x y, z and w.
The system (A) can be written in the matrix form as
31
11 12 1 1 1
21 22 2 2 2
1 2
n
n
m m mn n mm n
a a a x b
a a a x b
a a a x b
Again this, we can write as AX B ; where A is a matrix of order
m n also it is called as matrix of coefficient. X is a column matrix of
variables and B is a column matrix of constant.
Summary :
1) The m n zero matrix denoted by m n0 or simply 0, is the matrix
whose elements are all zero. Find x, y, z, t.
If 3
04
x y z
y z w
Solution :
Set : 0 3 0
4 0 0
x y z
y z w
4 4 3 3x , y , z , w
2) Show that for any matrix A , we have A A
Solution : ij m nA a
ijm,n
a
ij m,na
3) Fin x, y, z, w if 6 4
31 2 3
x y x x y
z w w z w
Solution : 3 4x x
3 6y x y
3 1z z w
3 2 3w w
2, 4, 1, 3x y z w
Theorem : Let M be the collection of an m n matrices over a field K
of scalars. Then for any matrices ijA a , ijB b , ijC c in M and
any scalars 1 2 K , K K .
i) A B C A B C
ii) 0A A
iii) 0A A
iv) A B B A
v) 1 1 1K A B K A K B
32
vi) 1 2 1 2K K A K A K A
vii) 1 2 1 2K K A K K A
viii) 1.A A
Problem
1. Show that 1 A A
Answer : Consider
1 1 1A A A A
1 1 A
0 . A
0
Thus, 1 0A A
Or -A + A +(-1)A = 0 - A
Or 1 A A ( - A + A = 0)
2. Show that 2 3A A A, A A A A
2 1 1A A
1 1A A
A A
Thus 2 A A A
3 2 1A A
2 1A A
A A A
3 A A A A
Matrix Multiplication :
The product of row matrix and a column matrix with the same
number of elements is their inner product as defined as :
1
1 1 1 n n n
n
b
a , ..., a a b ... a b
b
1
n
k kk
a b
3
8 4 5 2 8 3 4 2 1
1
+ (5)(-1) = 24+ (-8) – 5 = 11
Example 1 : 1 3
2 1A
and
2 0 4
3 2 6B
33
Answer : 2 2A & 2 3B then AB is defined as 2 3 .
2 0 41 3
3 2 62 1AB
1 2 3 3 1 0 3 2 1 4 3 6
11 6 14
Now 2 0 41 3
3 2 62 1
AB
11 6 14
2 2 1 3 0 2 2 4 1 6
11 6 14
1 2 4AB
2. Find AB if
2 1
1 0
3 4
A
& 1 2 5
3 4 0B
.
Solution :
1 8 10
1 2 5
9 22 15
AB
3. Find 1 6
3 5
2
7
.
Solution : The first factor is 2 2 and the second 2 1 i.e.
1 6 2 40
3 5 7 41
4. Theorem – Suppose that A , B , C are matrices and K is scalar then
i) AB C A BC associative law
ii) A B C AB AC left distributive law
iii) B C A BA CA right distributive law
iv) K AB KA B A KB
5. Let 1 6 4 0
3 5 2 1
A , B
than 16 6
2 5AB
4 24
5 7BA
Here AB BA i.e. matrix multiplication does not obey the
commutative law.
6. Show that A B C D AC AD BC BD
Solution : Using distributive laws :
A B C D A B C A B D
AC BC AD BD AC AD BC BD
34
2.4 TRANSPOSE OF A MATRIX :
The transpose of a matrix A , denoted by AT is the matrix obtained by
writing the rows of A , in order, as columns.
If ij m,nA a , then A
T = (aij)
T , if
1 2 3
4 5 6A
then
1 4
2 5
3 6
TA
.
Theorem : The transpose operations on matrices satisfieds.
i) T T TA B A B
ii) T
TA A
iii) is scalarT TKA KA K
iv) T T TAB B . A
2.5 DIAGONAL :
The diagonal of ijA a consists of the elements 11 22, , ..., nna a a where
A is n-square matrix.
Trace of an n-square matrix ijA a is the sum of its diagonal elements
i.e. 11 22 nnntr A a a ... a
viz. trace of the matrix
1 2 3
4 5 6
7 8 9
A
is 1 5 9 15
Property : Suppose ijA a and ijB b are n-square matrices and k
is any scalar. Then
i) tr A B tra A tra B ,
ii) tr kA k.tr A ,
iii) tr A .B tr B . A
Identity matrix : In is the n-square matrix with 1s on the diagonal and 0s
elsewhere.
Examples :
35
2
1 0
0 1I
is the identity matrix of order 2.
3
1 0 0
0 1 0
0 0 1
I
is the identity matrix of order 3 etc.
Kronecker delta : Kronecker delta is defined as
0 if
1 if ij
i j
i j
Accordingly, ijI
Note: Trace of nI n
Scalar Matrix kD : kD K . I
Example : Find the scalar matrices of orders 2, 3 and 4 corresponding to
the scalar 5k .
Solution : Put 5s on the diagonal and 0s elsewhere
55 0 0
5 0 5 0 5 0
0 5 50 0 5
5
, ,
Powers of Matrices : The non-negative integral powers of a square
matrix M may be defined recursively by
0 1 1 1 2r rM I , M M , M MM r , ,...
Property :
i) p q p qA A A
ii) If A and B are commutative then pA and qB are also commutative.
Note : If 1 2
0 1A
and 1
0 1k
KS
then 2K K KAS S A S .
Note : 1 2
0 1
n nA
Idempotent : A matrix E is idempotent if 2E E .
Note : i) identity matrix is idempotent 2I I
ii) zero matrix is idempotent 20 0
Example : Given matrix
2 2 4
1 3 4
1 2 3
E
is idempotent.
Nilpotent : A is a nilpotent of class p if pA 0 but 1 0pA .
36
Example – Show that
1 1 3
5 2 6
2 1 3
A
is nilpotent of class 3.
Answer : 2
0 0 0
3 3 9
1 1 3
A
3 2. 0A A A
Inverse Matrix : A square matrix A is invertible if there exists a square
matrix B such that AB BA I ; where I is identity matrix.
Note : 1
1A A
Example 1 : Show that 2 5
1 3A
and 3 5
1 2B
are inverses.
Answer : 1 0
0 1AB I
1 0
0 1BA I
Example 2 : When is the general 2 2 matrix a b
Ac d
invertible?
What then is its inverse?
Answer : Take scalars x, y, z, t such that
1 0
0 1
a b x y
c d z t
or
1 0
0 1
ax bz ay bt
cx dz cy dt
,
1ax bz 0ay bt
0cx dz 1cy dt
Both of which have coefficient matrix A . Set A ad bc . We know
that A is invertible if A 0 . In that case first and second system have
the unique solutions.
x d A , z c A ,
y b A , t a A .
Thus 1 1d A b A d bA
c A a A c aA
In other words, when A 0 the inverse of 2 2 matrix A is obtained by
i) interchanging the elements or the main diagonal,
ii) taking the negative of the other elements, and
37
iii) multiplying the matrix by 1
A
Example 1 : Find the inverse of 3 5
2 3A
Answer : 1 0A , 1A exists. Next, interchanging the diagonal
elements, take the negative of the other elements and multiply by 1
A.
1 3 5 3 51
2 3 2 3A
Example 2 : Find the inverse of 5 3
4 2A
Answer : 12 A , A exists.
1 2 3 1 3 21
4 5 2 5 22A
Example 3 : Find the inverse of 2 3
1 3A
1 3 3 1 3 1 31
1 2 1 9 2 99A
Definition : A square matrix „ A ‟ of order „n‟ A 0 . Inverse of A is
denoted and defined by 1 Adjoint of AA
A
Transpose of TA A
Adjoint of A = Transpose of co-factor of A
Co-factor of A = 1 A i j
Example :
i) Find inverse of the matrix 1 1
2 1A
Solution : 3 0A
1 1 3 1 3
2 3 1 3A
ii) If 3 1
2 2A
1 1 4 1 8
1 4 3 8A
38
iii) If
1 5 2
2 0 1
3 1 2
A
0A A is not invertible.
iv) If
1 5 2
2 1 1
3 1 2
A
Answer : 18 0 A , A exists. Take B = matrix of cofactors called
as Adjoint of A.
1 1 1 2 1 3
2 1 2 2 2 3
3 1 3 2 3 3
1 1 1 1 1 1
1 8 1 8 1 16
1 3 1 5 1 11
B
1 1 1
8 8 16
3 5 11
1 8 3
1 8 5
1 16 11
TB
1
1 8 31
1 8 58
1 16 11
TBA
A
1 8 1 3 8
1 8 1 5 8
1 8 2 11 8
v) Find whether matrix is invertible and find its inverse.
3 1 4
2 0 1
1 1 1
A
Solution : 4 0A . 1A exists.
1 1 2
3 1 2
1 5 2
B
1 3 1
1 1 5
2 2 2
TB
1
1 3 11
1 1 54
2 2 2
A
39
40
3 ELEMENTARY ROW OPERATIONS AND
GAUSS-ELIMINATION METHOD, ROW–
ECHELON FORM
Unit structure:
3.0 Objectives
3.1 Elementary row operations
3.2 Gauss Elimination Method to Solve AX=B
3.3 Matrix units and Elementary Matrices
3.4 Elementary Matrices
3.5 Linear Algebra System of Linear Equations
3.0 OBJECTIVES :
Object of this section is to develop a simple algorithm for
finding the inverse of an invertible matrix. In this section, we
give a systematic procedure for solving of systems of linear
equations; it is based on the idea of reducing the augmented
matrix to a form that is simple enough so that the system of
equations can be solved by inspection.
3.1 ELEMENTARY ROW OPERATIONS:
1E : Interchange the ith
row and jth
row : i jR R .
2E : Multiply the ith
row by a non-zero scalar K; i iR KR , 0K .
3E : Replace the ith
row by K time the jth
row plus the ith
row:
j iiR KR R
Explanation :
a) Interchanging the same two rows we obtain the original matrix that is
this operation is its own inverse
b) Multiply the ith
row by K and then by 1K , or by 1K and then by
K , we obtain the original matrix. In other words, the operations
i iR KR and 1i iR K R are inverses.
41
c) Applying the operations i j iR KR R and then operation
i j iR KR R ,
Elementary Matrices :
Let E be the matrix obtained by applying an elementary row operation to
the identity matrix I, i.e. let E e i . Then E is called the elementary
matrix corresponding to the row operations.
Column Operations, Matrix equivalence :
The elementary column operations.
1F : Interchange the ith
column and the jth
column : i jC C
2F : Multiply the ith
column by a non-zero scalar K : 0i iKC C K
3F : Replace the ith
column by K times the jth
column plus the ith
column
i j iC KC C
We first define elementary row operations on an mn matrix over .
The following operations on A n are called elementary row
operations.
i) Interchanging thi and thj row of A denoted by i jR R .
ii) Multiplying the thi row by a non-zero scalar λ (denoted by
i iR λR )
iii) Adding a scalar multiple of ith
row to thj row of A denoted by
j iR λR + RJ.
Two mn matrices are said to be row equivalent if one can be obtained
from the other by a succession of elementary row operations.
Consider the system of linear equations.
AX B , ij m nA a
1
2
m
b
bB
b
If the augmented matrix A,b is row equivalent to 1 1,A B , then the
solutions of the system AX b are same as the solutions of the system 1 1A X B .
Thus, to obtain the solutions of the system AX B , we try to reduce the
matrix A, B to a simple form. Again, our purpose is to eliminate
unknowns. We define a row-echelon matrix :
42
Definition : An mn matrix ijA a is called a row echelon matrix if
0A or if there exists an integer r. 1 min ,r m n and integers
1 1 2K K K r m s.t.
i) For each , 1 , 0iji i r a for j K i
ii) For each , 1 , 0iki i r a i , ika i is called the pivot element of
ith
row.
iii) For each , 1 , 0ski i r a i for s i .
iv) 0ija for all i r and for all j .
Note : Sometimes we consider the pivot elements are 1.
For example,
0 2 1 0 5 6
0 0 0 5 2 1
0 0 0 0 4 2
0 0 0 0 0 0
A
1 2, 2 5, 4, 3K K K r .
Theorem : Every matrix A, A M is row equivalent to a matrix in
row echelon form.
Example : Reduce the matrix
0 1 3 2
2 1 4 3
2 3 2 1
to row echelon form.
Answer :
0 1 3 2
2 1 4 3
2 3 2 1
A
1 2
2 1 4 3
0 1 3 2
2 3 2 1
R R
(bringing the left most non-zero
entry to 1st row)
23 3 1
2 1 4 3
0 1 3 2
0 2 6 4
R R R
(making entries below 1st pivot 0)
43
23 3 2
2 1 4 3
0 1 3 2
0 0 0 0
R R R
(making entries below 2nd
pivot 0)
Exercise : Reduce the following matrices to row echelon form :
i)
6 3 4
4 1 6
1 2 5
ii)
1 0 2
2 1 3
4 1 8
iii)
1 2 3 1
2 1 2 2
3 1 2 3
iv)
1 2 1 2 1
2 4 1 2 3
3 6 2 6 5
Note : In the system of equations AX B , where ij m nA a
and
1
m
x
X
x
, the elements in the thj column are coefficients of the
unknown jx (or the thj unknown) in each equation of the system.
We have already seen that a system of non-homogeneous linear
equations may not have a solution. However, when the system of linear
equations has a solution, we find the solution by reducing the augmented
matrix to row echelon form. We assign arbitrary values to unknowns
corresponding to non-pivot elements (if any). The equations are then
solved by the method of back-substitution.
3.2 GAUSS ELIMINATION METHOD TO SOLVE
AX B :
Algorithm : for augmented matrix A, B .
Step I : If the matrix consists entirely of zeros. STOP. It is in row-
echelon form.
Step II : Find the first column from the left containing non-zero entry.
If this is in ith
row, not is 1st row, perform 1 iR R .
Step III : Multiply that row by suitable numbers and subtract from
rows below it to make all entries below it 0.
44
Step IV : Repeat steps 1 – 3 on the matrix consisting of remaining
rows. The process stops when either no row remains at step
4 or all rows consists of zeros.
Call the leading non-zero entries of each row pivots. If there are
columns corresponding to non-pivot elements , we assign arbitrary values
to unknowns corresponding to them and we solve the system by back
substitution. The method breaks down when the pivots appear in last
column.
Exercises 1.1 :
i) Solve the given system of equations by Gauss-Elimination method.
1 2 3 4 52 3 7 5 2 2x x x x x
1 2 3 4 52 4 3 2x x x x x
1 3 4 52 4 2 7x x x x
ii) 1 2 3 42 2 1x x x x
1 2 3 4 2x x x x
1 2 3 47 5 3x x x x
iii) 1 2 3 42 3x x x x
1 2 3 4 2x x x x
1 2 3 42 3 9x x x x
1 2 3 42 2x x x x
iv) 1 2 3 42 4x x x x
2 3 43 4 2x x x
1 2 3 42 3 5 0x x x x
1 2 3 45 6 3x x x x
v) 1 2 3 43 8 3 14 1x x x x
1 2 3 42 3 2 2x x x x
1 2 3 42 10 3x x x x
1 2 3 45 2 12 1x x x x
vi) 5 2 142x y z
45
3 30
2 3 5
x y z
x y z
Ans : 39.2, 16.7, 18.97x y z
vii) 10 7 3 5 6x y z u
6 8 4 5x y z u
3 4 11 2x y z u
5 9 2 4 7x y z u
Ans : 1, 7, 4, 5u z y x
2. Problems : Which of the following matrices are in row echelon form.
i)
1 1 2
0 0 0
0 0 1
ii) 2 1 1 3
0 0 0 0
iii) 1 1
0 1
iv)
1 0 0 3 1
0 0 0 1 1
0 0 0 0 1
v)
0 0 1
0 0 1
0 0 1
3. Problems : Reduce the following matrices to row echelon form by a
sequence of elementary row operations.
i)
0 1 2 1 2 1 1
0 1 2 2 7 2 4
0 2 4 3 7 1 0
0 3 6 1 6 4 1
ii)
1 5 2 1 4 0
3 0 4 6 2 1
1 2 1 2 3 1
46
4. Theory : Show that any m n matrix over can be reduced to a
matrix in row echelon form by performing a finite number of
elementary row operations.
3.3 MATRIX UNITS AND ELEMENTARY MATRICES :
Let rsI denote an mn matrix which has entry 1 in the th
r, s
place and 0 elsewhere.
0 0 0
0 1 0
0 0 0
rsI
rsI are called matrix units. Let ij m nA a
. Then
1
1
1
1
0 0 0
0 1 0
0 0 0
n
r rn
rs
s sn
m mn
a a
a a
I A
a a
a a
r, s
1
0 0
0 0
s sna a
rth
row
rrI A A
0rs rs rr rr rrI .I , I .I I if r s
47
1
1
0 0
0
s sn
rs sr
rnr
a a
I I A
aa
rth
row
3.4 ELEMENTARY MATRICES :
We recall that there are three types of elementary row operations
for matrices.
i) Interchanging two rows.
ii) Multiplying a row by a non-zero scalars.
iii) Adding a scalar multiple of a row to another row.
The matrix E obtained by performing any of the elementary row
operations on the identity matrix is called an elementary matrix.
We shall denote –
i) The matrix obtained by exchanging ith
row and jth
row of identity
matrix by ijE .
ii) The matrix obtained by multiplying ith
row of identity matrix by
0λ by iE λ .
iii) The matrix obtained by adding λ times j th
row to i th
of identity
matrix by Eij λ
Theorem : Let A be an mn real matrix. Then,
i) ijE A is the matrix obtained by exchanging ith
row and jth
row of A.
ii) iE λ A is the matrix obtained by multiplying ith
row of A by
0λ λ .
iii) ijE λ A is the matrix obtained by adding λ times jth
row to ith
row of
A .
Thus, multiplying by an elementary matrix is same as performing
the corresponding elementary matrix.
1. Proposition : An elementary matrix is invertible.
sth
row
48
Proposition : Let A be an nn real matrix and B is row equivalent to A ,
then A is invertible if and only if B is invertible.
Proof : A is row equivalent to B .
B is obtained from A by a finite sequence of elementary row
operations.
1 2B E E Ek; where E1E2… Ek are elementary matrices. If A is
invertible, B is product of invertible matrices and is invertible (Each
elementary matrix is invertible). If B is invertible, then
1 1 1
1 2 1k kA E E ...E B E ...E B .
A is invertible.
Theorem : An nn matrix A is invertible if and only if A is row
equivalent to identity matrix. (Any upper triangular matrix with non-zero
diagonal elements is invertible).
Proof : If A is row equivalent to identity matrix clearly A is invertible,
conversely, suppose A is invertible, then the row echelon form of A does
not have any row consisting of zeros and A is row equivalent to
11 12 1
22 20
0 0
n
n
nn
b b b
b bB
b
0iib for 1 i n
Then multiplying ith
row by 1iib , we get a matrix C (row
equivalent to A),
12 1
2
1
0 1
0 0 1
n
n
c c
cC
i in nR C R gives
121 0
0 1 0
0 0 0 1
c
Similarly,
49
1 1i in nR C R makes 1th
n column
0
1
0
and so on and finally we get
identity matrix. Thus, an invertible matrix is row equivalent to identity
matrix. Further 1k nE ...E A I .
i.e. 1 1 1 11 1n k kA I E ... E E ... E which is a product of elementary
matrices.
Exercises 1.2 :
1. For each of the following elementary matrices describe the
corresponding elementary row operation and write the inverse.
i)
1 0 3
0 1 0
0 0 1
E
ii)
0 0 1
0 1 0
1 0 0
iii)
1 0 0
2 1 0
0 0 1
E
2. In each of the following, find an elementary matrix E such that
B EA .
i) 2 1
3 1A
,
2 1
1 2B
ii) 1 2
0 1A
, 1 2
0 1B
iii) 2 1
1 3A
,
1 3
2 1B
3. Show that an invertible 2 2 matrix can be expressed as a product of
at most 4 elementary matrices.
4. State which of the following are elementary matrices
i) 1 0
5 1
ii) 1 0
0 3
50
iii) 5 1
1 0
iv)
0 1 0
0 0 1
1 0 0
v)
1 1 0
0 0 1
0 0 0
vi)
2 0 0 2
0 1 0 0
0 0 1 0
0 0 0 1
5. Express A as a product of elementary matrices
i)
2 0 0 0
0 4 0 0
0 0 5 0
0 0 0 2
A
ii)
0 0 0 1
0 0 5 0
0 2 0 0
0 0 0 1
A
6. Let 1 0
5 2A
, find elementary matrices 1 2E , E such that
2 1E E A I .
Gauss – Elimination Method :
1) Solve the given system by Gauss – Elimination Method.
1 2 3 4 5
1 2 3 4 5
1 3 4 5
1 2 3 4 5
2 3 7 5 2 2
2 4 3 2
2 4 2 3
5 7 6 2 7
x x x x x
x x x x x
x x x x
x x x x x
Answer : The augmented matrix corresponding to the system is
2 3 7 5 2 2
1 2 4 3 1 2
2 0 4 2 1 3
1 5 7 6 2 7
2 2 1
3 3 2
4 4 1
1
2
1
2
R R R
R R R
R R R
2 3 7 5 2 2
1 1 10 0 1
2 2 2
0 3 3 3 1 5
7 7 70 1 6
2 2 2
51
3 3 2
4 4 2
6
7
R R R
R R R
2 3 7 5 2 2
1 1 10 0 1
2 2 2
0 0 0 0 1 1
0 0 0 0 0 1
4 4 3R R R
2 3 7 5 2 2
1 1 10 0 1
2 2 2
0 0 0 0 1 1
0 0 0 0 0 0
STOP
The unknown corresponding to columns without pivots are 3 4x , x
. The solution set is
12 2 1 1 1 0 2 0s t , s t, s,t, t , , , , s,t
i.e. 1 2 0 0 1 2 1 1 0 1 0S , , , , S , t , , , , s,t
3.5 LINEAR ALGEBRA SYSTEM OF LINEAR
EQUATIONS :
General form of the system of linear equation is
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x ... a x b
a x a x ... a x b
a x a x ... a x b
(A)
This is called as a system of m‟ equation in n-unknown variables.
Here ija ‟s and ib ‟s are real number and 1 2 nx , x ,... ,x are unknowns aij‟s
are called constants.
Row Echelon Form : A matrix in row echelon form has zeros below each
leading 1.
Example :
52
i)
1 4 3 7
0 1 6 2
0 0 1 5
ii)
1 1 0
0 1 0
0 0 0
iii)
0 1 2 6 0
0 0 1 1 0
0 0 0 0 1
iv)
1 1 3 4
0 1 3 4
0 0 1 5
0 0 0 2
Reduced row echelon form : A matrix in reduced row echelon form has
zeros below and above each leading 1.
Examples :
i)
1 0 0 4
0 1 0 7
0 0 1 1
ii)
1 0 0
0 1 0
0 0 1
iii)
0 1 2 0 1
0 0 0 1 3
0 0 0 0 0
0 0 0 0 0
iv) 0 0
0 0
v)
1 0 2 3
0 1 1 2
0 0 0 0
0 0 0 0
vi)
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
Example : Suppose that the augmented matrix for a system of linear
equations has been reduced by row operations to the given reduced row
echelon form. Solve the system.
a)
1 0 0 5
0 1 0 2
0 0 1 4
b)
1 0 0 4 1
0 1 0 2 6
0 0 1 3 2
53
c)
1 6 0 0 4 2
0 0 1 0 3 1
0 0 0 1 5 2
0 0 0 0 0 0
d)
1 0 0 0
0 1 2 0
0 0 0 1
Solution :
(a) The corresponding system of equations is
1
2
3
5
2
4
x
x
x
By inspection
(b) The corresponding system of equation is
1 4
2 4
3 4
4 1
2 6
3 2
x x
x x
x x
Since 1 2,x x and 3x correspond to leading 1‟s in the augmented
matrix, we call them leading variables. Solving for the leading variables
in terms of 4x gives
1 4
2 4
3 4
1 4
6 2
2 3
x x
x x
x x
Since 4x can be assigned an arbitrary value, say t, we have
infinitely many solutions.
The solution set is given by the formulas.
1 2
3 4
1 4 , 6 2 ,
2 3 ,
x t x t
x t x t
(c) 1 5 22 4 6x x x , 3 51 3x x , 4 52 5x x
1 2 3 4 52 4 6 , , 1 3 , 2 5 ,x t s x s x t x t x t
(d) 1 2 30 0 0 1x x x , therefore there is no solution to the system.
Elementary matrices :
54
A matrix obtained from a unit matrix I by means of one elementary
operation on I is called an elementary matrix.
For example, Let I be a unit matrix of order 4.
i.e.
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
I
then each of the following is an elementary matrix obtained from I by the
elementary row operations indicated :
i)
0 0 1 0
0 1 0 0
1 0 0 0
0 0 0 1
, by interchanges of 1R and 3R .
ii)
1 0 0 0
0 1 0 0
0 0 5 0
0 0 0 1
, by 35R .
iii)
1 0 0 0
0 1 3 0
0 0 1 0
0 0 0 1
, by 2 33R R .
Exercises 1.3 :
Find rank and inverses of the matrices if exists :
55
i)
1 4 3
1 2 0
2 2 3
A
Ans : 1A does not exists.
ii)
1 2 3
2 5 3
1 0 8
A
, 1
40 16 9
13 5 3
5 2 1
A
iii)
1 6 4
2 4 1
1 2 5
A
, 1A does not exists.
iv)
1 1 1
5 5 5
1 1 4
5 5 5
2 1 1
5 10 10
A
, 1
1 0 2
3 1 2
1 1 0
A
v)
cosθ sin θ 0
sinθ cosθ 0
0 0 1
A
, 1
cosθ sin θ 0
sinθ cosθ 0
0 0 1
A
Inverse of a matrix A :
Let
11 12 13
21 22 23
31 32 33
a a a
A a a a
a a a
be an 33 matrix. Then inverse of A, denoted
and defined by
1 Adjoint of AdjA AA
A A
Adj Cofactor ofT
A A ; T indicates transpose
56
Co-factor of A = Co-factor of each element of A
Co-factor of 22
11 1a
i ja
Some corrections on pages nos: 2, 9, 10, 17, 24, 30, 38, 41, 42, 43, 49,
57
57
4
VECTOR SPACES AND SUBSPACES Unit Structure:
4.0 Objectives
4.1 Introduction
4.1.1 Addition of vectors
4.1.2 Scalar multiplication
4.2 Vector spaces
4.3 Subspace of a vector space
4.4 Summary
4.5 Unit End Exercise
4.0 OBJECTIVES
This unit would make you to understand the following concepts :
Vectors and Scalars in plane and space
Addition and scalar multiplication of vectors
Various properties regarding addition and scalar multiplication of
vectors
Idea of vector space
Definition of vector space
Various examples of vector space
Definition of subspace
Examples of subspace
Results related to union and intersection of subspace
Linear Span
4.1 INTRODUCTION
The significant purpose of this unit is to study a vector space. A
vector space is a collection of vectors that satisfies a set of conditions.
We‟ll look at many of the important ideas that come with vector spaces
once we get the general definition of a vector and a vector space.
So, first we are going to revise the concept of „normal‟ vectors,
vector arithmetic and their basic properties. We will see the vectors in
plane (space 2 ) and space (space 3 ).
Our final aim is first to define vector space using the concept of
vectors and scalars and study various examples then to study some special
subsets of vector space known as subspaces with examples and their
properties.
58
Vectors : A vector can be represented geometrically by a directed line
segment that starts at a point A, called the initial point, and ends at a point
B, called the terminal point. Below is an example of a vector in the plane.
Remark : 1) Vectors are denoted with a boldface lower case letter. For
instance we could represent the vector above by v, w, a, b, etc. Also when
we‟ve explicitly given the initial and terminal points we will often
represent the vector as, v = AB
2) Vector has magnitude and direction.
3) In plane i.e. in IR2 we write vector v with initial point at origin and
terminal point at (x, y) as v = (x, y) and a vector w with initial point at (x1,
y1) and terminal point at (x2, y2) as w = (x2 – x1, y2 – y1). Similarly we
can express vectors in IR3 (space).
4) Zero vector is a vector having zero magnitude. – v is negative of vector
v having same magnitude as that of v but opposite direction.
5) Scalars do not have direction. All real numbers are scalars.
Now we quickly discuss arithmetic of vectors.
4.1.1 Addition of vectors :
The addition of the vectors v and w is shown in the following
diagram.
u v -v
u + v u -v v
u
Suppose that we have two vectors v and w then the difference of w
from v, denoted by v - w is defined to be, v - w=v + (-w).
4.1.2 Scalar multiplication:
Suppose that v is a vector and c is a non-zero scalar (i.e. c is a real
number) then the scalar multiple, cv, is the vector whose length is c times
the length of v and is in the direction of v if c is positive and in the
opposite direction of v if c is negative.
59
-2v
v 2v
4.1.3 Properties of vector addition and scalar multiplication:
If u, v, and w are vectors in 2 or 3 and c and k are scalars then,
(1) u + v = v + u
(2) u + (v + w) = (u + v)+w
(3) u + 0 = 0+u = u
(4) u-u = u + (-u) = 0
(5) 1u = u
(6) (ck)u = c(ku) = k(cu)
(7) (c + k)u = cu + ku
(8) c(u + v) = cu + cv
4. 2 VECTOR SPACES
So far we have looked at vectors in 2 and 3 as directed line
segments. We have also seen two major operations addition and scalar
multiplication defined on set of vectors having certain properties.
Now we will generalize this concept to any set (not necessarily set
of traditional vectors) with two operations satisfying all the conditions
which are satisfied by addition and scalar multiplication in normal vectors.
So we can call members of this set also as vectors and set as vector space.
Let us see the definition of vector space
Vector Space: Let V be a set on which addition and scalar multiplication
are defined (this means that if u and v are objects in V and c is a scalar
(real number) then we‟ve defined u+v and cu in some way).
If the following axioms are true for all objects u, v, and w in V and
all scalars c and k then V is called a vector space and the objects in V are
called vectors.
1) u + v is in V. This is called closed under addition.
2) cu is in V. This is called closed under scalar multiplication.
3) u + v = v + u (commutative)
4) u + (v + w) = (u + v) + w (associative)
5) There is a special element in V, denoted 0 and called the zero
vector, such that for all u in V we have u + 0 = 0 + u = u.
6) For every u in V there is another element v in V such that u + v =
v + u = 0.
60
7) c(u + v) = cu + cv
8) (c + k)u = cu + ku
9) c(ku) = (ck )u
10) 1u = u
Note 1) V can be any set. It need not be 2 or 3 .
2) Addition and scalar multiplication defined above can be any two
operations not necessarily the usual or standard addition and scalar
multiplication. How strange they may be, they must satisfy above axioms.
3) Hence a vector in vector space now, is a much more general
concept and it doesn‟t necessarily have to represent a directed line
segment in 2 or 3 .
4) The scalars actually can be any numbers, even complex numbers.
But here we specify that scalars must be real numbers only. We call V as a
vector space over . Whenever a vector space is mensioned it is a vector
space over .
Let us study some simple results.
Result 1: Let V be a vector space then (a) Additive identity is unique.
(b) The additive inverse of a vector is unique. (c) For vectors u, v, w in V
u + w = v + w u = v and w + u = w + v u = v i.e. Cancellation laws
hold
Proof: (a) We know that „0‟ is additive identity in V.
To show that it is the only identity in V.
Suppose „01‟ is another additive identity in V.
0 + 01 = 01 + 0 = 0 ( 01 is additive identity)
01 + 0 = 0 + 01 = 01 ( 0 is additive identity)
0 = 01
Additive identity is unique.
(b) Suppose u1 and u2 are two additive inverses of vector u in V.
u + u1 = u1 + u = 0 …..(*)
and u + u2 = u2 + u = 0 …..(**)
Now u1 = u1 + 0 (0 is additive identity)
= u1 + (u + u2) (from (**) )
= (u1 + u) + u2 (Associative property)
= 0 + u2 (from (*) )
= u2 (0 is additive identity)
61
Remark : The unique additive inverse of u is denoted by –u and known as
negative of u.
(c) u + w = v + w
(u + w) + (-w) = (v + w) + (-w)
u + (w + (-w)) = v + (w + (-w) ) (Associative property)
u + 0 = v + 0 ( -w is the additive inverse of w)
u = v (0 is the additive identity)
Hence u + w = v + w u = v Similarly we can prove that w + u = w + v u = v
Result : Suppose that V is a vector space, u is a vector in V and c is any
scalar. Then,
(a) 0u = 0
(b) c0 = 0
(c) (-1)u = -u
(d) If cu = 0 then either c = 0 or u = 0
Proof: (a) 0u is a vector. Hence – 0u is also a vector
Consider (0 + 0)u = 0u + 0u (from axiom 8) 0u = 0u + 0u
0u + (-0u) = 0u + 0u + (-0u)
b = 0u + 0
0 = 0u.
(b) Consider c (0 + 0) = c0 + c0 (from axiom 7)
c0 = c0 + c0 c0 + (-c0) = c0 + c0 + (-c0)
0 = c0 + 0 0 = c0.
(c) (-1)u = - (1u) (from axiom 9)
= -u (from axiom 10)
(d) Consider cu = 0, if c = 0 then we are through.
If c 0 then to show that u = b
c is a real number and c 0 1/c is a real number (1/c)(cu) = (1/c) (0)
((1/c)c) u = 0 1u = 0 u = 0.
62
In the following illustrations we will see that a vector can be a
matrix or a function or a sequence.
Example : Let V = n , where n is a natural number.
n = { (x1, x2, …..,xn)/ x1, x2, … ,xn are real numbers }
Define addition and scalar multiplication as follows
(x1, x2, …..,xn) + (y1, y2, …..,yn) = (x1 + y1, x2 + y2, …..,xn + yn)
c(x1, x2, …..,xn) = (cx1, cx2, …..,cxn), where c is a real number.
Since RHS of above addition and scalar multiplication are
members of n therfore these two operations are closed in n .
Clearly we can see that addition defined above is commutative and
associative. Further (x1, x2, …..,xn) + (0, 0,……,0) = (x1, x2, …..,xn) for all
(x1, x2, …..,xn) in n and (x1, x2, …..,xn) + (-x1, -x2, …..,-xn) =
(0, 0,……,0).
Hence first six axioms for vector space are satisfied, remaining 4
axioms can be easily verified. Hence n is a vector space. It is known as
the Euclidean Space. Members of n are vectors of this vector space.
Example 2: Let V = M2×3 = Set of all matrices of order 2×3.
So, V = {
654
321
aaa
aaa
/ 1a , 2a ,…, 6a are real numbers}. Addition and
scalar multiplication in V is usual addition and scalar multiplication of
matrices.
i.e.
654
321
aaa
aaa
+
654
321
bbb
bbb
=
665544
332211
bababa
bababa
and
654
321
aaa
aaa
=
654
321
aaa
aaa
Here the vectors are matrices.
Zero vector is
000
000
Negative of
654
321
aaa
aaa
is
654
321
aaa
aaa
.
Oher axioms can be easily verified. Hence V is a vector space.
Example 3: Let V = = Set of all real sequences
63
= {)( nx / nx are real numbers for all natural numbers n }
Define addition as )( nx + )( ny = )( nn yx and
scalar multiplication as )( nx = )( nx .
Under these operations V is a vector space. Vectors are sequences. Zero
vector is a zero sequence )( nx , where nx = 0 for all n and negative of
)( nx is )( nx .
Example 4: Let V = [x] = Set of all polynomials with coefficient from
= { )(xp = n
n xaxaxaa ......2
210 / naaaa ......,,, 210 are real
numbers}
Define addition and scalar multiplication as follows:
If )(xq = n
n xbxbxbb ......2
210 then
)(xp + )(xq = n
nn xbaxbaxbaba )(......)()( 2
221100 and
)(xp = n
n xaxaxaa ......2
210
V is a vector space. Polynomials are vectors. Zero vector is a zero
polynomial with all coefficient equal to zero. Negative of
)(xp = n
n xaxaxaa ......2
210 is n
n xaxaxaa ......2
210 .
Example 5: Let X be a non empty set. Let V = Set of all functions from X
to .
If f, g V then f + g and f are also functions from X to defined by
(f + g)(x) = f(x) + g(x) and ( f)(x) = f(x) for all x in X.
Thus addition and scalar multiplication are closed in V.
V is a vector space. Here functions are vectors. Zero vector is „o‟ function
defined by o(x) = 0 for all x in X. Negative of f in V is –f defined by
(-f)(x) = - f(x) for all x in X.
Example 6: Let V and W be vector spaces.
V × W = {(v, w)/ v V and w W}
For (v1, w1), (v2, w2) V × W define addition as follows
(v1, w1) (v2, w2) = (v1 + v2, w1 +‟ w2)
Where + and +‟ are addition in V and W respectively.
For and (v, w) V × W define scalar multiplication as follows
(v, w) = ( v, w)
Where and are scalar multiplication in V and W respectively.
64
Since V and W are vector spaces + and +‟ are closed in V and W
is closed in V × W.
Also and satisfy all the axioms of scalar multiplication in V and W
respectively
satisfy all the axioms of scalar multiplication in V × W.
Suppose 0 and 0‟ are additive identities in V and W respectively.
Then (v, w) (0, 0‟) = (v + 0, w +‟ 0‟) = (v, w) for all (v, w) V × W
(0, 0‟) is the additive identity in V × W.
If –v and –w are additive inverses of v and w in V and W respectively,
then
(v, w) (-v, -w) = (v + (-v), w +‟ (-w)) = (0, 0‟) (-v, -w) is the additive inverse of (v, w)
Hence V × W is a vector space.
Check your progress
1) 2 = {(x1, x2)/ x1, x2 are real numbers}
(x1, x2) + (y1, y2) = (x1 + y1, x2 + y2) and c(x1, x2) = (cx1, cx2)
Show that 2 is a vector space. (Verify all required axioms)
2) M2 2 = {
43
21
aa
aa/ 1a ,
2a , 3a and 4a are real numbers}
Show that M2 2 is a vector space under usual addition and scalar
multiplication of matrices.
3) If c is a real number and u and v are vectors in vector space V,
show that
c(u – v ) = cu – cv. (Use definition of u – v and axiom no 7)
4. 3 SUBSPACE OF A VECTOR SPACE
We have learned about vector space. We know that any set can
behave as a set of vectors if it satisfies certain axioms. Now the question is
„if we take a subset of a vector space then is it a set of vectors? i.e. Is it a
vector space?
Let us study following example :
We have seen that 2 is a vector space (ex.1 in check your progress).
Consider a subset H = { (x, 1) / x is a real number} of 2 .
(x, 1) + (y, 1) = (x + y, 2) H
65
Addition is not closed in H.
H is not a vector space.
Hence a subset of a vector space is not necessarily a vector space.
So, which subsets are?
Let us see the definition of a subspace of a vector space.
Subspace: Let W be a subset of a vector space V then W is a subspace of
V if and only if W is itself a vector space with respect to addition and
scalar multiplication defined in V.
Now, technically if we want to show that a subset W of a vector
space V is a subspace we need to show that all 10 of the axioms from the
definition of a vector space are valid, however, in reality that need to be
done. Many of the axioms (3, 4, 7, 8, 9, and 10) deal with how addition
and scalar multiplication work, but W is inheriting the definition of
addition and scalar multiplication from V. Therefore, since elements of W
are also elements of V the six axioms listed above are guaranteed to be
valid on W.
The only ones that we really need to worry about are the remaining
four, all of which require something to be in the subset W. The first two
(1 and 2) are the closure axioms that require that the sum of any two
elements from W is also in W and that the scalar multiple of any element
from W will be also in W. Note that the sum and scalar multiple will be in
V we just don‟t know if it will be in W. We also need to verify that the
zero vector (axiom 5) is in W and that each element of W has a negative
that is also in W (axiom 6).
As the following theorem shows however, the only two axioms
that we really need to check about are the two closure axioms. Once we
have those two axioms valid, we will get the zero vector and negative
vector for free.!
Theorem 1: W is a subspace of V if and only if
i) W is non empty.
ii) W is closed under addition. x, y W x + y W iii) W is closed under scalar multiplication. , x W x W
Proof: If W is a subspace of V then it satisfies all 10 axioms so nothing to
prove.
Conversely if we have given three axioms then axioms 3, 4, 7, 8, 9 and 10
are anyway satisfied.
From iii) 0 , x W 0x = 0 W and
-1 , x W (-1)x = -x W Hence axioms 5 and 6 are valid.
66
W is a subspace of V.
Let us reduce these three conditions to two conditions with the
help of following theorem.
Theorem 2: A subset W of a vector space V is a subspace of V if and only
if
i) W is non empty
ii) x, y W, , x + y W
iii)
Proof: Suppose W is a subspace of V then W is non empty.
Also , x W x W and , y W y W
Hence x + y W .
Now suppose W satisfies two given conditions.
Take = 1 and = 1
x, y W, , x + y W x + y W
Now take = 1 and = 0
x, y W, , x + y W x W.
Hence W is a subspace of V (From theorem 1).
Remark : A set {0} consisting of zero vector of any vector space V and V
itself are subspaces of V.
We can now state whether the given subset is a subspace of a vector space.
Following are some interesting examples
Example 7: Let L = {(x ,y) / y = mx, m is fixed real number. x }
(0, 0) L, L is nonempty subset of 2 .
Now TST a, b L, , ba L
Let a = (x1, y1), b = (x2, y2) y1 = mx1 and y2 = mx2 …..(*)
ba = (x1, y1) + (x2, y2) = ( x1 + x2, y1 + y2)
y1 + y2 = mx1 + mx2 = m ( x1 + x2) (from (*) )
Hence ba L L is a subspace of 2 .
67
Remark : L is a line passing through origin in 2 .
Example 8: Let L’ = {(x ,y, z) / x = kx0, y ky0, z = kz0, ; (x0, y0, z0) is
fixed vector in 3 and k }
i.e. L’ = {( kx0, ky0, kz0) / (x0, y0, z0) is fixed vector in 3 and k }.
(0, 0, 0) L’, L’ is nonempty subset of 3 0 0 00 0 0 0 0 0, , x , y , z
.
Now TST a, b L’, , ba L’
Let a = ( kx0, ky0, kz0), b = ( k‟x0, k‟y0, k‟z0), where k, k‟
ba = ( kx0, ky0, kz0) + ( k‟x0, k‟y0, k‟z0)
= ( kx0 + k‟x0, ky0 + k‟y0, kz0 + k‟z0)
= (( k + k‟) x0, ( k + k‟) y0, ( k + k‟) z0), k + k‟
Hence ba L’. L’ is a subspace of 3 .
Remark : L’ is a line passing through origin and (x0, y0, z0) in 3 .
Example 9: Let P = {(x ,y, z) / 2x + 3y + z = 0, x, y, z } i.e. P = {(x, y, -2x – 3y)/ x, y }.
(0, 0, 0) P, P is nonempty subset of 3 .
Now TST a, b P, , ba P Let a = (x1, y1, z1), b = (x2, y2, z2) 2x1 + 3y1 + z1 = 0, 2x2 + 3y2 + z2 = 0
ba = (x1, y1, z1) + (x2, y2, z2) = ( x1 + x2, y1 + y2, z1 +
z2)
Consider 2( x1 + x2) + 3( y1 + y2) + ( z1 + z2)
= (2x1 + 3y1 + z1) + (2x2 + 3y2 + z2)
= 0 + 0 = 0.
Hence ba P. P is a subspace of 3 .
Remark : P is a plane through origin in 3 .
In general = {(x ,y, z) / ax + by + cz = 0, a, b, c } is a subspace of 3 and is a plane passing through origin.
68
Example 10: Let W = { )( nx / )( nx is convergent sequence of real
numbers}
Since zero sequence is convergent. So W is nonempty.
If )( nx,
)( ny
are convergent sequences then
)( nn yx
is
convergent.
Hence )( nx + )( ny
W for , W is a subspace of
.
Example 11: Let W = = {
2
1
0
0
a
a/ 1a ,
2a are real numbers}. W is a set
of all diagonal matrices of order 2.
00
00
W, W is nonempty.
Let A, B W. and ,
2
1
0
0
a
aA
,
2
1
0
0
b
bB
2
1
2
1
0
0
0
0
b
b
a
aBA W
ba
ba
22
11
0
0
Hence W is a subspace of set of all matrices of order 2.
Example 12: Let Pn[x] be set of all polynomials of degree n with real
coefficients then W is a subspace of x .
Pn[x] is non empty since it contains zero polynomial.
Let )(xp , )(xq Pn[x] and ,
)(xp = n
n xaxaxaa ......2
210 and
q x = n
n xbxbxbb ......2
210
)(xp + )(xq =
n
nn xbaxbaxbaba )(......)()( 2
221100
Hence )(xp + )(xq Pn[x].
Therefore Pn[x] is a subspace of x .
69
Example 13: Let W be set of all continuous real valued functions defined
on [a, b] then W is a subspace of vector space of all real valued functions
on [a, b]
„o‟ function defined by o(x) = 0 for all x in [a, b] is in W.
Further if f and g are continuous functions defined on [a, b] then f + g
is also continuous function on [a, b]. Hence f + g W.
Example 14: Let A = [aij] be a matrix of order m× n and X = [xj] be a
matrix of order 1×n. Then AX = O, where O is 1×n zero matrix is a
homogeneous system of m linear equations in n unknowns.
Let S be set of solutions of this system.
i.e. S = { X n / AX = O}
S is a subset of n .
Since AO = O
O =
0
0
0
S
S is non empty.
For X, Y S and ,
AX = O and AY = O
A( X + Y) = A X + A Y = (AX) + (AY) = ×0 + ×0 = 0
X + Y S
Hence S is a subspace of n
Check your progress
1) W1 = {(0, y) / y is a real number}. Show that W1 is a subspace of 2 .
2) W2 = {(x, 0, y)/ x, y are real numbers}. Show that W2 is a
subspace of 3 .
3) W3 = Set of upper triangular matrices of order 2
70
= { 321
3
21,,/
0aaa
a
aa
are real numbers }.Show that W3 is a
subspace of M22.
We have seen variety of examples of subspaces. Now the question
is about the intersection and union of subspaces. What will we get again a
subspace or only subset of a vector space? The following theorem gives
you the answer.
Theorem 3: If W1 and W2 are subspaces of a vector space V then
W1W2 is a subspace of V.
Proof: W1 and W2 are subspaces
0 W1 and 0 W2
0 W1W2 . Hence W1
W2 is nonempty.
Now to show that if a, b W1W2 ; , then a+ b
W1W2
Let a, b W1W2 and ,
a+ b W1 and a + b W2 ( W1 and W2 are subspaces and
by thm 3)
a+ b W1W2 and , . Hence W1W2 is a subspace
of V.
Remark : W1 = {(0, y) / y is a real number} and W2 = {(x, 0) / x is a real
number} are subspaces of 2 . W1W2 = {(x, 0), (0, y)/ x, y are real
numbers}.
If W1W2 is a subspace of 2 then addition must be closed in W1W2.
(2, 0) W1, (0, 3) W2 but (2, 0) + (0, 3) = (2, 3) W1W2. Hence
W1W2 is not a subspace of 2 .
In general union of two subspaces need not be a subspace. But
with some condition it is.
Theorem 4: If W1 and W2 are subspaces of a vector space V then
W1W2 is a subspace of V if and only if either W1 W2 or W2 W1.
Proof: Suppose W1 W2 or W2 W1.
71
If W1 W2 then W1W2 = W2. If W2 W1 then W1
W2 = W1. Since
W1 and W2 are subspaces of V, W1W2 is a subspace of V.
Now let W1W2 be a subspace of V.
To show that W1 W2 or W2 W1
Suppose W1 W2 and W2 W1
there exist x W1 such that x W2 and there exist y W2 such that y W1
But x W1 x W1W2, y W2 y W1
W2
W1 W2
W1W2 is a subspace x + y W1
W2 x + y W1 or x + y
W2
Which is not possible hence contradiction to the assumption. either W1 W2 or W2 W1.
Theorem 5: If W1 and W2 are subspaces of a vector space V and W1 + W2
is a subset of V defined as W1 + W2 = { x + y / x W1 and y W2}
then W1 + W2 is a subspace of V.
Proof: W1 and W2 are subspaces
0 W1 and 0 W2
0 W1 + W2. Hence W1 + W2 is nonempty.
Now to show that if a, b W1 + W2 ; , β then a+ βb W1 +
W2
Let a, b W1 + W2 ; ,β
a W1 + W2 a = x1 + y1
b W1 + W2 b = x2 + y2,
where x1, x2 W1 and y1, y2 W2 (by definition of W1 + W2)
a+ βb = (x1 + y1) + β(x2 + y2)
y
X
72
= x1 + y1 + β x2 + β y2
= ( x1 + β x2) + ( y1 + β y2)
W1 and W2 are subspaces x1 + β x2 W1 and y1 + β y2 W2.
a+ βb W1 + W2
Hence W1 + W2 is a subspace of V
Definition: If W1 and W2 are subspaces of a vector space V the the sum
W1 + W2 is called the direct sum of W1 and W2 if W1 W2 = . We
denote this sum by W1 W2.
4. 4 SUMMARY :
In this unit we have learned following
Vector space over whose elements are vectors and real numbers
are scalars. These vectors under addition and scalar multiplication
have the properties which are similar to the properties of usual
vectors in the plane.
Various examples of vector spaces like set of matrices, set of
sequences, set of continuous functions etc.
Subspace of a vector space
Condition for a subset of a vector space to be a subspace
Intersection of subspaces is a subspace
Union of subspaces is not always a subspace
If one subspace is contained in other subspace then there union is a
subspace.
Sum of two subspaces is a subspace
4. 5 UNIT END EXERCISE :
Theory:
1) Define a vector space.
2) Define a subspace of a vector space
3) State the condition for a subset W to be a subspace of a vector
space V.
Problems:
1) Let V = {(x, y) / x, y are real numbers}. Addition and scalar
multiplication in V is defined as (x1, y1) + (x2, y2) = (x1 + y1, x2
+ y2) and k(x, y) = (kx, 0) respectively. Show that V is not a
vector space. Which axiom is not satisfied by V?
2) Show that following are subspaces of 2
73
i) W1 = {(x, x)/ x is a real number}
ii) W2 = {(2x, 3y)/ x, y are real numbers}
iii) W3 = {(x, y)/ x + y = 0 and x, y are real numbers}
iv) W4 = {(x, y)/ 2x - 3y = 0 and x, y are real numbers}
3) Show that following are subspaces of 3
i) W1 = {(x, x, x)/ x is a real number}
ii) W2 = {(x, 0, 0) / x is a real number}
iii) W3 = {(x, 2y, 3z)/x, y, z are real numbers}
iv) W4 = {(x, x, z)/ x, z are real numbers}
v) W5 = {(x, y, z)/ x + y = z, x, y, z are real numbers}
vi) W6 = {(x, y, z)/ 3x + 2y + 4z = 0, x, y, z are real numbers}
vii) W7 = {( y, y+ z, z)/ y, z are real numbers}
4) Let V = Set of all real valued functions defined on [a, b].
Show that following are subspaces of V.
i) W1 = Set of all differentiable functions in V
ii) W2 = Set of all even functions in V = {f V/ f(-x) = - f(x) for
all x }
5) If W = {(x, 4)/ x is a real number}. Show that W is not a
subspace of 2 .
(Hint: show that addition is not closed in W)
6) If W = {(x, y, z)/ x + y + z = 4; x, y, z are real numbers}. Show
that W is not a subspace of 3 . (Hint: show that scalar
multiplication is not closed in W)
7) If W‟ = {(x, y, z)/ x 0; x, y, z are real numbers}. Show that
W is not a subspace of 3 . (Hint: show that scalar
multiplication is not closed in W)
8) If W1, W2, ….,Wn are subspaces of vector space.
Show that W1W2 ….Wn is a subspace of V. (Hint: Use
induction)
74
5
LINEAR SPAN, LINEARLY DEPENDENT
AND INDEPENDENT SETS Unit Structure:
5.0 Objectives
5.1 Introduction
5.2 Linear combination of vectors
5.3 Linear span
5.4 Convex sets
5.5 Linearly dependent and linearly independent sets
5.6 Summary
5.7 Unit End Exercise
5. 0 OBJECTIVES
This chapter would make you to understand the following concepts:
Linear combination of vectors
Linear span of a set
Spanning set or generating set of a vector space
Linearly dependent set
Linearly independent set
5. 1 INTRODUCTION
In the vector space there are two operations addition and scalar
multiplication. Operating these operations on elements of vector space and
scalars (real numbers) we get an element of a vector space known as a
linear combination of vectors.
Set of all linear combination of elements of some subset of vector
space V has various properties. It is a subspace of V. The most important
is that it can cover V. Hence the concept of generators is introduced.
Suppose S is a subset of a vector space such that no element of S is
linear combination of other elements of S. We can say that elements of S
are not depend on other elements of S. We call S as linearly independent
set. Subsets of V which are not like S are linearly dependent.
In this unit we define and elaborate all these concepts by studing
various examples and properties.
5. 2 LINEAR COMBINATION IN A VECTOR SPACE
Definition: Let V be a vector space. Let v1, v2,….vn be vectors in V and
1, 2,…
n be scalars i.e. real numbers. Then the expression
1v1 + 2v2 + …. nvn is an element of V, known as linear combination
of v1, v2,…vn. There are infinite linear combinations of v1, v2,….vn.
75
Example 1: (1, 2), (2, 3) are vectors in vector space 2 .
2(1, 2) + 5(2, 3) is a linear combination of (1, 2) and (2, 3). In general
1(1, 2) + 2(2, 3) is a linear combination of (1, 2) and (2, 3) where 1
and 2 are real numbers.
Example 2: u = (-1, 2) and v = (4, -6) are vectors in 2 .
Is (-12, 20) a linear combination of u and v?
If yes then there exist real numbers 1 and 2 such that w = 1u + 2v
Suppose w = 1u + 2v
Then (-12, 20) = w = 1(-1, 2) + 2(4, -6)
(-12, 20) = (- 1 + 4 2, 2 1 - 6 2)
-12 = - 1 + 4 2 and 20 = 2 1 - 6 2
By solving these equations simultaneously we get 1 = 4 and 2 = -2
which are real numbers.
w is a linear combination of u and v.
Example 3: u = (2, 1, 0) and v = (-3, -15) are vectors in 2
Is w=(1,-4) a linear combination of u and v?
If yes then there exist real numbers 1 and 2 such that w = 1u + 2v
Suppose w = 1u + 2v
Then (1, -4) = 1(2, 10) + 2(-3, -15)
1 = 2 1 - 3 2 and -4 = 10 1 – 15 2
Multipling first equation by 5 we get
5 = 10 1 – 15 2
By comparing with second equation we get 5 = -4, which is not true.
we could not find 1 and 2 such that w = 1u + 2v
Hence (1, -4) is not a linear combination of (2, 10) and (-3, -15)
Check your progress
1) Is (2, 3) linear combination of (-1, 0) and (0, 4)?
2) Is (5, 6) linear combination of (2, 3) and (8, 12)?
3) Is (1, 2, 3) linear combination of (0, 2, 0), (4, 0, 0) and (0, 0, 1)?
4) Let S = {(1, 0), (3, 5)}. Express (4, 5) and (-3, 7) as a linear
combination of elements of S.
5) Let S = {x – 2, x2 + 1, 5}. Express 3x
2 – 2x + 4 as a linear
combination of elements of S.
76
Answers
1) Yes, (2, 3) = -2(-1, 0) + (3/4)(0, 4)
2) No, We cannot find real numbers a and b such that
2a + 8b = 5 and 3a + 12b = 6
3) Yes, (1, 2, 3) = 1(0, 2, 0) + (1/4)(4, 0, 0) + 3(0, 0, 1)
4) (4, 5) = 1(1, 0) + 1(3, 5)
(-3, 7) = (-36/5)(1, 0) + (7/5)(3, 5)
5) 3x2 - 2x + 4 = -2(x – 2) + 3 (x
2 + 1) + (-3/5)(5)
5. 3 LINEAR SPAN
Definition : Let S be a subset of a vector space V. The linear span of S,
denoted by L(S) is defined as set of all linear combinations of elements
of S.
i.e. L(S) = { 1v1 + 2v2 + …. nvn / v1, v2,….vn S and 1, 2,… n
, n }
By convention L(φ) = {0}
Example 4: Consider a vector space 2 . Let S = {(1, 2), (2, 3)}. Then
L(S) = { 1(1, 2) + 2(2, 3) / 1 and 2 are real numbers}
Note: Subset S in V may be finite or infinite. In L(S) we take all possible
linear combinations of finite elements in S. Hence L(S) is always infinite
except for S = φ
Now the question is „for any subset S of V, is L(S) subspace of V?‟. We
will get the answer in the following theorem.
Theorem 1: Let S be a subset of vector space V. Then L(S) is a subspace
of V containing S. Moreover it is the smallest subspace of V containing S.
Proof: If S is empty then L(φ) = {0} which is the smallest subspace of V
containing φ.
Let S be nonempty.
Then S has at least one element say x. x L(S) for any real number . 0 x = 0 L(S). Hence L(S) is nonempty.
Now to show that x, y L(S); , β x+ βy L(S)
77
x, y L(S) x = 1v1 + 2v2 + …. nvn and y = β1u1 + β2u2 + …….
Βnvn
where v1, v2,….vn , u1,u2,…un S and 1, 2,… n , β1, β2,…, βn
x+ βy = ( 1v1 + 2v2 + …. nvn ) + β (β1u1 + β2u2 + ……. Βnvn)
= 1v1 + 2v2 + …. nvn + β β1u1 + β β2u2 + ……. β βnvn
= ( 1)v1 + (
2)v2 + ….+ ( n)vn + (ββ1)u1 +( ββ2)u2 + ……+
(ββn)vn
1,
2 ,…., n , ββ1 , ββ2 ,…,,, ,ββn
x+ βy L(S)
Hence L(S) is a subspace of V.
For any x in S, 1x = x L(S) L(S) contains S.
To show that L(S) is the smallest subspace containing S.
Let W be a subspace of V containing S.
v1, v2,….vn S v1, v2,….vn W
Hence for 1, 2,… n , 1v1 + 2v2 + …. nvn W (W is a
subspace)
But 1v1 + 2v2 + …. nvn L(S)
L(S) is a subset of W
Thus, L(S) is the s smallest subspace of V containing S.
Remark : L(S) is a linear span of S, which is known as a subspace
generated or spanned by S. If L(S) = V then S is known as generating or
spanning set of V.
Example 5: Let V = 3 . S = {(1, 0, 0), (0, 1, 0)} is a subset of V.
Then, L(S) = { (1, 0, 0) + β (0, 1, 0)/ , β }
= { ( , 0, 0) + (0, β, 0) / , β }
= { ( ,β, 0) / , β IR } = {(x, y, 0) / x, y }
L(S) is X-Y plane in 3 .
Example 6 : Let V = M2 2 , Set of all 22 matrices.
78
Let S = {
00
01 ,
10
00 }
L(S) = {
00
01, β
10
00/ , β }
L(S) = {
00
0,
0
00 / , β }
L(S) = {
0
0 / , β }
L(S) is a subspace of all diagonal matrices in M2 2 .
Example 7: Let V = 3
and S = {(1, 1, 0), (2, 0, 2)} Let us check whether
(5, 2, 3) and (4, 1, 5) are in L(S).
If (5, 2, 3)L(S) then (5, 2, 3) = (1, 1, 0) + β(2, 0, 2) for some
, β .
i.e. (5, 2, 3) = ( + 2β, , 2β )
i.e. 5 = + 2β, 2 = , 3 = 2β
Solving these equations simultaneously, we get = 2 and β = 3/2.
Hence (5, 2, 3) = 2(1, 1, 0) + (3/2) (1, 0, 1)
So, (5, 2, 3) L(S).
Similarly if (4, 1, 5) L(S) then (4, 1, 5) = (1, 1, 0) + β(2, 0, 2) for
some , β .
i.e. (4, 1, 5) = ( + 2β, , 2β )
i.e. 4 = + 2β, 1 = , 5 = 2β = 1 and β = 5/2, but + 2β = 1 + 2(5/2) ≠ 4
Hence we cannot find and β such that (4, 1, 5) = (1, 1, 0) +
β (2, 0, 2).
(4, 1, 5) L(S).
Check your progress 5.3.1:
1) Let V be a vector space. Let S and T be subsets of V such that T S. Then show that L(T) L(S).
2) Let S = { 1, x} be a subset of x . Find L(S).
3) Let S = {(1, 1), (2, 3)}. Check whether (4, 6) L(S).
4) Find the span of a set { (0,1, 0,1) , (1, 0, -1, 0)}
79
Answers
2) L(S) = {a + bx/ a and are real numbers}
3) Yes, Since (4, 6) = 0(1,1) + 2(4, 6)
4) Span of the set is {(x, y, -x, y)/ x, y are real numbers} or
{(x, y, z, w)/ z = -x, w = y, x and y are real numbers}
5. 4 CONVEX SETS
We have defined lines in 2 and
3 . We now define a line in a vector
space.
Definition : A line l (w, v1) in a vector space V passing through a point w
in V and having direction v1 0 is defined as
l (w, v1) = {w + tv1 / t }
If w = 0 then l (0, v1) = {tv1/ t } = L({v1})
Definition : The line segment in V from w to u is defined as the set
{(1- t)w + tu/ 0 1t }
Definition : A parallelogram spanned by two non zero vectors u and v in a
vector space V is defined as {t1u + t2v/ 0 1 2, 1t t }
If v = u then parallelogram is degenerated (does on exist)
v + u
v
A parallelogram spanned by u and v
t1v
u
t2u
Definition : A subset S of a vector space V is said to be convex if
P, Q S (1 – t) P + tQ S for 0 1t
I.e. The line segment between P and Q is entirely contained in S
PpPPppP
1S 2S
80
1S is convex. 2S is 1
Example 8: The parallelogram S spanned by u, v (u u) for any real
number ina vector space V is a convex set
1 2 1 2/ 0 , 1S t u t v t t
Let P,Q S then P = t1u + t2v and Q = t1‟u + t2‟v
0 t1, t2 1, 0 t1‟, t2‟ 1
Consider (1 – t)P + tQ where 0 t 1
(1 – t)P + tQ = (1 – t)( t1u + t2v) + t(t1‟u + t2‟v)
= [(1 – t)t1 + t t1‟] u + [(1 – t) t2 + t t2‟]v ……….(*)
Since 0 t 1 0 1 – t 1
Also 0 t1, t1‟ 1
0 (1 – t) t1 + t t1‟ (1 – t) + t = 1
0 (1 – t) t1 + t t1‟ 1
Similarly 0 [(1 – t) t2 + t t2‟ 1
From (*) (1 – t) P + tQ S
Hence S is convex.
Theorem 2: If S1 and S2 are convex subsets of a vector space V then
S1S2 is convex if S1S2
Proof: S1S2 if S1S2 is singleton set then it is convex.
If not, then let P,Q S1S2
P, Q S1 and P,Q S2
Since S1 and S2 are convex,
(1 – t)P + tQ S1 and (1 – t)P + tQ S2
(1 – t)P + tQ S1S2
S1S2 is convex.
Check your progress :
1) Show that S = {(x, y) 2 / x + y 1} is a convex set.
2) Let S be a convex set in vector space V. Then show that
cS = {cx/ xS} is convex.
81
5.5 LINEARLY DEPENDENT AND LINEARLY
INDEPENDENT SETS:
In the previous section we have seen that a vector in a
vector space can be express as a linear combination of vectors of some set.
Let us study following example.
Let S = {(1, 1, 0), (2, 1, 1), (1, 0, 1)} be a subset of 3 . We will write a
vector
(4, 2, 2) as a linear combination of elements of S. By observing elements
of S, we get
(4, 2, 2) = 2(1, 1, 0) + 0(2, 1, 1) + 2(1, 0, 1). But one can also write,
(4, 2, 2) = 1(1, 1, 0) + 1(2, 1, 1) + 1(1, 0, 1) or
(4, 2, 2) = 0(1, 1, 0) + 2(2, 1, 1) + 0(1, 0, 1)
Hence linear combination for (4, 2, 2) is not unique.
If a subset S is a generator of vector space V then is it possible to
express a vector as a linear combination of elements of S uniquely?
Linearly dependent set
Definition : Let V be a vector space. Set S is said to be linearly dependent
if and only if for any v1, v2,….vn in S there exist scalars a1,
a2,……..an not all zero such that
a1 v1 + a2 v2 +……..anvn = 0
Note: 1) {0} is always linearly dependent. Since 10 = 0 where
1≠ 0.
2) Any subset S of V containing 0 is linearly dependent.
Since, if S = {0, v2,….vn } then for any nonzero real number a1 , a1 0 +
0v2 +……..0vn = 0
Example 9: Let S = {(1, 0), (-1, 2), (2, -4)} be a subset of 2 .
Let v1 = (1, 0), v2 = (-1, 2) and v3 = (2, -4)
Then, Since for a1 = 0, a2 = -2 and a3 = 1, not all zero such that a1 v1 + a2
v2 + a3v3 = 0
Hence S is linearly dependent.
Linearly independent set
Definition : A subset of a vector space is linearly independent if and only
if it is not linearly dependent. i.e. Set S = { v1, v2,….vn } is said to be
linearly independent if
82
a1 v1 + a2 v2 +……..anvn = 0 then a1 = a2 = ….. an = 0.
Empty set is defined as a linearly independent set.
An infinite subset S of V is said to be linearly independent if and only if
every finite subset of S is linearly independent.
Example 10: A subset S = {(1, 0), (0, 1)} of 2 is linearly independent.
Since a1(1, 0) + a2(0, 1) = (0, 0)
(a1, a2) = (0, 0)
a1 = 0 and a2 = 0.
Example 11: Let S = {(1, 7, -4), (1, -3, 2), (2, -1, 1)}.
To find whether S is linearly dependent or independent, consider
a(1, 7, -4) + b(1, -3, 2) + c(2, 1, 1) = (0, 0, 0) where a, b, c
(a + b + 2c, 7a - 3b + c, -4a + 2b + c) = (0, 0, 0)
a + b + 2c = 0 …….(i)
7a - 3b + c = 0 …….(ii)
-4a + 2b + c = 0 …….(iii)
We solve these three equations simultaneously.
By adding (ii) and (iii) we get 3a – b + 2c = 0 …..(iv)
By subtracting (iv) from (i) we get 2a - 2b = 0 a b From (i) 2a + 2c = 0 c a From (iii) -4a + 2a – a = 0 -3a = 0 a = 0 b = 0 and c = 0
Hence a(1, 7, -4) + b(1, -3, 2) + c(2, 1, 1) = (0, 0, 0) a = 0, b = 0, c = 0.
S is linearly independent set.
Example 12 Every nonzero singleton set is linearly independent.
Let v be a nonzero element of a vector space V,
We will show that { v } is linearly independent.
Consider av = 0 where a .
Since v ≠ 0 av = 0 a = 0
Because if a ≠ 0 then av = 0 a-1
(av) = a-1
0
(a-1
a)v = 0
1v = 0
v = 0. But v ≠ 0
83
Hence av = 0 a = 0
{ v } is linearly independent.
Example 13: Let V be a vector space of all differentiable functions.
S ={ f1, f2 } where f1(t) = et, f2(t) = e
2t is linearly independent in V
Now. S = { et, e
2t }
Let a et + b e
2t = 0 for all t , where a, b ……(i)
Differentiating equation (i) with respect to t, we get
aet + 2be
2t = 0 for all t …..(ii)
By putting t = 0 in (i) and (ii) we get
a + b = 0 and a + 2b = 0
By solving these equations simultaneously we get a = b = 0.
Hence a et + b e
2t = 0 a = b = 0.
S is linearly independent set.
Example 14: Let V = C[0, π] be a vector space of differentiable functions
defined on [0, π].
S ={ f1, f2 } where f1(t) = cost, f2(t) = sint is linearly independent in V
Now S = {cost, sint}
Let a(cost) + b(sint) = 0 for all t and a, b
For t = 0 we get a(cos 0) + b(sin 0) = 0 a = 0
cos 0 = 1 and sin 0 = 0.
For t = π/2 we get a(cos π/2) + b(sin π/2) = 0 b =0
cos(π/2) = 0 and sin(π/2) = 0
Hence a(cost) + b(sint) = 0 a = b = 0
S is linearly independent set.
Example 15: Let V = P2[x] be a vector space of all polynomials of degree
≤ 2 with real coefficients. S = { x2 + 1, 2x – 1, 3} is linearly independent.
Consider a, b, c such that a(x2 + 1) + b(2x – 1) + c(3) = 0.
ax2 + a + 2bx – b + 3c = 0
ax2 + 2bx + (a – b + 3c) = 0 a = 0, 2b = 0, a – b + 3c = 0
Hence a(x2 + 1) + b(2x – 1) + c(3) = 0 a = 0, b = 0, c = 0
S is linearly independent set.
84
Example 16: Let V be a vector space. If a subset {x, y, z} of V is linearly
independent then the subset {x + y, y + z, z + x} is also linearly
independent.
Now to show that {x + y, y + z, z + x} is linearly independent.
Consider a(x + y) + b(y + z) + c(z + x) = 0 where a, b, c
ax + ay + by + bz + cz + cx = 0
(a + c)x + (a + b)y + (b + c)z = 0
Since {x, y, z} is linearly independent a + c = 0, a + b = 0 and b + c = 0
a = b = c = 0
{x + y, y + z, z + x} is linearly independent.
Note : Let v1 and v2 be two non zero vectors in 2 . If v1 and v2 are
linearly dependent i.e. { v1, v2} is linearly dependent. Then there exist real
numbers a and b such that av1 + bv2 = 0 where a and b both are non zero.
Because if a = 0 then
bv2 = 0 gives b = 0.
v1 = (-b/a) v2.
If k = (-b/a) then v1 = kv2
Similarly if m = (-a/b) then v2 = mv1
v1 and v2 are on the same line through origin
i.e.
v1 v2 0
Note : Let v1, v2, v3 be three non zero vectors in 3 . If v1,v2, v3 are
linearly dependent then there exist real numbers a, b and c not all zero
such that av1 + bv2 + cv3 = 0
Let a ≠ 0 then v1 = (-b/a)v2 + (-c/a)v3
If (-b/a) = k1 and (-c/a) = k2 then v1 = k1v2 + k2v3
If v2 and v3 are linearly dependent then v2 and v3 are on the same line
through origin. In this case v1, v2, v3 are collinear.
i.e.
v2
v1 v3 0
85
If v2 and v3 are linearly independent then they are not on same line
through origin.
v1 – k1v2 – k2v3 = 0 v1 lies on the plane passing through v2, v3 and
origin.
i.e. v1, v2, v3 are coplanar.
v2 v1
0 v3
Check your progress :
1) Show that {(1, 2), (3, 4)} is linearly independent in 2 .
2) Show that {(1, 1, 2, 0), (0, 1, 4, 9)} is linearly independent in 4 .
3) If {x, y} is linearly independent in a vector space V then show that
{x + ay, x + by} is linearly independent where a and b are real
number which are not same.
Theorem 3: A subset of a linearly independent set in a vector space is
linearly independent.
Proof: Let S be a linearly independent set in a vector space V.
Let T S. To show that T is linearly independent
If T then by definition T is linearly independent.
So let T .
Suppose T is linearly dependent
Suppose S is finite.
Let S = { v2,….vn }
Without loss of generality assume that T = { v2,….vk } , k ≤ n
T is linearly dependent there exist real numbers a1, a2, … . ak not all
zero such that a1v1 + a2v2 + ….+ akvk = 0
a1v1 + a2v2 + ….+ akvk + 0vk+1 + 0vk+2 + ….+ 0vn = 0, where
a1, a2,.. ak are not all zero.
S is linearly dependent set.
Contradiction since S is linearly independent.
T is linearly independent.
86
If S is infinite then by definition every finite subset of S is linearly
independent. If T is infinite then a finite subset of T is a subset of S
therefore linearly independent. Hence T is linearly independent.
Theorem 4: A superset of a linearly dependent set is linearly dependent.
Proof: Let S be a linearly dependent set in a vector space V.
Let T be a superset of S i.e. T S . To show that T is linearly dependent
Suppose T is linearly independent. Then since S is a subset of T, by the
previous result, S is linearly independent.
Contradiction. Since S is linearly dependent.
T is linearly dependent.
Remark : A suset of a linearly independent set is linealy independent but
a superset of linearly independent can be a linearly dependent set.
Similarly superset of a linearly dependent set is linearly dependent but a
subset of a linearly dependent set may be a linearly independent set.
Theorem 5: Let V be a vector space. Let S be a finite linearly independent
subset of V and x V. Then x is linear combination of elements of S if
and only if S { x } is linearly dependent. i.e. x L(S) if and only if
S { x } is linearly dependent.
Proof: S is finite. Let S = { v1, v2,….vn} and let L be linearly independent.
Suppose x L(S) x = a1v1 + a2v2 + ….+ anvn, where a1, a2, …,an . a1v1 + a2v2 + ….+ anvn – x = 0 a1v1 + a2v2 + ….+ anvn + (-1)x = 0, where -1 ≠ 0 { v1, v2,….vn, x} = S { x } is linearly dependent.
Conversely, suppose S { x } is linearly dependent. To show that x L(S)
S { x } = { v1, v2,….vn, x} is linearly dependent
There exist real numbers a1, a2,….an, b not all zero such that
a1v1 + a2v2 + ….+ anvn + bx = 0 …….(i)
If b = 0 then a1v1 + a2v2 + ….+ anvn = 0 where not all a1, a2, …,an are zero.
{ v1, v2,….vn} = S is linearly dependent.
Contradiction since S is given to be linearly independent.
b ≠ 0
b-1
= 1/b exists
From (i) bx = a1v1 + a2v2 + ….+ anvn
x = b-1
( a1v1 + a2v2 + ….+ anvn)
87
x = (b-1
a1)v1 + (b-1
a2)v2 + ……..+ (b-1
an)vn, where b-1
a1, b-1
a2,…
b-1
an
x L(S).
Remark : We know that if P Q if and only if ~Q ~ P
So we can state above theorem also as If S is a finite linearly independent
subset of V and x V. Then S { x } is linearly independent if and only
if x L(S).
Theorem 6: If S = { v1, v2,….vn} be a linearly dependent set with nonzero
elements in a vector space V. Then there exist some i (2 ≤ i ≤ n) such that
vi can be expressed as a linear combination of vectors v1, v2,….,vi-1.
Proof: Since S is linearly dependent, there exists real numbers a1, a2, …an
not all zero such that a1v1 + a2v2 + ….+ anvn = 0. ……….(i)
Let i be the largest index (i = 2,3,…n) such that ai ≠ 0.
i.e. ai ≠ 0 but ai+1 = ai+2 = ….= an = 0.
a1v1 + a2v2 + …..aivi + 0vi+1 + 0vi+2 +…..+ 0vn = 0 (from (i)) a1v1 + a2v2 + …..aivi = 0
Here i > 1. Because if i = 1 then we will get a1v1 = 0 which implies v1 = 0
but v1 ≠ 0.
aivi = a1v1 + a2v2 + ……+ai-1vi-1
vi = ai-1(a1v1 + a2v2 +…….+ ai-1vi-1) (ai ≠ 0 )
vi = (ai-1a1)v1 + (ai-1a2)v2 + …….+ (ai-1ai-1)vi-1
Hence vi can be expressed as a linear combination of {v1, v2, ….vi-1}.
Theorem 7: A subset S ={ v1, v2,….vn} is linearly independent in a vector
space V if and only if every vector in L(S) can be uniquely expressed as a
linear combination of elements of S.
Proof: Let v L(S)
There exist real numbers a1, a2, …..an such that
v = a1v1 + a2v2 + …..anvn …….(i)
To show that this expression for v is unique
Suppose v can be also written as
v = b1v1 + b2v2 + ….bnvn …….(ii), where b1, b2,…bn .
a1v1 + a2v2 + …..anvn = b1v1 + b2v2 + ….bnvn (From (i) and (ii))
88
a1v1 + a2v2 + …..anvn - b1v1 + b2v2 + ….bnvn = 0
(a1 – b1)v1 + (a2 – b2)v2 + ………+ (an – bn )vn = 0
a1 – b1 = a2 – b2 = ……….= an – bn = 0. ( S is linearly independent)
a1 = b1, a2 = b2, …….., an = bn
Hence the expression for v is unique.
Remark : If L(S) = V then every element of vector space V is uniquely
expressed as a linear combination of elements of S.
5. 6 SUMMARY
In this unit we have defined span of a set, linearly independent and
linearly dependent set in a vector space.
The major results for a vector space V, we have proved are:
If S is a subset of V then L(S) is the smallest subspace containing S
Subset of a linearly independent set is linearly independent
Superset of linearly dependent set is linearly dependent
For x V, S {x} is linearly dependent if and only if x L(S)
Every element of linearly dependent set can be expressed as a
linear combination of other elements of the set
If S is linearly independent then every element of L(S) has unique
expression
If L(S) = V the S is a generating set of V
5. 7 UNIT END EXERCISE
Theory:
1) Define linearly dependent, independent set and convex set in vector
space V.
2) If S is a convex set in a vector space V and w V. Then show that
w + S = {w + x/ xS} is convex set.
3) If { v1, v2,….vn} is a linearly independent subset of a vector space V
then { v1 – v2, v2 – v3,….vn-1 – vn, vn} is also linearly dependent.
Problems:
1) Determine whether u and v are linearly dependent or independent.
(Hint: u and v are linearly dependent if one is multiple of other)
i) u = (3, 4) v = (1, -3)
ii) u = (2, -3), v= (6, -9)
89
iii) u = (4, 3, -2), v = (2, -6, 7)
iv) u =
103
421 , v =
206
842
v) u = 2 – 5t + 6t2 – t
3, v = 3 + 2t – 4t
2 + 5t
3
2) Check whether the following sets are linearly dependent or
independent
i) {(1, 1, 1), (0, 1, -2), (1, 3, 4)} in 3 .
ii) {(2, -3, 7), (0, 0, 0), (3, -1, -4)} in 3
iii)
1 1 1 0 1 1,
1 1 0 1 0 0
in 2 2M
iv) {(a, b), (c, d)} in 2 where ad – bc 0
v) {t3 – 4t
2 + 2t + 3, t
3 + 2t
2 + 4t – 1, 2t
3 – t
2 – 3t + 5} in P3[t]
3) If {x, y, z} is linearly independent set in V then show that {x+ y, x – y,
x – 2y + z} is linearly independent.
Answers
1) (i) linearly independent (ii) linearly dependent (iii) linearly
independent (iv) linearly dependent (v) linearly independent
2) (i) linearly independent (ii) linearly dependent (iii) linearly
independent linearly independent (v) linearly independent
90
6
BASIS AND DIMENSION
Unit Structure:
6.0 Objectives
6.1 Introduction
6.2 Basis of a vector space
6.3 Dimension of a vector space
6.4 Rank of a matrix
6.5 Summary
6.6 Unit End Exercise
6.0 OBJECTIVES
This chapter would make you to understand the following concepts:
Basis of a vector space
No of bases of a vector space
Dimension of a vector space
Finitely generated vector space
Finite dimensional vector space
Relation between the dimension of vector space and dimension of
its subspace
Column space and row space of a matrix
Column rank and row rank of a matrix
Rank of a matrix
6. 1 INTRODUCTION
In units 4 and 5 we have learnt about vector spaces, subspaces,
linear span which tell us about linear combination of vectors then linearly
dependent and linearly independent sets in a vector space. We also know
that if for a subset S, L(S) = V then S is generating set of V or S spans V.
Linearly independent sets and generating sets of V are very
important sets in V since they tell us many fascinating things about vector
space.
Suppose V is a finitely generated vector space. Therefore there
exist a finite subset S of V such that L(S) = V. So every element of vector
91
space V is expressed as a linear combination of elements of S. If S is
linearly independent then at the end of the previous unit we have seen that
every element of vector space V is uniquely expressed as a linear
combination of elements of S.
So, this S is a very special subset of V. Moreover S is finite. If T is
a subset of S then T is linearly independent. Now question is that is
L(T) = V?
Consider the following example.
We know that S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is linearly independent set
in 3 .
Since (x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1)
, ,x y z L S for all 3, , inx y z
3L S
Let T = {(1, 0, 0), (0, 1, 0)}
Clearly , ,x y z L T
L(T) 3 .
Hence the above example shows that we cannot reduce elements of S. So
number of elements in S plays very important role.
Now we ask following questions.
Is such subset S in V is unique?
If not then we get subset S‟ having same property which S has. since the
number of elements of S is very important then does number of elements
in S and S‟ same?
In this unit we will find answers of these questions.
Let us define a very special subset of a vector space
6. 2 BASIS OF A VECTOR SPACE
Definition: A subset B = { v1, v2,….vn} of a vector space V is said to a
basis of V if and only if it satisfies following conditions
1) B is linearly independent
2) L(B) = V
Example 1: 3Let V
Let B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}
Consider a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = (0, 0, 0)
92
(a, b, c) = (0, 0, 0) a = 0, b = 0, c = 0 B is linearly independent.
Since (x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1)
(x, y, z) L(B) for all (x, y, z) in 3
L(B) = 3 .
Hence B is a basis of 3 .
Example 2: Let B‟ = {(1, 1, 0), (1, 0, 1), (0, 1, 1)}
Consider a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (0, 0, 0) where a, b, c
(a + b, a + c, b + c) = (0, 0, 0)
a + b = 0, a + c = 0, b + c = 0
b – c = 0, b + c = 0
b = c = 0
a = 0
B‟ is linearly independent.
Now we find a, b, c such that
Let (x, y, z) = a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1)
(x, y, z) = (a + b, a + c, b + c)
x = a + b, y = a + c, z = b + c
b – c = x – y , b + c = z
b = (x – y + z)/2, c = (y + z – x)/2
a = (x + y – z)/2
a, b, c
L(B‟) = 3
Hence B‟ is a basis of 3 .
Example 3: Consider a vector space 2 2M , Set of all 2 2 matrices.
Let B = {
00
01 ,
00
10 ,
01
00 ,
10
00}
Consider a1
00
01 + a2
00
10 + a3
01
00 + a4
10
00 =
00
00
93
00
01a +
00
0 2a +
0
00
3a+
40
00
a=
00
00
43
21
aa
aa =
00
00
a1 = 0, a2 = 0 , a3 = 0, a4 = 0
B is linearly independent.
Now let
wz
yx
2 2M .
Since
wz
yx = x
00
01 + y
00
10+ z
01
00+ w
10
00
L(B) = 2 2M
B is a basis of 2 2M
Example 4: Consider a vector space Pn[x], of all polynomials of degree
≤ n with real coefficients. Let B = { 1, x, x2, ……, x
n} be a subset of Pn[x]
Let a0(1) + a1(x) + ……an(xn) = 0 where a0, a1,….an IR
a0(1) + a1(x) + ……an(xn) = 0 + 0x + …….ox
n
a0 = a1 = ……= an = 0
B is linearly independent.
Let f(x) = a0 + a1x + ……anxn Pn[x]
Clearly f(x) L(B)
L(B) = Pn[x].
Hence B is a basis of Pn[x].
Example 5: B = {(1, 1, 0), (-1, 0, 0)} is a subset of 3 .
Consider a(1, 1, 0) + b(-1, 0, 0) = (0, 0, 0)
(a – b, a, 0) = (0, 0, 0)
a – b = 0, a = 0 b = 0
So B is linearly independent.
(1, 1, 1) 3
If there exist a, b such that (1, 1, 1) = a(1, 1, 0) + b(-1, 0, 0)
Then (1, 1, 1) = (a – b, a, 0)
1 = a – b, 1 = a, 1 = 0
94
But 1 0
(1, 1, 1) L(B)
L(B)3
B is not a basis of 3 .
Note : In example 1 and example 2 we have seen that B and B‟ are two
different basis of 3 .
Which shows that a basis of a vector space is not unique.
Infact a vector space has infinitely many bases.
Note : If B = { v1, v2,….vn} is a basis of a vector space V then by
definition B is linearly independent and L(B) = V. So every element of V
is uniquely expressed as linear combination of v1, v2,….vn. If v V then v
= x1v1 + x2v2 + ……+ xnvn and this expression is unique. We call
(x1, x2,….,xn) coordinates of v with respect to the basis B.
Check your progress
1) Prove that {(-1, 1, 0), (0, -1, 1), (0, 1, -1)} is a basis of 3 .
2) Prove that {(1, 2), (3, 4)} is a basis of 2
3) Prove that {
11
11 ,
01
11 ,
00
11 ,
00
01} is a basis of
2 2M
4) Find the coordinates of (x, y, z) with respect to a basis
{(1, 1, 1), (1, 1, 0), (1, 0, 0)} of 3 .
Basis B of a vector space V is linearly independent and it
generates V. Now if we take superset of B or subset of B then is it a basis
of V? More precisely Is a superset of a basis linearly independent? Can
subset of B generates V?
Let us see following theorems.
Theorem 1: B = { v1, v2,….vn} is a basis of a vector space V if and only if
B is maximal linearly independent set in V.
Proof: Suppose B = { v1, v2,….vn} is maximal linearly independent set
in V.
If B B‟ then B‟ is not linearly independent…….(i)
To show that B is a basis
i.e. To show that L(B) = V since B is given to be linearly independent.
Suppose not. i.e. L(B) V
95
There exist v V such that v L(B)
B { v } is linearly independent.
{ v1, v2,….vn, v} is linearly independent and B { v1, v2,….vn, v}
Contradiction to (i)
L(B) = V.
Conversely let B is a basis of V.
To show that B is maximal linearly independent set in V
Since B is a basis of V therefore B is linearly independent.
Now to show that B is maximal
Suppose B is not maximal linearly independent set in V
there exist a linearly independent set B‟ such that B B‟.
Let x B‟ such that x B
B‟ is linearly independent, x, v1, v2,….vn are linearly independent.
x cannot be written as a linear combination of v1, v2,….vn
xL(B)
But L(B) = V x L(B)
Contradiction
B is maximal linearly independent set in V.
Theorem 2: B = { v1, v2,….vn} is a basis of a vector space V if and only if
B is a minimal set of generators of V.
Proof: Suppose B = { v1, v2,….vn} is a minimal set of generators.
No subset of B generates V. ……(i)
Also L(B) = V
To show that B is a basis of V
i.e. to show that B is linearly independent.
Suppose B is linearly dependent. Then some vi in B can be expressed as a
linear combination of v1, v2,….vi -1
L(B) = L{ v1, v2,….vn} = L{ v1, v2,…,vi -1, vi + 1, ….vn} = V.
{ v1, v2,…,vi -1, vi + 1, ….vn} is a generator set of V.
But { v1, v2,…,vi -1, vi + 1, ….vn} { v1, v2,….vn} = B
Contradiction to (i) B is linearly independent.
B is a basis of V.
Conversely let B be a basis of V. L(B) = V
To show that B is a minimal set of generators of V.
96
Suppose B is not a minimal set of generators of V. There exists a subset B1 of B such that B1 also generates V.
i.e. L(B1) = V B1 B therefore there exists x B such that x B1 x V, x L(B1) B1 { x } is linearly dependent.
But B1 { x } is a subset of B and B is linearly independent. Contradiction since subset of linearly independent set is linearly
independent. B is a minimal set of generators of V.
We combine theorem 1 and theorem 2 and get the following theorem.
Theorem 3: Let B = { v1, v2,….vn} be a subset of a vector space V. Then
following statements are equivalent
(i) B is a basis of V
(ii) B is maximal linearly independent set in V
(iii) B is a minimal set of generators of V.
Theorem 4: Let V be a vector space. If B = { v1, v2,….vn} be a basis of
V. Let w1, w2,….wm V and m > n then { w2,….wm} is linearly
dependent.
Proof: Since B is a basis of V, L(B) = V w1 = a11v1 + a12v2 + ……+ a1nvn
w2 = a21v1 + a22v2 + ……+ a2nvn
.
.
.
Wm = am1v1 + am2v2 + ……+ amnvn
For x1, x2, …..xm in
consider x1w1 + x2w2 + ….+ xmwm
= x1(a11v1 + a12v2 + ……+ a1nvn) + x2(a21v1 + a22v2 + ……+ a2nvn) +
…………..+xm(am1v1 + am2v2 + ……+ amnvn)
= (x1a11 + x2a21 + ….+ xmam1) v1 + (x1a12 + x2a22 + ….+ xmam2) v2 +
…………..+ (x1a1n + x2a2n + ….+ xmamn) vn
= j
n
j
m
i
jij vxa )(1 1
x1w1 + x2w2 + ….+ xmwm = j
n
j
m
i
jij vxa )(1 1
…………….(i)
97
Hence if x1w1 + x2w2 + ….+ xmwm = 0,
then j
n
j
m
i
jij vxa )(1 1
= 0
B = { v1, v2,….vn} islinearly independent
m
i
jij xa1
= 0 , j = 1, 2,……n …………..……(ii)
Now (ii) is a homogeneous system of n linear equations in m unknowns
m > n this system has non trivial solution say (c1, c2, …..,cm), where at
least one cj is non zero.
m
i
jij ca1
= 0, where at least one cj is non zero.
c1w1 + c2w2 + …..+ cnwn = j
n
j
m
i
jij vca )(1 1
(from 1)
= j
n
j
v)0(1
= 0
c1w1 + c2w2 + …..+ cnwn = 0, where at least one cj is non zero.
{w1, w2, …….,wm} is linearly dependent.
Note : From above theorem it is clear that If B is a basis of containing n
elements then any linearly independent set in V contains at the most n
elements.
Corollary 1: Let V be a finitely generated vector space. If B1 and B2 are
two different bases of V then B1 and B2 contain same number of elements
Proof: Let n(B1) = No of elements in B1 = n and n(B2) = m B2 is a basis of V, B2 is linearly independent
But since B1 is a basis m n …………………(i)
Now B1 is linearly independent B2 is a basis of V n m …………………(ii)
Hence from (i) and (ii) m = n.
Remark : Basis of a vector space is not unique, but number of elements in
all bases of a vector space is same. So the size of a basis is very important
for a vector space.
98
6. 3 DIMENSION OF A VECTOR SPACE
Definition : Number of elements in a basis of a vector space is known as
the dimension of a vector space. If V = {0} then V has dimension 0. We
denote the dimension of V by dim V.
Note: 1) If a basis is finite then the vector space is said to be finite
dimensional vector space otherwise it is infinite dimensional vector space.
2) n(A) denotes no. of elements in A.
Remark : If dim V = n, then 1) if a set S in V is linearly independent then
n(S) n.
2) If a set S in V generates V i.e. L(S) = V then n(S) n .
Example 6: Since {(1, 0), (0, 1)} is a basis of 2 , dim
2 = 2.
Example 7: Since {
00
01 ,
00
10 ,
01
00 ,
10
00}
is a basis of
2 2M , dim 2 2M = 4.
Since the size of every linearly independent set in a finite
dimensional vector space is at the most equal to the dim V then if the size
is equal to dim V then it is a basis of V. But if the size is smaller than dim
V then can we get basis by adding some elements in to it?
Theorem 5: Let V be a finite dimensional vector space. Then every
linearly independent set in V can be extended to a basis of V.
Proof: Let dim V = n. Let S be a linearly independent set in V.
If n(S) = n then L(S) = V then S is a basis of V.
If n(S) < n then L(S) V
There exists v1 V such that v1 L(S)
S { v1 } is linearly independent.
If L(S { v1 }) = V then S { v1 } is a basis of V.
If not then There exists v2 V such that v2 L(S { v1 })
S { v1, v2 } is linearly independent.
If L(S { v1, v2 }) = V then S { v1, v2 } is a basis of V.
If not continue this process.
dim V = n, any linearly independent set in V contains maximum n
elements.
The above process must stop in maximum n – 1 steps which gives a
basis of V.
Hence every linearly independent set in V can be extended to a basis of V.
99
Theorem 6: Let V be a finite dimensional vector space. Then any set of
generators of V contains a basis of V.
Proof: Let S be a subset of a finite dimensional vector space V such that
L(S) = V.
To show that S contains a basis of V.
Let S = { v1, v2,….vm }
If S is linearly independent then S itself is a basis of V.
Suppose S is linearly dependent
a1v1 + a2v2 + ….+ amvm = 0 where not all a1, a2, ..,am are zero.
Without loss of generality we can assume that am 0
vm = (- a1/am)v1 + (- a2/am) v2 + …..+ (-am – 1/am)vm – 1.
vm L{v1, v2, …., vm – 1}
L{v1, v2, …., vm – 1} = L(S) = V
If {v1, v2, …., vm – 1} is linearly independent then it a basis of V in S
If not then we continue the process by removing elements of S, which are
in L(S).
This process must stop since {v1} is linearly independent.
We have defined dimension of a vector space. Now we will discuss the
dimension of a subspace of a vector space.
Theorem 7: Let V be a finite dimensional vector space and W be a
subspace of V then dim W dim V. If dim W = dim V then W = V.
Proof: W is a subspace of V.
If W = { 0 }, dim W = 0 dim V
If W { 0 }, there exists w1 W (w1 0)
{ w1 } is linearly independent set in W
It can be extended to a basis {w1, w2, …., wr} of W
dim W = r
Now {w1, w2, …., wr} is linearly independent in W and W V
{w1, w2, …., wr} is linearly independent in V, which can be extend to
basis of V.
r dim V
dim W dim V
If dim W = dim V = r then any set of r + 1 elements in V is linearly
dependent
{w1, w2, …., wr} is maximal linearly independent set in V
100
{w1, w2, …., wr} is a basis of V.
L({w1, w2, …., wr}) = V
But {w1, w2, …., wr} is a basis of W
L({w1, w2, …., wr}) = W
W = V.
Note: From above theorem we can deduce that any subspace of 3 is
either a zero space or a line through origin or a plane through origin or 3
itself.
For, If W is any subspace of 3 then since dim
3 = 3 dim W 3 dim W = 0, 1, 2 or 3
i) If dim W = 0 then W is a zero space.
ii) If dim W = 1 then
{(a1, a2, a3)} is a basis of W for some (a1, a2, a3) (0, 0, 0) W = L({(a1, a2, a3)} = {t(a1, a2, a3)/ t } W represents a line through origin.
iii) If dim W = 2 then
{a, b} is a basis of W for some a, b 0; a = (a1, a2, a3), b =
(b1, b2, b3) W = L({a, b}) W = {xa + yb/ x, y }
This represents a plane passing through origin and normal to a b .
iv) If dim W = 3 then W = 3 .
Theorem 8: Let V be a finite dimensional vector space. Let W1 and
W2 be subspaces of V. Then dim(W1 + W2) = dimW1 + dim W2 – dim
(W1 W2)
Proof: Let dim(W1 W2) = r
Let {w1, w2, …., wr} be a basis of W1 W2.
{w1, w2, …., wr} is linearly independent in W1 W2.
{w1, w2, …., wr} is linearly independent in W1 as well as W2.
It can be extended to a basis {w1, w2, …., wr, u1, u2, …um} of W1
and a basis
{w1, w2, …., wr, v1, v2, ….vs} of W2.
dim W1 = r + m and dim W2 = r + s
Let B = {w1, w2, …., wr, u1, u2,…, um, v1, v2,…vs}
101
Claim: B is a basis of W1 + W2
W1 + W2 = { x + y/ x W1, y W2}
Let w W1 + W2
w = x + y where x W1, y W2
x = a1w1 + a2w2 + ….+ arwr + b1u1 + b2u 2 + , ,,,,,+ bmum
y = c1w1 + c2w2 + ….+ crwr + d1v1 + d2v2 + ….+ dsvs
x + y = a1w1 + a2w2 + ….+ arwr + b1u1 + b2u2 + ,,,,,,+ bmum + c1w1 +
c2w2 +
+ …+.crwr + d1v1 + d2v2 + ….+ dsvs
x + y = (a1 + c1)w1 + (a2 + c2)w2+ ….(ar + cr)wr + b1u1 + b2u 2 +
,,,,,,+ bmum +
d1v1 + d2v2 + ….+ dsvs
x + y = w L(B)
L(B) = W1 + W2
Now to show that B is linearly independent
Consider a1w1 + a2w2 + ….arwr + b1u1 + b2u2 + ,,,,,,+ bmum + c1v1 +
c2v2 + ….+ csvs = 0 ………(i)
c1v1 + c2v2 + ….+ csvs = - a1w1 - a2w2 - ….- arwr - b1u1 - b2u2 -
…..- bmum
Now - a1w1 - a2w2 - ….- arwr - b1u1- b2u2 - …..- bmum W1
c1v1 + c2v2 + ….+ csvs W1
{w1, w2, …., wr, v1, v2, ….vs} is a basis of W2, c1v1 + c2v2 + ….+
csvs W2
c1v1 + c2v2 + ….+ csvs W1 W2.
c1v1 + c2v2 + ….+ csvs = d1w1 + d2w2 + ….+ drwr ({w1, w2, …., wr}
is a basis of W1 W2)
c1v1 + c2v2 + ….+ csvs – (d1w1 + d2w2 + ….+ drwr) = 0
c1v1 + c2v2 + ….+ csvs – d1w1 –d2w2 - ….. – drwr = 0
{w1, w2, …., wr, v1, v2, ….vs} is a basis of W2. linearly
independent
c1 = c2 = …cs= d1 = d2 = …= dr = 0
From (i), a1w1 + a2w2+ ….arwr + b1u1+ b2u2+ ,,,,,,+ bmum = 0
{w1, w2, …., wr, u1, u2, …um} is a basis of W1. linearly
independent
a1 = a2 = …= ar = b1 = b2 = …= bm = 0
Hence a1w1 + a2w2+ ….arwr + b1u1+ b2u2 + ,,,,,+ bmum + c1v1 + c2v2 +
….+ csvs = 0 a1 = a2 = …= ar = b1 = b2 = …= bm = c1 = c2 = …cs
= 0
102
B is linearly independent.
B is a basis of W1 + W2
dim (W1 + W2) = r + m + s = (r + m) + (r + s) – r
= dim W1 + dim W2 – dim (W1 W2).
Note : If W1 W2 = { 0 } then dim (W1 W2 ) = 0
W1 + W2 = W1 W2
dim (W1 W2) = dim W1 + dim W2
Example 8: Let W be subspace of 3 given by
W = {(x, y, z)/ x + y + z = 0 }
W = {(x, y, -x – y)/ x, y }
W = {(x, 0, -x) + (0, y, -y)/ x, y }
W = {x(1, 0, -1) + y(0, 1, -1)/ x, y }
W = L({(1, 0, -1), (0, 1, -1)})
{(1, 0, -1), (0, 1, -1)} generates W ………….(i)
Let a(1, 0, -1) + b(0, 1, -1) = (0, 0, 0)
(a , b, -a –b) = (0, 0, 0)
a = 0, b = 0
{(1, 0, -1), (0, 1, -1)} is linearly independent ………….(ii)
From (i) and (ii) {(1, 0, -1), (0, 1, -1)} is a basis of W
dim W = 2.
Example 9: Let U and W be subspaces of 4 given by
U = {(x, y, z, w)/ x + y + z = 0} and
W = {(x, y, z, w)/ x + w = 0, y = 2z}
We find dim W, dim U, dim (UW) and dim (U+W)
U = {(x, y, z, w)/ x + y + z = 0}
U = {(x, y, -x – y, w)/ x, y, w }
U = {(x, 0, -x, 0) + (0, y, -y, 0) + (0, 0, 0, w)/ x, y, w }
U = {x(1, 0, -1, 0) + y(0, 1, -1, 0) + w(0, 0, 0, 1)/ x, y, w }
W = L({(1, 0, -1, 0), (0, 1, -1, 0), (0, 0, 0, 1)})
{(1, 0, -1, 0), (0, 1, -1, 0), (0, 0, 0, 1)} generates W ………….(i)
Let a(1, 0, -1, 0) + b(0, 1, -1, 0) + c(0, 0, 0, 1) = (0, 0, 0, 0)
103
(a , b, -a –b, c) = (0, 0, 0, 0)
a = 0, b = 0, c = 0
{(1, 0, -1, 0), (0, 1, -1, 0), (0, 0, 0, 1)} is linearly independent
………….(ii)
From (i) and (ii) {(1, 0, -1, 0), (0, 1, -1, 0), (0, 0, 0, 1)} is a basis of U
dim U = 3 …………(iii)
Now W = {(x, y, z, w)/ x + w = 0, y = 2z}
W = {(x, 2z, -x, z)/ x, z }
W = {x(1, 0, -1, 0) + z(0, 2, 0, 1)/ x, z }
{(1, 0, -1, 0), (0, 2, 0, 1)} generates W
It can be easily proved that {(1, 0, -1, 0), (0, 2, 0, 1)} is linearly
independent.
{(1, 0, -1, 0), (0, 2, 0, 1)} is a basis of W
dim W = 2 …………….(iv)
Now (x, y, z, w) UW if and only if (x, y, z, w) U and (x, y, z,
w) W
UW = {(x, y, z, w)/ x + y + z = 0 and x + w = 0, y = 2z}
UW = {(x, y, z, w)/ x + 2z + z = 0 and x + w = 0, y = 2z}
UW = {(x, y, z, w)/ x + 3z = 0 and w = -x, y = 2z}
U W = {(x, y, z, w)/ x = -3z, w = 3z, y = 2z}
U W = (-3z, 2z, z, 3z)/ z }
U W = {z(-3, 2, 1, 3)/ z }
{(3, 2, 1, -3)} generates UW
Since (3, 2, 1, -3) (0, 0, 0, 0)
{(3, 2, 1, -3)} is linearly independent.
{(3, 2, 1, -3)} is a basis of U W
dim (UW) = 1 ……………(v)
Since dim(U + W) = dimU + dim W – dim (U W)
dim(U + W) = 3 + 2 – 1 = 4 = dim 4
U + W = 4
Example 10: Let S = {(1, 2, 1, 0), (0, 0, 1, 1)} be a subset of 4 .
Since dim IR4 = 4 and S contains two elements S is not a basis of
4 .
It can be easily verified that S is linearly independent.
104
Now we extend S to a basis of 4
Consider (1, 0, 0, 0) in 4
If (1, 0, 0, 0) L(S)
Then there exist real numbers a and b such that
(1, 0, 0, 0) = a(1, 2, 1, 0) + b(0, 0, 1, 1)
a = 1 also 2a = 0, not possible
(1, 0, 0, 0) L(S)
S {(1, 0, 0, 0)} is linearly independent. Now it contains 3
elements.
Consider (0, 1, 0, 0) in 4
If (0, 1, 0, 0) L(S {(1, 0, 0, 0)})
Then there exist real numbers a, b and c such that
(0, 1, 0, 0) = a(1, 2, 1 0) + b(0, 0, 1, 1) + c(1, 0, 0, 0)
a + c = 0, 2a = 1, a + b = 0, b = 0
a = ½ and a = 0, not possible
(0, 1, 0, 0) L(S {(1, 0, 0, 0)})
S {(1, 0, 0, 0)} {(0, 1, 0, 0)} is linearly independent
{(1, 2, 1, 0), (0, 0, 1, 1), (1, 0, 0, 0), (0, 1, 0, 0)} is linearly
independent
It a basis of 4 since it has 4 elements.
Check your progress
1) Find the dimension of following subspaces
(i) W = {(x, y) 2 / x = 0}
(ii) W = {(x, y) 2 / y = 0}
(iii) W = {(x, y) 2 / x + y = 0}
(iv) W = {(x, y, z) 3 / x = 0}
(v) W = {(x, y, z) 3 / y = 0}
(vi) W = {(x, y, z) 3 / z = 0}
(vii) W = {(x, y, z) 3 / x + y = 0}
2) Extend {(1, -1)} to a basis of 2
3) Extend {(1, 0, 2), (0, 1, 2)} to a basis of 3
Answers:
1) (i) 1 (ii) 1 (iii) 1 (iv) 1 (v)1 (vi) 1 (vii) 2
105
2) {(1, -1), (1, 0)} or a set containing (1, -1) and any element of 2
which is not a multiple of (1, -1)
3) {(1, 0. 2), (0, 1, 2), (0, 0, 1)} or a set containing (1, 0, 2) and
(0, 1, 2) and any element of 3 which is not a linear combination
of {(1, 0. 2), (0, 1, 2)}
6.4 RANK OF A MATRIX
Every matrix of order mn has m rows and n columns.
Let A =
mnmm
n
n
aaa
aaa
aaa
21
22221
11211
then (a11, a12, …,a1n) n ,
(a11,a21,…am1) m
Hence every row is an element of n and every column is an element
of m .
Set of rows is a subset of n and set of columns is a subset of
n .
In the following section we will relate dimensions of span of these two
sets with the rank of the matrix.
Row rank and Column rank of a matrix
Definition : Let A = [aij] Mm n Let Ai = (ai1, ai2, …,ain) denote the
ith row of A
( 1 i m). Then L({A1, A2,…Am}) is known as the row space of A
and dim(row space of A) is known as the row rank of A.
Let Aj =
mj
j
j
a
a
a
2
1
1 j n denote the jth
column of A. Then L({A1,
A2,…A
m}) is known as the column space of A and dim(column space
of A) is known as the column rank of A.
Note: L({A1, A2,…Am}) is a subspace of n ( every row is an element
of n )
dim(L({A1, A2,…Am})) n
106
{A1, A2,…Am} is generating set of L({A1, A2,…Am})
dim(L({A1, A2,…Am})) m
dim(row space of A) min{m, n}
row rank of A min {m, n}
Similarly L({A1, A
2,…A
m}) is a subspace of
m
column rank of A min {m, n}
In the earlier units we have defined elementary row operations.
We now define elementary column operations on a matrix
(i) Exchanging two columns of a matrix
(ii) Multiplying a column by a non zero scalar
(iii) Adding a scalar multiple of a column to another column
We have the following theorem.
Theorem 9: Elementary row and column operations do not change row
rank or column rank of a matrix.
Proof: Let A = [aij] Mm n
We first show that elementary row operations do not change row rank
of A
L({A1, A2,…Am}) is a row space of A
(i) Interchanging ith and jth row of A, we obtained a matrix B
whose rows are A1, A2, …,Aj,,,,Ai,,,,Am
L({A1, A2,…,Ai,…,Aj,…,Am}) = L({A1, A2,…,
Aj,….,Ai,….Am})
row space of A = row space of B
row rank of A = row rank of B
(ii) Multiplying ith row of A by a nonzero scalar λ, we obtained a
matrix B whose rows are A1, A2,…, λ Ai,…,,Am
L({A1, A2,… λ Ai,…,Am}) = L({A1, A2,…,Ai,…,Am})
row space of A = row space of B
row rank of A = row rank of B
(iii) Adding λ times ith row to jth row of A, we obtained a matrix B
whose rows are A1, A2,…, Ai,…,Aj + λAi,..,Am
Any linear combination of A1, A2,…, Ai,…,Aj + λAi,..,Am is
1A1+ 2 A2 + …+ i Ai +…,+ j(Aj + λAi) +..+ mAm
= 1A1+ 2 A2 + …+( i + λ j) Ai +…,+ jAj +..+ mAm
L({A1, A2,…Am})
107
Similarly Any linear combination of A1, A2,…, Ai,…,Aj, ..,Am
is 1A1+ 2 A2 + …+ i Ai +…,+ jAj +..+ mAm
= 1A1+ 2 A2 + …+( i - λ j) Ai +…,+ j(Aj + λAi) +..+
mAm
L({A1, A2,…, Ai,…,Aj + λAi,..,Am})
L({A1, A2,…Am}) = L({A1, A2,…, Ai,…,Aj + λAi,..,Am})
row space of A = row space of B
row rank of A = row rank of B
Thus elementary row operations do not change row rank of A
We next show that elementary row operations do not change column
rank of A
(i) Interchanging ith and jth row of A, we obtained a matrix
B =
mnm
ini
jnj
n
aa
aa
aa
aa
..................
..................
..................
...............
1
1
1
111
The Columns of B are B1, B
2, …B
n.
1B1 + 2B
2 + ….+ nB
n = 0
1ak1 + 2ak2 + ….+ nakn = 0 for all k = 1, 2, …m
1A
1 + 2A
2 + ….+ nA
n = 0, where A
1, A
2, …A
n are
columns of A.
Column rank of B = column rank of A
(ii) Multiplying ith row of A by a nonzero scalar λ, we obtained a
matrix
B =
mnm
ini
n
aa
aa
aa
..................
..................
...............
1
1
111
, 0
108
1B
1 + 2B
2 + ….+ nB
n = 0
1
1
1
11
.
m
i
a
a
a
+ 2
12
2
2
i
m
a
a .
a
+… + n
mn
in
n
a
a
a
.
1
= 0
1 ai1 + 2 ai2 + ……+ n
ain = 0 for i = 1, 2, …m
( 1ai1 + 2ai2 + ……+ nain) = 0 for i = 1, 2, …m 1ai1 + 2ai2 + ……+ nain = 0 for i = 1, 2, …m
1A1 + 2A
2 + ….+ nA
n = 0, where A
1, A
2, …A
n are
columns of A. Column rank of B = column rank of A
(iii) Adding λ times ith row to jth row of A, we obtained a matrix
B =
mnm
injniji
ini
n
aa
aaaa
aa
aa
.........
.............
........
.......
1
1
1
111
1B
1 + 2B
2 + ….+ nB
n = 0
1ak1 + 2ak2 + ……+ nakn = 0 for k = 1, 2, …m and k j
And 1(aj1+ ai1) + 2(aj2 + ai2) + ……+ n(ajn + ain) = 0
for k = j
1 ak1 + 2 ak2 + ……+ n akn = 0 for k = 1, 2,
…m and k j
And ( 1aj1 + 2aj2 + ……+ najn) + ( 1ai1 + 2ai2 +
……+ nain ) = 0 for k = j 1aj1 + 2aj2 + ……+ najn = 0 1ak1 + 2ak2 + ……+ nakn = 0 for k = 1, 2, …m
1A1 + 2A
2 + ….+ nA
n = 0, where A
1, A
2, …A
n are
columns of A. Column rank of B = column rank of A
109
Thus elementary row operations do not change column rank of A.
Since rows of a matrix A are columns of the matrix At (Transpose of A)
row rank of A = column rank of A
t and column rank of A = row rank of
At
Now as any elementary column operation on A is an elementary row
operation on At which do not change row and column rank of A
t.
elementary column operation on A do not change row and column rank
of A.
Theorem 10: Let A be an m n matrix of row rank r. Then by elementary
row and column operations A can be reduced to the matrix (echelon form)
0000
0000
0.........1....00
0.........0....10
0.........0...01
(r 0)
In particular row rank of A = column rank of A
Proof: If A = 0, row rank of A = column rank of A = 0
Suppose A 0, then A has a non zero entry at ijth position.
By interchanging row A1 with row Ai and column A1 with column A
j,
We get a11 0. If a21, a31,…,am1 are non zero then elementary row
operation which add (-ai1/a11) multiple of row A1 to row Ai, give ai1 = 0 for
all i = 2, 3,..m.
Similarly a12, a13, …,a1n are non zero then elementary column operation
which add (-a1i/a11) multiple of column A1 to column Ai, give a1i = 0 for
all i = 2, 3,..m.
Hence we get A equivalent to
mnmm
n
aaa
aaa
a
.....0
0
.....0
0.....00
32
22322
11
110
Let A* =
mnmm
n
aaa
aaa
.....
.....
32
22322
A* is (m -1) (n -1) matrix
Performing elementary row and column operations on rows
i = 2, 3,…m and columns j = 2, 3,…n of A is same as performing
elementary row and column operations on A* do not change 1st row or 1
st
column of A.
Hence Performing elementary row and column operations on A* as we
performed on A, we get A* i.e. A equivalent to
mnmm
n
aaa
aaa
a
.....0
0
.....0
0.....00
32
33332
22
By repeating this process we get A equivalent to
0000
0000
0.............00
0.........0....0
0.........0...0
22
11
rra
a
a
By multiplying rows A1, A2,…Ar by (1/a11), (1/a22),…(1/arr) respectively
we get
A =
0000
0000
0.........1....00
0.........0....10
0.........0...01
where rows Ar+1, Ar+2,…Am are zero.
row rank of A = column rank of A.
Now we define rank of the matrix.
111
Definition : Let A = [aij] be m n matrix of real numbers. Then the row
rank (dimension of row space of A) or the column rank of A(dimension of
a column space of A) is called rank of A. It is denoted by rank A.
Note : rank A = row rank of A = No of nonzero rows in row echelon form
of A
Let A be a square matrix of order n.If A is invertible then A can
be reduced to identity matrix of order n. Hence A is invertible then rank
A = n.
Example 11: Let A =
2363
3121
Row space of A = L({(1, 2, -1, 3), (-3, -6, 3, -2)}) 4
row rank of A min {4, 2}
Column space of A = L({(1, -3), (2, -6), (-1, 3), (3, -2)}) 2
Column rank of A min {2, 4}
Consider the set {(1, -3), (2, -6), (-1, 3), (3, -2)}
It is linearly dependent ( subset of 2 )
We reduce this set to a basis of column space of A
{(1, -3), (3, -2)} {(1, -3), (2, -6), (-1, 3), (3, -2)})
Also {(1, -3), (3, -2)} is linearly independent (check it)
Column rank of A = 2
rank A = 2
Example 12: We reduce A =
3101
1123
1011
to echelon form
A1,A
2, A
3 and A
4 are columns of A
We perform following elementary column operations
By A2 + (-1)A
1, A
4 + A
1 on A , we get
112
A ~
4111
4113
0001
By (-1)A2, we get
A ~
4111
4113
0001
By A1 + 3A
2, A
3 + A
2, A
4 + 4A
2 we get
A ~
0012
0010
0001
Now we perform following elementary row operations
By A3 + 2A1 and A3 + A2 we get
A ~
0000
0010
0001
Hence rank A = 2
Check your progress
1) Find the rank of the matrix A where
(i) A =
16
73
12
(ii) A =
32
15
20
31
113
(iii) A =
241
312
012
321
2) Determine which of the following matrices are invertible by
finding rank of the matrix.
(i)
512
510
211
(ii)
1513
1012
311
Answers:
1) (i) 2 (ii) 2 (iii) 3
2) (i) Invertible, rank = 3 (ii) invertible, rank = 3
6. 5 SUMMARY
We have define basis of a vector space. Following results are
studied in this unit
A basis of a vector space is maximal linearly independent set
A basis of a vector space is minimal set of generators
Every vector in a vector space is uniquely expressed as a linear
combination of elements of basis.
A vector space has infinitely many bases.
No of elements in a basis is a dimension of a vector space
Every linearly independent set in a finitely generated vector space
can be extended to a basis of a vector space
A basis can be obtained from a set of generators of finitely
generated vector space
The dimension of a subspace can not exceed dimension of vector
space
There is a relation between the dimension of two subspaces,
dimension of their intersection and dimension of their sum
For a matrix of order m n, we have defined row space and column
space. Their dimensions are respectively known as row rank and column
114
rank. The rank of the matrix is the value of row rank which is same as
column rank.
6.6 UNIT END EXERCISE
Theory:
1) Define basis of a vector space.
2) Define dimension of a vector space.
3) Define row rank, column rank and hence rank of a matrix.
Problems:
1) Find the coordinate vector of v relative to the basis
{(1, 1, 1), (1, 1, 0), (1, 0, 0)} of 3 where (i) v = (4, -3, 2)
(ii) v = (1, 2, 3)
2) Let V be a vector space of matrices of order 2. Find the coordinate
vector of the matrix A in V relative to the basis
{
11
11,
01
10,
00
11,
00
01} where A =
74
32
.
3) Find the dimension of the following subspaces
(i) W = { (x, y)/ 2x + y = 0} of 2
(ii) W = { (x, y, z)/ x = 2y, z = 5y} of 3
(iii) W= { (x, y, z)/ x – 3y = 0, y – z = 0} of 3
(iv) W = { (x, y, z)/ 2x + 3y + z = 0} of 3
(v) W = {
dc
ba/ a = d and b = c} of M2
4) If W1 = {(x, y, z)/ x + y + z = 0} and W2 = {(x, y, z)/ x – y + z
= 0} are subspaces of 3 then find the dimension of W1 W2.
5) If W1 = {(x, y, z)/ x = y } and W2 = {(x, y, z)/ 2x + 4y + 2z = 0}
are subspaces of 3 the find the dimension of W1, W2, W1W2,
W1 + W2.
6) If U and W are subspaces of 8 such that dim U = 3, dim W = 5
and U + W = 8 . Show that U W = { 0 }.
7) Extend {(1, 0, 2)} to a basis of 3 .
Answers
1) (4, -3, 2) = 2(1, 1, 1) + (-1) (1, 1, 0) + 2(1, 0, 0)
115
(1, 2, 3) = 3(1, 1, 1) + (-1) (1, 1, 0) + (-1)(1, 0, 0)
2)
74
32 = (-7)
11
11 + 11
01
10+ 14
00
11 + (-5)
3) (i) 1 (ii) 1 (iii) 1 (iv) 2 (v) 2
4) W1 W2= { (x, y, z)/ x + z = 0, y = 0} So dimW1 W2 = 1
5) dim W1 = 2, dim W2 = 2, dim W1 W2 = 1 dim W1 + W2 = 3
7) A basis of 3 containing (1, 0, 2) is {(1, 0, 0), (0, 1, 0), (1, 0, 2)}
116
7
INNER PRODUCT SPACES
Unit Structure:
7.0 Objectives
7.1 Introduction
7.2 Inner product
7.3 Norm of a vector
7.4 Summary
7.5 Unit End Exercise
7.0 OBJECTIVES
This chapter would make you to understand the following concepts
Inner product of vectors of vector space
Inner product space
Norm of a vector
Unit vector
Cauchy Schwarz inequality
Triangle inequality
7. 1 INTRODUCTION
By defining vector space we have generalized vectors together
with their addition and scalar multiplication. The definition of vector
space does not include product of the vectors. The product of the vectors
(dot product) gives the definition of angle between two vectors and length
of the vector. We could not learn various geometrical properties of
vectors without the notion of angle and length.
In this unit we wish to extend these ideas to the vectors of vector
space. We introduce a generalization of the dot product of the vectors of
the vector space which we call as inner product. As the length of a vector
is expressed in terms of dot product, we introduce length of vector of a
vector space in terms of inner product which we call as the norm of a
vector.
117
7. 2 INNER PRODUCT ON A VECTOR SPACE
We recall the definition of dot product of vectors in IR2.
Let x = (x1, x2) and y = (y1, y2). The dot product of x and y (x . y) is
defined as x . y = x1y1 + x2y2
We introduce now dot product in IRn
Let x = (x1, x2,…..,xn), y = (y1, y2, ….,yn) IRn. We define dot product of
x and y as x . y = x1y1 + x2y2 + …….+ xnyn =
n
i
ii yx1
Since x, y n n and x . y , dot product „ . „ is a real valued
function defined on n n .
Result : The dot product in IRn has the following properties
For x = (x1, x2,…..,xn), y = (y1, y2, ….,yn), z = (z1, z2, …..zn) n ,
i) x . x 0 and x . x = 0 if and only if x = 0
ii) x . y = y . x
iii) ( x) . y = (x . y)
iv) (x + y) . z = x . z + y . z
Proof: For x, y, z n and
i) x . x = x1x1 + x2x2 + …..+ xnxn
= x12 + x2
2 + …..+ xn
2 0
x . x = 0 x12 + x2
2 + …..+ xn2 = 0
x12 = x2
2 = …..xn
2 = 0
x1 = x2 = ……= xn = 0
(x1, x2,…..,xn) = (0, 0, ….,0)
x = 0
ii) x . y = x1y1 + x2y2 + …….+ xnyn
= y1x1 + y2x2 + …..+ ynxn
= y . x
iii) 1 2, ,.....,x x x xn
1 21 2. .....x x x x nny y y y
1 1 2 2 ....... n nx y x y x y
1 1 2 2 ..... n nx y x y x y
118
.x y
iv) x + y = (x1 + y1, x2 + y2, …….,xn + yn)
(x + y) . z = (x1 + y1)z1 + (x2 + y2)z2 + …….+ (xn + yn) zn
= x1z1 + y1z1 + x2z2 + y2z2 + …..+ xnzn + ynzn
= (x1z1 + x2z2 + …..+ xnzn) + (y1z1 + y2z2 + ….+ ynzn)
= x . z + y . z
Now we generalise the concept of dot product to any vector space
V. This dot product must satisfy the properties stated in the above result.
We call this dot product on V as an inner product.
Definition : An inner product (generalised dot product) on a vector space
V is defined as a real valued function < , > defined on VV satisfying
following properties.
For x, y, z V and
(i) <x, x> 0 and <x, x> = 0 if and only if x = 0
(ii) <x, y> = <y, x>
(iii) < x, y> = <x, y>
(iv) <x + y, z> = <x, z> + <y, z>
(V,< , >) is called an inner product space over .
Clearly n is an inner product space where <x, y> = x . y (usual dot
product)
Note : From the definition of an inner product we observe following
(i) <x, 0> = <0, x> = 0 for all x in V
Since <0, x> = <00, x> = 0<0, x> = 0
(ii) <z, x + y> = <z, x> + <z, y>
Since <z, x + y> = <x + y, z>
= <x, z> + <y, z>
= <z, x> + <z, y>
(iii) <x, y> = <x, y>
Since <x, y> = < y, x> = <y, x> = <x, y>
(iv) <x, y + z> = <x, y> + <x, z>
Since <x, y + z> = <x, y> + <x, z>
= <x, y> + <x, z>
Let us study some examples of inner product spaces
119
Example 1: Let V = 2
Let x = (x1, x2), y = (y1, y2)
Define <x, y> = 2x1y1 + 5x2y2 ……………(*)
Let x, y, z 2
x = (x1, x2), y = (y1, y2), z = (z1, z2)
(i) <x, x> = 2x1x1 + 5x2x2
= 2x12 + 5x2
2 0
<x, x> = 0 2x12 + 5x2
2 = 0
x12 = 0, x2
2 = 0
x1 = 0, x2 = 0
<x, x> = 0 x = 0
(ii) , <x, y> = 2x1y1 + 5x2y2
= 2y1x1 + 5y2x2
= <y, x>
(iii) For , x = ( x1, x2) < x, y> = 2 x1y1 + 5 x2y2
= (2x1y1 + 5x2y2)
= <x, y>
(iv) x + y = (x1 + y1, x2 + y2) <x + y, z> = 2(x1 + y1)z1 + 5(x2 + y2) z2
= (2x1z1 + 5x2z2) + (2y1z1 + 5y2z2)
= <x, z> + <y, z>
Hence 2 is an inner product space under the inner product < , > is
defined in (*)
Note : From Example 1 it is clear that different inner products can be
defined on the same vector space.
Example 2: We know that C [a, b] is a vector space of all continuous
functions defined on [a, b] (a, b ; a < b)
Let f, g C[a, b], define <f, g> = b
a
tdtgtf )()(
For f, g, h C[a, b]
(i) <f, f> = b
a
tdtftf )()(
120
= b
a
tdtf 2))(( 0 for all t [a, b] (Since (f(t))2 = f
2(t) 0)
Now if f = 0 then <f, f> = 0
If <f, f> = 0 and f 0 then there exist t0 [a, b] such that f(t0) 0
f2(t0) 0
Since f is continuous there exist an interval (t - , t + )
around t such that f2(t) >0 for all t (t - , t + ).
t
t
tdtftf )()( > 0
0 = <f, f> = b
a
tdtftf )()( = t
a
tdtftf )()( +
t
t
tdtftf )()( +
b
t
tdtftf
)()( > 0
Contradiction
Hence <f, f> = 0 f = 0 <f, f> = 0 f = 0
(ii) <f, g> = b
a
tdtgtf )()(
= b
a
tdtftg )()(
= <g, f>
(iii) For , < f, g> = b
a
tdtgtf )()()(
= b
a
tdtgtf )()(
= b
a
tdtgtf )()(
= <f, g>
(iv) <f+ g, h> =
b
a
tdthtgf )()()(
=
b
a
tdxhtgtf )())()((
121
= b
a
tdthtf )()( + b
a
tdthtg )()(
= <f. g> + <g, h>
Thus C[a, b] is an inner product space.
Example 3: Let V = C, a vector space of complex numbers
Let z , w C . Define < z , w > = Re( z w ) (Real part of complex number
z w )
For z , w , t C
(i) If z = a + i b , z z = a2 + b
2
< z , z > = Re( z z ) = a2 + b
2 0
< z , z > = 0 Re( z z ) = 0
a2 + b
2 = 0
a = 0, b = 0
z = 0
(ii) z = a + i b, w = c + i d, z = a - i b, w = c - i d
< z , w> = Re( z w ) = ac + bd,
<w, z> = Re( w z ) = ac + bd < z , w > = < w , z >
(iii) For IR, z = a + i b
< z , w > = Re( z w )
= ac + bd
= (ac + bd)
= < z , w >
(iv) < z + w , t > = Re(( z + t ) w )
= Re( z w + t w )
= Re( z w ) + Re( t w )
= < z , t > + < w , t >
Hence C is an inner product space.
Example 4: M2 : Set of all 2 2 matrices with real entries
Let A =
43
21
aa
aa then tr(A) = Trace of A = a1 + a4.
If A, B M2 , define <A, B> = tr(ABt)
122
Let B =
43
21
bb
bb then B
t =
42
31
bb
bb and AB
t =
44332413
42312211
babababa
babababa
tr(ABt) = a1b1 + a2b2 + a3b3 + a4b4
<A, B> = a1b1 + a2b2 + a3b3 + a4b4
For A, B, C M2
(i) <A, A> = tr(AAt) = a1a1 + a2a2 + a3a3 + a4a4
= a12 + a2
2 + a3
2 + a4
2 0
<A, A> = 0 a12 + a2
2 + a3
2 + a4
2 = 0
a12 = a2
2 = a3
2 = a4
2 = 0
a1 = a2 = a3 = a4 = 0
43
21
aa
aa = O
A = O
(ii) <A, B> = tr(ABt)
= a1b1 + a2b2 + a3b3 + a4b4
= b1a1 + b2a2 + b3a3 + b4a4
= tr(BAt) (Verify)
(iii) For IR, tr( A) = tr(A)
< A, B> = tr(( A)Bt)
= tr( (ABt))
= tr(ABt)
= <A, B>
(iv) <A + B, C> = tr((A + B)Ct)
= tr(ACt + BC
t)
= tr(ACt) + tr(BC
t)
= <A , C> + <A, B>
Example 5: We know that P2[x] = { a0 + a1x + a2x2/ a0, a1, a2 } is a
vector space.
For p(x) = a0 + a1x + a2x2 and q(x) = b0 + b1x + b2x
2 in P2[x] define
<p(x), q(x)> = p(0)q(0) + p(1)q(1) ……(*)
Does this definition give inner product? i.e. Is P2[x] inner product space?
Let p(x) = x – x2
Clearly p(x) P2[x]
123
Now p(0) = 0, p(1) = 0 <p(x), p(x)> p(0)p(0) + p(1)p(1) = 0 + 0 = 0 but p(x) 0 <p(x), p(x)> = 0 p(x) = 0 is not true.
<p(x), q(x)> defined in (*) is not an inner product
Hence P2[x] is not an inner product space
Check your progress
Show that following are inner product spaces over .
1) (2 , < , >), Where <x, y> = x1y1 + 2x2y2
x = (x1, x2), y = (y1, y2)
2) (2 , < , >), where <x, y> = 3x1y1 + 4x2y2
x = (x1, x2), y = (y1, y2)
3) (3 , < , >), where <x, y> = x1y1 + 2x2y2 + 3x3y3
x = (x1, x2, x3), y = (y1, y2, y3)
4) (P2[x], < , >), where <p(x), q(x)> = a0b0 + a1b1 + a2b2
For p(x) = a0 + a1x + a2x2 and q(x) = b0 + b1x + b2x
2
7. 3 NORM OF A VECTOR
Definition : Let V be an inner product space. Let v V. Norm of v(length
of v ) is denoted by v and defined as v = vv,
Example 6: We know that IRn is an inner product space with usual dot
product.
For ),......,,( 21 nxxxx x = xx, = 22
2
2
1 .... nxxx
Note: Since v IR Norm is a function from V to IR known as the
norm function.
Example 7: Let us find norm of (3, 4) 2 with respect to usual inner
product i.e. dot product and with respect to the inner product given in
Examlpe 1.
(i) The usual inner product in 2 is <x, y> = x1y1 + x2y2 , for x =
(x1, x2),
y = (y1, y2)
x = xx, =
2
2
2
1 xx
)4,3( = 22 43 = 5
124
(ii) The inner product given in example 1 is <x, y> = 2x1y1 + 5x2y2
x = xx, = 2
2
2
1 52 xx
)4,3( = )4(5)3(2 22 = 98
Theorem 1: Let V be an inner product space. The norm function has the
following properties. For x V,
(i) x 0 and x = 0 x = 0
(ii) x = x
Proof: For x V,
(i) x = xx, 0 (< x , x > 0)
x = 0 xx, = 0 ,x x = 0 x = 0
(ii) x = xx , = xx, = xx,2 = xx,
= x
Definition : Let x ( 0) be in an inner product space V.
Then since x
x =
x
x = 1,
x
x is said to be a unit vector in the
direction of x .
Theorem 2: (Cauchy – Schwarz inequality) : Let V be an inner product
space.
If x , y V then yx, x y
The equality holds if and only if x y or y x for some in .
Proof: Let ,x y V.
If 0y then , ,0 0x y x and x y = x 0 = 0
equality holds.
Suppose 0y
Define f : by f( t ) = ytx
0f t for all t
Also f t = < x + t y , x + t y >
125
= < x , x > + < x , t y > + < t y , x > + < t y , t y >
= < x , x > + t < x , y > + t < y , x > + t 2< y , y >
f t = < x , x > + 2 t < x , y > + t 2< y , y > ……………..(i)
Differentiating with respect to t we get,
f t = 2< x , y > + 2 t < y , y >
If t0 is an extremum (maximum or minimum) the f „( t
0) = 0 2< x , y > + 2 t
0<y , y > = 0
t0 =
yy
yx
,
,, < y , y > 0 ……………….(ii)
f t = 2< y , y > > 0 for all t
f is minimum at t = t0
0 f( t0) f( t ) for all t
Putting the value of t0 from (ii) in (i), we get
< x , x > + 2
yy
yx
,
,< x , y > + (
yy
yx
,
,)2 < y , y > 0
x - 2
yy
yx
,
, 2
+
yy
yx
,
, 2
0
2
x - 2
2,
y
yx 0
2, yx
2x
2y
yx, x y
The equality holds if and only if f( t0) = 0
i.e. 0 0, 0x t y x t y
i.e 0 0x t y
i.e 0x t y
Theorem 3: (Triangle inequality): Let V be an inner product space.
If ,x y V then yx x + y
Proof: Consider 2yx
= yxyx ,
= xx , + xy , + yx , + yy ,
= 2
x + 2 | yx , | + 2
y
126
( x + 2 x2
y +2
y )2
(By Cauchy
Schwarz inequality)
2yx
( x + 2
y )2
By taking non negative square roots of both sides
yx x + y .
Note : yx = x + y if and only if x = y or y = x for some
in .
Since 2yx =
2x + 2 | yx , | +
2y
By Cauchy Schwarz inequality yx, = x y if and only if x =
y or y = x for some in .
Corollary : For x , y V, yxyx
Proof: For , ,x y V x x y y
By Triangle inequality
x = yyx )( yx + y
x - y yx ……………(i)
Now y y x x
y = xxy )( xy + x
y - x xy
- ( x - y ) ( xy = )( yx = yx ) ………….(ii)
yxyx
Note : yx denotes the distance between vectors x and y in the inner
product space V.
Check your progress
1) Find the norm of a matrix
11
21 with respect to the norm given in
the example 4.
127
2) Find the norm of the function f (x ) = x2 + x with respect to the norm
given in the example 2
Answers:
1) 7
2) 6/5
7. 4 SUMMARY
In this unit we have defined inner product in a vector space. This
concept is similar to the concept of dot product of the vectors.
Using inner product we have defined norm of a vector which is
nothing but length of a vector and a unit vector.
We have studied following results
Cauchy Schwarz inequality
Triangle inequality
7. 5 UNIT END EXERCISE
1) Show that following are inner product spaces over .
(i) (2 , < , >), Where <x, y> = 2x1y1 + x1y2 + x2y1 + x2y2
x = (x1, x2), y = (y1, y2)
(ii) (2 , < , >), Where <x, y> = x1y1 - x1y2 - x2y1 + 3x2y2
x = (x1, x2), y = (y1, y2)
(iii) (C[a, b], < , >), where <f, g> = b
a
tdtwtgtf )()()(
Where w(t) 0 for all t in [a, b] and w c[a, b]
(iv) (C[1 , e], < , >), <f, g> = e
tdtgtft1
)()(log
2) If V is an inner product space and x, y V then prove following
(i) <x, y> = 0 || x + y || = || x – y ||
(ii) <x, y> = 0 || x + y ||2 = || x ||
2 + || y ||
2
(iii) <x, y> = 0 || x + cy || || x || for c
(iv) <x + y, x – y> = 0 || x || = || y ||
(v) 4 <x, y> = || x + y ||2 - || x – y ||
2
(vi) || x + y ||2 + || x – y ||
2 = 2|| x ||
2 + 2|| y ||
2
128
3) Let V be an inner product space and x, y V. If || x || = 3, || x + y
|| = 4,
|| x – y || = 6, find || y ||
Answers:
1) 17
129
8
ORTHOGONALITY
Unit Structure:
8. 0 Objectives
8.1 Introduction
8.2 Angle betweennon zero vectors
8.3 Orthogonal Prrojection on to a line
8.4 Orthogonal vectors
8.5 Orthogonal and Orthonormal sets
8.6 Gram Schmidt Orthogonalisation Process
8.7 Orthogonal Complement of a set
8.8 Summary
8.9 Unit End Exercise
8. 0 OBJECTIVES
This chapter would make you to understand the following concepts
Angle between vectors in an inner product space
Orthogonal vectors
Orthonormal vectors
Orthogonal projection on to a line
Orthogonal set
Orthonormal set
Orthogonal basis
Orthonormal basis
Orthogonal compliment of a set
Gram Schmidt method to find orthogonal basis
8. 1 INTRODUCTION
In unit VII we have define inner product on vector space. Using
inner product we have define length of a vector i.e. norm of a vector. The
definition of norm gives the distance between two vectors.
In this unit we see how to define an angle between two non zero
vectors. Once the angle is defined we able to learn perpendicular vectors
130
i.e. orthogonal vectors. We introduce orthonormal vectors using
orthogonal and unit vectors.
8. 2 ANGLE BETWEEN NON ZERO VECTORS
Definition : Let V be an inner product space. If x and y are non zero
vectors then by Cauchy Schwarz inequality is ,x y x y
yx
yx ,
1
- 1 yx
yx ,
1
There exist unique in [0, ] such that ,x y
Cosx y
This is known as the angle between vectors x and y.
Example 1: The angle between vectors (1, 1, -1) and (0, -1, -1) in IR3
with usual inner product (dot product) is given by
Cos =)1,1,0()1,1,1(
)1,1,0(.)1,1,1(
= 222222 )1()1(0)1(11
)1()1()1(101
= 0
= /2
8. 3 ORTHOGONAL VECTORS
Two vectors x, y in an inner product space are perpendicular if and only if
the angle between them / 2 .
i.e. if and only if Cos = 0
i.e. if and only if yx
yx ,= 0
i.e. if and only if ,x y = 0
Definition : Two vectors x, y in an inner product space are said to be
orthogonal (perpendicular) if and only if ,x y = 0.We denote this by
x y .
Note: We can verify that vectors (1, 1, -1) and (0, -1, -1) in 3 are
orthogonal with respect to usual inner product i.e. dot product.
131
Now we check the orthogonality of these two vectors with respect to
inner product defined as
<x, y> = x1y1 + 2x2y2 + 3x3y3, where x = (x1, x2, x3); y = (y1, y2, y3)
…………..(i)
Then x = (1, 1, -1); y = (0, -1, -1) gives
<x, y> = (1)(0) + 2(1)(-1) + 3(-1)(-1) = 1 0
(1, 1, -1) and (0, -1, -1) are not orthogonal with respect to the inner
product defined by (i).
We now give some geometrical applications of orthogonal vectors.
Theorem 1: (Pythagoras Theorem): Let V be an inner product space.
x and y are orthogonal vectors in V if and only if 2
yx = 2x +
2y
Proof: 2yx = yxyx ,
= xx , + xy, + yx , + yy,
= 2x + 2 yx , +
2y
Thus 2yx =
2x + 2
y if and only if yx , = 0
Hence 2yx =
2x + 2
y if and only if x and y are orthogonal
vectors in V.
Theorem 2: The sum of the squares of the diagonals of a parallelogram is
equal to the sum of the squares of its sides.
Proof: To prove this result let us translate it in to the language of linear
algebra.
Let three vertices of the parallelogram be end point of the vectors 0, x
and y.
x x + y
A B
O y C
So, by vector addition fourth vertex is the end point of the vector x + y.
To show that OB2 + AC
2 = 2OA
2 + 2OC
2
132
The length of the side OA is x and that of side OC is y and diagonal
OB is yx . The end points of diagonal AC are x and y. So length of
AC is yx
To show that 2yx + 2yx = 2
2x + 2 2
y
Now, 2yx =
2x + 2 yx , + 2
y
2yx =
2x - yx , + 2
y
By adding these two equations we get the result.
Check your progress :
1) Find whether the following vectors are orthogonal with respect to
the usual inner product (dot product) in 3 .
(i) (1, 0, 1), (0, 1, 0) (Ans: Yes)
(ii) (1, 2, 3), (0, -6, 4) (Ans: Yes)
(iii) (2, -3, 4), (-1, 3, 5) (Ans: No)
2) Find whether the following vectors are orthogonal with respect to
the inner product in 3 defined as
<x, y> = x1y1 + 2x2y2 + 3x3y3, where x = (x1, x2, x3); y = (y1, y2,
y3)
(i) (1, 2, 1), (-3, 3, 4) (Ans : No)
(ii) (1, 2, 3), (-3, -2, -1) (Ans : No)
8. 4 ORTHOGONAL PROJECTION ON TO A LINE
x
u (x.u)u
While studying vectors in plane 2 , we have seen that If x is
any vector and u is an unit vector then the orthogonal projection of x on u
is (x . u) u = (|x| Cos )u where is the angle between x and u.
Now we generalize this to vectors of an inner product space
Definition : Let V be an inner product space and u be an unit vector in V.
The projection of v V along u is defined as <v, u>u and is denoted by
Pu(v).
133
Theorem 3: Let V be an inner product space and u be an unit vector. Then
for any v V the distance between v and Pu(v). is smaller than the
distance between v and u for all .
i.e. )(vPv u uv for all .
The equality holds if and only if Pu(v) = u.
Proof: Consider < )(vPv u , u > = < v , u > - < )(vPu , u >
= < v , u > - << v , u >u , u >
= < v , u > - < v , u > < u , u >
= < v , u > - < v , u > u
= < v , u > - < v , u > (u is an unit vector,
u = 1)
= 0 < )(vPv u , u > = 0
)(vPv u is orthogonal to u
)(vPv u is orthogonal to u for all in
Since )(vPu is along u )(vPu - u is also along u
< )(vPv u , )(vPu - u > = 0 …………….(i)
2uv = 2)()( uvPvPv uu
= 2)( vPv u +
2)( uvPu (From (i) and Pythagoras
theorem)
2uv
2)( vPv u for all in .
)(vPv u uv for all in .
Note: For any vector w in V )(vPw = www
wv
,
,
Check your progress
1) Find the orthogonal projection of (1, 1) along (1, -2) with respect
to the usual inner product. Ans : 1 25 5
,
2) Find the shortest distance of point (1, 1, 1) from (3, 0, 0)
8. 5 ORTHOGONAL AND ORTHONORMAL SETS
We know that every finite dimensional vector space has a basis.
Now we look for a basis having some additional properties.
134
Definition : A subset S of an inner product space V is said to be an
orthogonal set if and only if S does not contain zero and <x, y> = 0 for all
x y in S..
S is an orthogonal basis of V if and only if S is a basis of V and S is an
orthogonal set
Example 2: We have seen in unit VI that the subset
S = {(2, 0, 0), (0, 2, 0), (0, 0, 2)} is a basis of 3 .
Further, <(2, 0, 0), (0, 0, 2)> = <(2, 0, 0), (0, 2, 0)> = <(0, 2, 0),
(0, 0, 2)> = 0
S is an orthogonal set S is an orthogonal basis of V.
Definition : A subset S of an inner product space V is said to be an
orthonormal set if and only if
(i) S does not contain zero
(ii) <x, y> = 0 for all x y in S.
(iii) || x || = <x, x>1/2
= 1 for all x in S
S is an orthonormal basis of V if and only if S is a basis of V and S is an
orthonormal set
Example 3: We have seen in unit VI that the subset S = {(1, 0, 0),
(0, 1, 0), (0, 0, 1)} is a basis of 3 .
Further, <(1, 0, 0), (0, 0, 1)> = <(1, 0, 0), (0, 1, 0)> = <(0, 1, 0),
(0, 0, 1)> = 0
And || (1, 0, 0) || = || (0, 1, 0) || = || (0, 0, 1) || = 1
S is an orthonormal set S is an orthonormal basis of V.
Let V be a finite dimensional inner product space and {v1, v2,…., vn} be a
basis of V.
If v V then v = 1v1 + 2v2 + ….+ nvn and this expression for v is
unique.
But we do not have any clue about the values of 1, 2, …., n.
However for the orthonormal basis the story is different.
Theorem 4: Let V be a finite dimensional inner product space.
If {v1, v2,…., vn} is an orthonormal basis of V and x = 1v1 + 2v2 + ….+
nvn.
Then i = < x, vi> for 1 i n, and
135
|| x ||2 = <x, v1>
2 + <x, v2>
2 + …..+ <x, vn>
2 =
n
i
ii vvx1
,
Proof: x = 1v1 + 2v2 + ….+ nvn
For 1 j n,
Consider <x, vj> = < 1v1 + 2v2 + ….+ nvn, vj>
= 1<v1, vj> + 2<v2, vj> + ….+ j<vj, vj> + ….+ n<vn, vj>
= j <vj, vj > ( {v1, v2,…., vn} is orthonormal <vi, vj> = 0 if i j
and <vj, vj> = 1)
= j
Thus, x = <x, v1>v1 + <x, v2>v2 + ……+ <x, vn> vn =
n
i
ii vvx1
,
|| x ||2 = <x , x>
= <
n
i
ii vvx1
, ,
n
j
jj vvx1
, >
= (
n
i
ivx1
,
n
j
jvx1
, ) vjvi,
= (
n
i
ivx1
,
n
i
ivx1
, ) ii vv , ( vjvi, = 0 if i j )
= (
n
i
ivx1
, )2 ( ii vv , = 1)
Theorem 5: Let V be an inner product space. An orthogonal subset of V
is linearly independent.
Proof: Let S be an orthogonal subset of V. If v1, v2, …vn S and 0
Then for 1 j n, (j fixed)
n
i
n
i
ijiijii avvavva1 1
,,
(Since 0, ji vvji )
But 01
n
i
ii va
0,0 jv
0jv for each j
S is linearly independent.
136
Note : We know that If u V then u
u is the unit vector along u.
Let { kuuu ......,,, 21 } be an orthogonal set in V.Then
1
1
u
u=
2
2
u
u=……..=
k
k
u
u= 1
Also jiifuu
uu
u
uj
u
u
ji
ji
ji
i
0,
,
= jiifuu
uu
ii
ii
1
,
{k
k
u
u
u
u
u
u........,,,
2
2
1
1} is an orthonormal set.
Check your progress
1) Prove that following sets are orthogonal
(i) {(2, -4), (4, 4)} in IR2
(ii) {(1, 1, 1), (-1, 1, 0), (1, 1, -2) in IR3
8.6 GRAM SCHMIDT ORTHOGONALISATION
PROCESS
Every orthogonal set in an inner product space is linearly
independent and every vector space has a basis.
So for a inner product space we want to construct an orthogonal
basis from the given basis. This construction is known as Gram – Schmidt
orthogonalisation process.
The proof of the following theorem gives this process.
Theorem 6: If { v1, v2, …,vk} is a linearly independent subset of an inner
product space V then there exist an orthogonal set {u1, u2, ….uk} such that
L({ v1, v2, …,vk}) = L({u1, u2, ….uk})
Proof: Proof is by induction
For k = 1, we take u1 = v1
Being a singleton set, {u1} is orthogonal.
Also L({u1}) = L({v1})
137
For k = 2, we take 1
11
12
22,
,u
uu
uvvu
Then
1
11
12
2121,
,,, u
uu
uvvuuu
=
11
11
12
21 ,,
,, uu
uu
uvvu
= 1221 ,, uvvu
= 0
Now if u2 = 0 then 1
11
12
22,
,u
uu
uvvu
= 0
1
11
12
2,
,u
uu
uvv
= cu1 = cv1 (Since u1 = v1)
{ v1, v2} is linearly dependent
Since { v1, v2, …,vk} is linearly independent, its subset {v1, v2} is
linearly independent Contradiction. Hence u2 0 . {u1, u2} is an orthogonal set. Also L({u1, u2}) = L({v1, v2})
Suppose the result is true for i (1 i k – 1)
i.e. If { v1, v2, …,vi} is a linearly independent subset of an inner product
space V then there exist an orthogonal set {u1, u2, ….ui} such that
L({ v1, v2, …,vi) = L({u1, u2, ….ui})
Now we define
1
1 ,
,k
i
i
ii
ik
kk uuu
uvvu
Then for 1 j k - 1, (j fixed)
1
1
,,
,,
k
i
ji
ii
ik
kjk uuuu
uvvuu
=
1
1
,,
,,
k
i
ji
ii
ik
jk uuuu
uvuv
=
jj
jj
jk
jk uuuu
uvuv ,
,
,,
138
Since 0i ju ,u if i j
= jkjk uvuv ,,
= 0.
uk 0 , otherwise vk L({u1, u2, …..,uk-1}) = L({v1, v2, …..vk-1})
Contradiction, since {v1, v2, …..vk} is linearly independent.
Further, uk L({u1, u2, ….,uk-1, vk}) = L({v1, v2, ……vk-1, vk}) L({u1, u2, ….,uk-1, uk}) L({v1, v2, ……vk-1, vk})
Similarly L({v1, v2, ……vk-1, vk}) L({u1, u2, ….,uk-1, uk}) L({v1, v2, ……vk-1, vk}) = L({u1, u2, ….,uk-1, uk})
Thus by induction the result is true for all k.
Note: Gram Schmidt process is to get an orthogonal set {u1, u2,..,uk} from
linearly independent set {v1, v2,….vk}. Where u1 = v1 and
1
1 ,
,k
i
i
ii
ik
kk uuu
uvvu
for k = 2, 3, …..n
Corollary : A finite dimensional inner product space has an orthonormal
basis.
Proof: Let V be a finite dimensional inner product space. V has a finite basis.
Let {v1, v2, …..vn} be a basis of V.
We apply Gram Schmidt process to get an orthogonal set {u1, u2,…,un}
such that
L({v1, v2, ……, vk}) = L({u1, u2, ….,, uk}) = V {u1, u2, ….un} is an orthogonal basis of V.
From Note 8.4.1 { n
n
u
u
u
u
u
u........,,,
2
2
1
1 } is orthonormal basis
of V.
Example 4: We now apply Gram Schmidt process to convert linearly
independent set {(0, 1, -1), (1, 2, 1), (1, 0, 1) } of 3 to an orthogonal set
Let {v1, v2, v3} = {(0, 1, -1), (1, 2, 1), (1, 0, 1) } …………(i)
Let u1 = v1 = (0, 1, -1) ………(ii)
139
1
11
12
22,
,u
uu
uvvu
Now <u1, u1> = <(0, 1, -1), (0, 1, -1)> = 0 + 1 + 1 = 2
<v2, u1> = <(1, 2, 1), (0, 1, -1)> = 2 + (-1) = 1
u2 = (1, 2, 1) - (1/2)(0, 1, -1) = (1, 2, 1) – (0, 1, -1/2) = (1, 1, 3/2)
u2 = (1, 1, 3/2) ………………(iii)
2
22
22
1
11
12
33,
,
,
,u
uu
uvu
uu
uvvu
)2/3,1,1()2/3,1,1(,)2/3,1,1(
)2/3,1,1(,)1,0,1()1,1,0(
)1,1,0(,)1,1,0(
)1,1,0(,)1,0,1()1,0,1(3
u
)2/3,1,1(17
10)1,1,0(
2
1)1,0,1(3 u
)17
15,
17
10,
17
10()
2
1,
2
1,0()1,0,1(3 u
)34
32,
34
3,
17
10(3 u
{u1, u2, u3 } = {(0, 1, -1), (1, 1, 3/2), )34
32,
34
3,
17
10( } is an
orthogonal set.
Example 5: We now apply Gram Schmidt process to find an orthonormal
basis for the space of solutions of the linear equation 3x – 2y + z = 0
Let W = { (x, y, z) 3 / 3x – 2y + z = 0}
W is a subspace of 3 and it is the space of solutions of the linear equation
3x – 2y + z = 0
Now W = {(x, y, z)/ z = - 3x + 2y}
W = {(x, y, - 3x + 2y) / x, y }
W = {(x, 0, - 3x) + (0, y, 2y) / x, y }
W = { x(1, 0, -3) + y(0, 1, 2) /x, y }
W = L({(1, 0, -3) + (0, 1, 2)})
Clearly S = {(1, 0, -3), (0, 1, 2)} is linearly independent.
S is a basis of W
140
Now we find an orthonormal basis of W from S by Gram Schmidt process
Let v1 = (1, 0, -3) and v2 = (0, 1, 2)
u1 = v1 = (1, 0, -3)
1
11
12
22,
,u
uu
uvvu
=)3,0,1(
)3,0,1(,)3,0,1(
)3,0,1(,)2,1,0()2,1,0(
= )3,0,1(10
6)2,1,0(
= )5
1,1,
5
3(
{(1, 0, -3), )5
1,1,
5
3( } is an orthogonal basis of W
{ ,)3,0,1(
)3,0,1(
)5
1,1,
5
3(
)5
1,1,
5
3(
} = { ,10
)3,0,1(
5
35
)5
1,1,
5
3(
}
= { )10
3,0,
10
1(
, )
35
1,
35
5,
35
3( } is an orthonormal basis
of W.
8. 7 ORTHOGONAL COMPLEMENT OF A SET
Let x V. Let x = { y V / <x, y> = 0}
Since <x, 0> = 0, x is non empty.
Let y, z x and a, b
Since y, z x , <x, y> = <x, z> = 0
Consider <x, ay + bz> = <x, ay> + <x, bz> = a<x, y> + b<x, z> = 0
ay + bz x
Hence x is a subspace of V
Definition: Let V be an inner product space. If x V then
x = { y V / <x, y> = 0} is known as the orthogonal complement of x
141
Example 6: Let x = (a, b) 2 . Let x 0.
x = {v =(x, y)/ <x, v> = ax + by = 0}
Thus x is a line ax + by = 0 passing through origin and perpendicular to
(a, b).
Definition : Let W be a subspace of an inner product space V.
Then W = {x V/ <w, x> = 0 for all w W} is called the orthogonal
complement of W.
Theorem 7: If W is a subspace of an inner product space V then W is a
subspace of V.
Proof: Since <0, w> = 0 for all w W, 0 W
Let , , ,x y W a b <x, w> = 0 and <y, w> = 0 for all w in W
To show that ax + by W
< ax + by, w > = <ax, w> + <by, w>
= a <x, w> + b <y, w>
= 0
ax + by W
HenceW is a subspace of W.
Theorem 8: If W is a subspace of an inner product space V then
W = W
Proof: W = {x V/ <x, w> = 0 for all w W }
If x W then <x, w> = 0 for all w W
x W
W W
Now let x W then <x, w> = 0 for all w W
x W
W W
Hence W = W .
Theorem 9: Let V be a finite dimensional inner product space and W be a
subspace of V. Then V = W W
142
Proof: To show that V = W W
i.e. to show that V = { x + y / x W and y W } and also W W
= { 0}
Since W is a subspace of finite dimensional inner product space
dim W dim V W it self is a finite dimensional inner product space It has an orthonormal basis
Suppose { e1, e2, …,ek} be an orthonormal basis of W
Let v V
Let
k
i
ii eevw1
,
So w W.
Let w‟ = v – w v = w + w‟, w W ………….(i)
Claim: w‟ W
To show that <w‟, w1> = for all w1 W
Since {e1, e2, …., ek} is a basis of W It is enough to show that <w‟, ej> = for j = 1, 2 …k
< w‟, ej > = < v – w , ej >
= < v -
k
i
ii eev1
, , ej >
= < v, ej > -
k
i
jii eeev1
,,
= < v, ej > - < v, ej > < ej, ej > (< ei, ej > = 0 for i j)
= < v, ej > - <v, ej > = 0
w‟ W
v = w + w‟ where w W and w‟ W V = W + W
Now to show that W W = { 0}
Clearly { 0} W W
Suppose x W W such that x 0
Since x W , < x, w > = 0 for all w in W
Now x W < x, x > = 0
143
x = 0 W W = {0}
V = W W .
8. 8 SUMMARY
In this unit we have defined the angle between two vectors using
inner product.
The we have defined very special orthogonal and orthonormal
sets in an inner product set. Orthogonal and orthonormal basis are basis
which are orthogonal and orthonormal.
We have studied following results
Every vector is uniquely expressed as linear combination of vectors
of orthonormal basis and the values of the coefficients are expressed
in terms of inner products
Every orthogonal set is linearly independent
Gram Schmidt process of orthogonalisation which converts a
linearly independent set of generators to an orthogonal set set of
generators.
Definition of orthogonal complement. A vector space is direct sum
of its subspace and orthogonal complement of that subspace.
8. 9 UNIT END EXERCISE:
Theory:
1) Define orthogonal set and orthonormal set in an inner product space
2) Define orthogonal basis and orthonormal basis of an inner product
space.
3) Define orthogonal projection of a vector along an unit vector.
4) Define orthogonal projection of a vector along any vector.
5) How to obtain an orthogonal set from a linearly independent set in an
inner product space?
6) Define an orthogonal complement of a set.
Problems:
1) In 3 , with respect to usual inner product convert the linearly
independent set { (1, 5, 7), (-1, 0, 2)} to an orthogonal set using Gram
Schmidt process. 62 65 5875 75 75
Ans : , , ,
144
2) In 3 , with respect to usual inner product convert the linearly
independent set
{ (1, 1, 0), (1, 0, 1), (0, 1, 1)} to an orthogonal basis using Gram Schmidt
process. 1 1 1 2 22 2 3 3 3
Ans : 1,1,0 , , ,1 , , ,
3) Use Gram Schmidt process to find an orthonormal basis of 3 from an
linearly independent set { (0, 1, 1), (1, -1, 0), (2, 0, 1)}
1 1 2 1 1 1 1 1 43 2 2 3 3 32 2 2
Ans : 0, , , 1, , , , ,
4) Consider the inner product space M2 with respect to inner product
< A, B > = tr( ABt). Transform the following linearly independent set in to
orthogonal basis using Gram Schmidt process.
i) {
00
11,
01
01 ,
10
10,
10
01 }
1 1 1 11 1
3 3 2 22 21 1 1
3 2 2
1 0Ans : , , ,
11 0 1 0
ii) {
10
11 ,
11
01,
10
01,
00
01}
1 2 1 2 13 3 5 5 2
1 2 1 13 5 5 2
01 0Ans : , , ,
1 01 1
5) Find the projection of
31
21 along
21
10 in the inner product
space given in problem 4) 0 1
Ans :1 2
6) Find the cosine of angle between (1, -3, 2) and (2, 1, 5) in 3 with
respect to usual inner product. 9
Ans :14 30
7) Find the cosine of angle between
13
12and
32
11 in M2
with respect to inner product given in problem 4). Ans :
145
9
LINEAR TRANSFORMATIONS
Unit Structure :
9.0 Objectives
9.1 Introduction
9.2 Rank Nullity theorem
9.3 The space L(U,V) of all linear transformations from U to V
9.4 Summary
9.0 OBJECTIVES :
This chapter would help you understand the following terms.
A linear transformation mapping vectors to vectors, characterized by
certain properties.
Natural projection mappings from n to m n m as linear
transformations.
Rotations and Reflections in a plane, also stretching and shearing as
linear mappings from n to n .
Orthogonal projections in n as a linear transformation with extra
properties.
Linear transformation from n to , which is called as a functional.
Every linear transformation is uniquely determined by it‟s action on
basis of a domain vector space.
Algebra of linear transformations. The space L U,V of linear
transformations from U to V.
The dual space V as the set of all functionals on a vector space V.
9.1 INTRODUCTION :
146
Say we have the vector 1
0
in n , and we rotate it through 90
degrees (anticlockwise), to obtain the vector 0
1
. We can also stretch a
vector U to make it 2U, for example 2
3
becomes 4
6
or, it we look at
the projection of one vector onto the x-axis, extracting it‟s x-component.
E.g. 2
3
to 2
0
, these all are examples of mapping between two vectors,
and are all linear transformations. A linear transformation is an important
concept in mathematics, because many real world phenomenon can be
approximated by linear models.
Definition : Given real vector spaces U and V, a linear transformation
T :U V is a function that maps vectors in U to vectors in V, also
satisfying the following properties.
(1) T preserves Scalar multiplication :
. . T u T u For all vectors u U & scalars .
(2) T preserves vector addition :
T u u T u T u for all vectors u, u in U.
Examples :
(1) Let us consider the projection of vectors in 2 to vectors on the x-
axis. 2 2:T , and T maps x
y
to 0
x
clearly this is linear map.
(x, y)
(x, 0)
Y
X
We shall check for both the properties of preserving scalar multiplication
and vector addition.
i) 1 1 1 1, , , T x y x y T x x y y
1, 0x x
147
1, 0 , 0x x
1 1, , T x y T x y
ii) , , , 0T x y T x y x
. , 0 . , x T x y
Clearly, T preserves the null vectors.
0, 0 0, 0T .
(2) Let V be the space of all continuous functions from into .
Define a function :T V V by
0
x
T f x f t dt
then T satisfies the properties of linear transformation.
For, continuous functions , :f g
0
x
T f g f g t dt
0
x
f t g t dt (By properties of Riemann Integration)
0 0
x x
f t dt g t dt
T f T g
For any scalar ,
0 0
.
x x
T f f t dt f t dt
0
x
f t dt
. T f (By properties of Riemann Integration)
So that T is a linear transformation.
(3) Let 1 2, , ..., nv v v be a basis of a finite n – dimensional vector
space V over . Define a map : nT V by associating to each
148
element V , it‟s unique co-ordinate vector with relative to this basis of
V.
If 1 1 2 2 ... , n n j
then 1 2, , ..., nT .
You can check that T is a linear transformation.
Note : Whenever :T U V is a linear transformation, T sends null vector
of U.
To null vector of V, because of the following :
u u u uT O O T O T O
(By addition property)
u v u uT O O T O T O
u u uO O O
u vT O O (By cancellation on both the sides)
Exercises :
Check whether the following transformations satisfy the properties of a
linear transformation or not.
i) 3 21 2 3 1 2: , , , , 0T T x x x x x
ii) 2 2:T
, cos sin , sin cosT x y x y x y
iii) 3 2 2: , , , , 0, 0T T x y z x
iv) Let A be mn matrix over . Define a function : n mAL by
AL X AX where
1
2
n
x
X x
x
a column vector in n .
(Hint : Use properties of matrices)
v) Let V be any vector space, the identity mapping :I V V , I ,
the null mapping :O V V , O O where O V is the null
vector.
149
In the next theorem we shall find a way of checking whether a
given function :T U V is a linear transformation or not.
Theorem : Let U, V be real vector spaces, then :T U V is a linear
transformation iff. 1 1 . . T u u T u T u for all , and
vectors 1 u,u U .
Proof : Assume that :T U V is a linear transformation, then for
1 , , u,u U .
1 1T u u T u T u (By addition property)
1 T u T u (By Scalar multiplication property)
Conversely, assume that 1 1 T u u T u T u .
Put 1 10 0 . . T u O T u T u
1 1 . T u T u
Similarly for 1 ,
1 11 . 1 . 1 . 1 . T u u T u T u
1 1T u u T u T u
this implies that T is a linear transformation. The next theorem gives you
the important properties of a linear transformation.
Theorem : If T is a linear transformation from U to V then
(i) u vT O O
(ii) – – T u T u
(iii) 1 1– –T u u T u T u
Proof :
(i) u u uO O O
u u u vT O T O T O O
u vT O O (By adding uT O on both the sides)
150
(ii) – uO u u
– uT O T u T u
– vO T u T u
– – T u T u OR – T u T u
(iii) 1 1 1– – – T u u T u u T u T u
1–T u T u [By (ii)]
Given a linear transformation T :U V . We saw that 0 0T , in other
words, T fixes the null vector, T induces some important subspaces of U
& V in view of the following definitions.
Definition : Let U, V be real vector spaces. :T U V be a linear
transformation. Then define the set 0u U T u to be the kernel of
T, denoted by ker T and the set T u V u U to be the image of T,
denoted by ImT.
Examples :
(1) The projection of a vector x
y
in a plane on the x-axis given by
, , 0T x y x .
In this example all the vectors on y-axis constitute the kernel of T.
2, , 0, 0KerT x y T x y
2, , 0 0, 0x y x
2, 0x y x
20, y y
= Y-axis
Clearly ImT consists of all vectors on Real line (x-axis) in a plane.
(2) Consider the linear map.
2 2:T defined by , – , – T x y x y x y
In this example,
151
kerT is the line passing through the origin in a plane having the
equation y x .
Y
XO
Ker T
Image of T is just a line in 2 with co-ordinates of the form
, – r r for scalar r .
Both of these sets KerT and ImT are subspaces of U & V
respectively. We shall check this in the following theorem.
Theorem : Let :T U V be a linear transformation, then
(i) ker T is a subspace of U.
(ii) ImT is a subspace of V.
Proof :
(i) Let , and 1, keru u T then we need to show that
1 keru u T .
For that consider 1 1 . . T u u T u T u
. 0 . 0 0 0 0
1 keru u T
kerT is a subspace of U.
(ii) Let 1, ImT there exist 1, u u U such that T u and
1T w .
Let , be scalars we need to show that 1 Imd T .
Consider 1 1 1T u u T u T u T u T u
1
152
1 1T u u
1 ImT 1u u U
Given a linear transformation :T U V , where U and V are real vector
spaces. kerT helps us to check whether T is a one-to-one linear
transformation or not, as stated in the following theorem.
Theorem : Let :T U V be a linear transformation, then T is injective
map iff ker 0T .
Proof : Assume that T is injective map.
Let ker 0u T T u .
0T u T and injectivity of T implies that 0u .
ker 0T
Conversely, Assume that ker 0T . We need to show that T is one-to-
one, for assume that
1T u T u for 1, u u U
1 10 , 0T u T u T u u
1– keru u T but ker 0T
1– 0u u
1u u , this shows that T is injective map.
Given a linear transformation :T U V , where U is a finite
dimensional vector space over .
Suppose that 1 2, , ..., mB u u u is a basis for U and we know the
values of T on 1 2, , ..., mu u u , then we can determine the action of a linear
transformation T naturally on any vector u U .
For example consider a linear transformation 3 3:T , that
maps a natural standard basis 1, 0, 0 0, 1, 0 0, 0, 1 of 3 as follows.
1, 0, 0 0, 0, 1T
0, 1, 0 1, 0, 0T
0, 0, 1 0, 1, 0T
153
In this example can we determine the action of T on any vector
, , x y z in 3 ? The answer is yes. Because for scalars , , x y z , we
can write , , 1, 0, 0 0, 1, 0 0, 0, 1x y z x y z
, , 1, 0, 0 0, 1, 0 0, 0, 1T x y z xT yT zT
, , 0, 0, 1 1, 0, 0 0, 1, 0T x y z x y z
0, 0, , 0, 0 0, , 0 , , x y z y z x
Therefore, T can be defined uniquely as follows.
, , , , T x y z y z x for 3, , x y z .
You can check that T satisfies both the properties of a linear
transformation.
We shall try summarise the understanding in the following
theorem.
Theorem : Let U and V be vector spaces 1 2, , ..., nu u u be a basis of U.
Let 1 2, , ..., n be any vectors in V then there exists a unique linear
transformation :T U V such that i iT u for all I, 1 i n .
Proof : Let u U be any vector since 1 2, , ..., nu u u is a basis of U.
There are unique scalars 1 2, , ..., n in such that
1 1 2 2 ... n nu u d u u .
Define a mapping :T U V as follows.
1 1 2 2 ... n nT u
We shall show that T is a linear map.
For ,c d & 1, u u U .
Suppose 1 1 2 2 ... n nu u u u and 1 1 1 11 1 2 2 ... n nu u a u a u
for scalars 1 11, ..., n ,
then 1 1 11 1 2 2 1 ... ... n n n n ncu du c u u d d u u
154
1 11 1 1 ... n n nc d u c d u
1 1 11 11
... n n nT cu du T c d u c d u
1 11 1 1 ... n n nc d c d
1 11 1 1 1 ... ... n n n nc d
1 1T cu du cT u dT u
T is a linear transformation
Also, we can write 1 20 . 0 . ... 1 . ... 0 . i i nu u u u u
10 . ... 1 . ... 0 . i i nT u T u T u T u
10 . ... 1 . ... 0 . i n i
i iT u for all i, 1 i n
T is the required linear map.
To prove uniqueness of T, Assume that :M U V is another
linear transformation that sends iu to i for all i, 1 i n , then
1 1 2 2 1 1 2 2 ... ... n n n nM u M u u u
T u
M T on U
Now, it‟s your turn to solve few problems.
Check your progress :
1. Check whether 2:F , defined by , F x y xy is a linear
transformation or not.
(Hint : Take 1, 2 , 3, 4x y in 2 . Find T x y &
T x T y )
2. Let 3 3:T be a linear transformation defined by
, , – 2 , 2 , – – 2 2T x y z x y z x y x y z . Find ker , ImT T .
Find a basis for each of them and their dimensions.
155
Hint
: To find kerT , solve the system of equations.
2 0x y z
2 0x y
– – 2 2 0x y z
Using matrix form, after reducing the matrix
1 – 1 2
2 1 0
– 1 – 2 2
to row
reduced echelon form
1 – 1 2
40 1 – 3
0 0 0
.
4
3y z and 2–
3zx .
ker –2, 4, 3T
To find ImT see that
3
, , 2 –1, 1, – 2 2, 0, 22
xT x y z x y z
Because 3
1, 2, –1 2 –1, 1, – 2 2, 0, 22
Im – 1, 1, – 2 2, 0, 2 , T
In this case ImT is a two-dimensional subspace of 3 generated by
vectors – 1, 1, – 2 and 2, 0, 2 .
3. Show that the following maps are linear.
i) 3 3:T , , , , , 0T x y z x y
ii) 3:M , , , 2 – 3 5M x y z x y z
iii) :I V V , I for all V
iv) :O V V , 0O for all V
4. Show that the following maps are not linear
i) 2:Q , , Q x y y x
ii) 3 2:T , 2, , , 0T x y z x
156
5. Let 4 3:T be defined by
, , , – , 2 – , T x y z w x y z w x z w x z w . Find dimensions
of ImT and kerT.
(First check that T is a linear transformation)
6. Let V be a vector space of all nn matrices over . Let B be a fixed
nn matrix over .
Let :T V V be defined by –T A AB BA , verify that T is a
linear transformation from V into V.
(Hint : Use properties of matrices)
7. Let U, V be vector spaces, and :T U V be a linear transformation.
If 1 2, , ..., mu u u are linear by independent vectors in U then show that
1 2, , ..., mTu Tu Tu are linearly independent vectors in V.
(Hint : Assume that 1
0m
j jj
Tu
and show that all scalars
1 2, , ..., m are bound to be zero.)
In example (2) above dim kerT = 1 and dim ImT = 2 therefore
dim kerT + dim ImT = 3 = dim 3 .
9.2 RANK-NULLITY THEOREM :
If U is a finite dimensional vector space then given a linear
transformation :T U V we can establish a relation between kerT and
ImT and vector space U in terms of their dimensions which we shall see
in the following theorem, which is called as Rank-nullity theorem.
Definition : If :T U V is linear transformation then the dimension of
ImT is called as rank T and the dimension of kerT is called as nullity T.
For example, consider a linear transformation 2:T defined by
, T x y x y then dimkerT nullity = 1 and dimIm 1T .
(Here ker 1, – 1T )
Theorem : Let U, V be vector spaces and :T U V be a linear
transformation then
dim dim ker dim ImU T T
= rank T + nullity T
157
Proof : If Im 0T then kerT U and dim dim ker dim ImU T T .
Assume that 1 2, , ..., m be a basis of ImT . Let
1 2, , ..., mu u u be a set of vectors in U such that i iTu for all i,
1 i m .
Let 1 2, , ..., pk k k be a basis of kerT. In order to prove the result we need
to show that 1 2 1 2, , ..., , , , ..., p mk k k forms of basis for U.
Let Imu U T u T , hence there are scalars 1 2, , ..., m such that
1 1 2 2 ... m mT u
1 1 2 2 ... m mTu Tu Tu
1 1 2 2 ... m mT u u u
1 1 2 2– ... kerm mu u u u T
This means that there are scalars 1 2, , ..., p such that
1 1 2 2 1 1 2 2– ... ... m m p pu u u u k k k .
In other words,
1 1 2 2 1 1 2 2= ... ... m m p pu u u u k k k
Here we have proved that U is generated by the set of vectors
1 2 1 2, , ..., , , , ..., m pu u u k k k .
We have to show that this is a linearly independent set. So consider
scalars 1 2 1 2, , ..., , , , ..., m p such that
1 1 2 2 1 1 2 2 ... ... 0m m p pu u u k k k
1 1 2 2 1 1 2 2 ... ... 0 0m m p pT u u u k k k T
1 1 2 2 1 1 2 2 ... ... 0m m p pTu Tu Tu Tk Tk Tk
Here 1, ..., ker 0p jk k T Tk for all j, 1 j p
1 1 ... 0m m
158
Since 1, ..., m are linearly independent vectors.
1 2 ... 0m
Similarly, 1 2, , ..., p are all zero.
Thus dim dim ker dim Im U p m T T
= nullity T + rank T
Corollary : If U, V are vectors spaces such that dim dim U V and if
:T U V is given linear transformation, then T is one-to-one iff T is
onto.
Proof : If T is one-to-one then ker 0T .
dim ker 0 dim dim Im dim T U T V
dim Im dim T V T is onto Im T V
Conversely, assume that T is onto.
ImT V
dim dim ker dim Im U T T
dim ker dim T V
But dim dim dim ker 0U V T
ker 0T
Therefore T is injective map.
Note : If dim dim U V the above corollary implies that any linear
transformation :T U V is bijective.
Definition : Let U, V be two vector spaces over . Define , L U V to be
the set consisting of all linear transformations from U to V.
On a set , : is a linear transformationL U V T U V T , define two
operations + and as follows :
For , , S T L U V &
, S T L U V and S T u S u T u u U
159
, S L U V and . S u S u u U
Check that S T and S are linear transformations from U to V.
1S T u u
1 1S u u T u u
1 1S u T u S u T u
1S T u S T u for any 1, u u U
Similarly, , S T u S T u u U
1 1 1s u u S u u S u S u
1 S u S u
1 S u S u
Similarly,
, , S cu C S u u U C
Both S + T and S are linear maps from U into V.
As we compose functions, we can also find composite of linear
transformations.
Theorem : Let U, V, W be real vector spaces.
Let :T U V and :S V W be both linear transformations, then the
composite map.
:SOT U W is also a linear transformation.
Proof : Let 1, u u U and , then
1 1SOT u u S T u u
1S T u T u (T is linear)
1S T u S T u (S is linear)
1S T u S T u
160
1SOT u SOT u
This proves that SOT is a linear transformation.
Example :
1) Let 3 3:T be defined by , , , , 0T x y z x y and 3 3:S
be defined by , , , , 0S x y z x z , then 3 3:SOT is defined by
, , , , 0 , 0, 0SOT x y z S x y x
whereas , , , , 0 , , 0TOS x y z T x z x z .
Here in this example SOT TOS
2) If T is a linear transformation from a vector space V to V then we call
T as an operator. If T is an operator, denote TOT by 2T in general the
operator
n times
... TOTO OT by nT .
Define T Identity map = I
Given a linear map :T U V we may find it‟s inverse linear map
as given in the following theorem.
Theorem : Let :T U V be a linear transformation. If T is bijective
then there is a linear map :S V U such that VTOS I and USOT I
where VI and UI denote the identity maps on V and U respectively.
Proof : Since :T U V is a bijective linear map, it‟s inverse 1 :T V U exists, denote 1 T by S.
We shall show that :S V U is a linear map.
Let 1 2, V V V , then 1 2 , u u U such that 1 1v Tu and 2 2v Tu
or 1 1u S and 2 2u S .
1 2 1 2 . .S a b S a Tu b Tu
1 2S T au bu
11 2T T au bu
1 2 1 2au bu aS bS for ,a b
161
S is a linear transformation..
By definition of inverse function, it follows that VT S I and
US T I .
Definition : A linear map :T U V which has inverse is called
invertible or non singular transformation or an isomorphism.
Example : 3 3:F defined by
, , 2 , 2 , 2 2 3F x y z x y z x y z x y z is invertible linear
transformation.
Here, , , kerx y z F
2 0x y z
2 0x y z
2 2 3 0x y z
This system can be written as
1 1 2 0
1 2 1 0
2 2 3 0
x
y
z
.
The coefficient matrix
1 1 2
1 2 1
2 2 3
has it‟s row reduced echelon form
as
1 0 0
0 1 0
0 0 1
.
The system is equivalent to
1 0 0 0
0 1 0 0
0 0 1 0
x
y
z
0x y z is the only solution of this system.
ker 0F
F is invertible.
Check your progress :
162
1) Define 3 3:T by , , , 2 , 3 5T x y z x y z x y z x y z .
Find if T is non-singular. If not, find 0u in 3 such that 0Tu .
(Hint : Solve the system
0
2 0
3 5 0
x y z
x y z
x y z
, , kerx y z T
3 , 2x z y z
the solution is 3, 2,1 ,z z .
ker 0T , if 3, 2,1u then 0Tu )
2) Let U, V, W be vector spaces over . Let :T U V be a linear
transformation and let 1 2,S S be two linear transformations of V into
W . Then show that 1 2 1 2S S T S T S T .
(Hint : For u U , find 1 2S S T u )
3) In above example prove that if , then 1 1S T S T .
(Hint : For 1,u U S T u )
4) Find a map 3 4:F whose image is spanned by 1, 2, 0, 4 and
2, 0, .
(Hint : We know by the theorem that if 1, ..., n is a basis of V, for
arbitrary dements iw of 1v a unique linear map 1:T V V s.t.
1 i n .
i iT w
Here Im p 1, 2, 0, 4F S an
Let 1 21, 2, 0, 4 , 2, 0, 1, 3w w
We want 3w such that 1 2 3 1 2p , , p ,s an w w w s an w w
3 0, 0, 0, 0w is an obvious choice.
1 2 3, ,F x y z F xe ye ze
1 2 3 1, 2, 0, 4 2, 0, 1, 3xw yw zw x y
2 , 2 , , 4 3x y x y x y .)
163
5) Show that 2 2:T defined by , ,T x y x y x y is
invertible.
6) Let :L V V be a linear map s.t. 2 2 0L L I . Show that L is
invertible.
7) Let V be a vector space and :T V V be a linear map such that
2T T show that ker ImV T T .
(Hint : V can be written as T T )
8) Let V and 1V be two vector spaces over . Show that there is a
bijective linear transformation 1:T V V iff 1dim dimV V .
9) Let ,A B be linear maps of V into itself. If ker 0 kerA B , show
that ker 0AOB .
10) Let A be a linear map of V to itself such that 2 0A A I . Show
that A is invertible, and 1A I A .
9.3 THE SPACE L(U,V) OF ALL LINEAR
TRANSFORMATIONS FROM U TO V :
We know that the space of linear transformations from U to V is
denoted by ,L U V where U & V are both real vector spaces. We can
investigate for the dimension of the vector spaces then ,L U V in terms
of dimensions of vector spaces U and V. If U, V both are finite
dimensional vector spaces then ,L U V is a finite dimensional vector
space over , we prove this is the next theorem.
Theorem : Let U, V be both finite dimensional vector spaces over ,
with dimU n and dimV m . Then the space ,L U V of linear maps
from U into V is finite dimensional and dim ,L U V mn .
Proof : Let 1 2, , ..., nB u u u and 11 2, , ..., mB be ordered bases
for U & V respectively. For each pair of integers ,p q with 1 p m ,
1 q n , define a linear transformation
, :p qE U V as follows.
164
,p q j pE u if j q
0 if j q
Then there exists such a unique linear transformation from U into
V satisfying these conditions according to the theorem for existence and
uniqueness of a linear transformation.
We claim that , 1 , 1p qE p m q n forms a basis for
,L U V .
Let T be any linear transformation from U to V. Suppose that
1 11 1 21 2 1... m mTu a a a
2 12 1 22 2 2m mTu a a a
1 1 2 2 ...n n n mn mTu a a a
We show that ,1 1
m n
pq p qp q
T a E
Consider ,1 1
m n
pq p q jp q
a E u
,1 1
m n
pq p q jp q
a E u
1 1
m n
pq jq pp q
a
1
m
pj jp
a Tu
, ,1 1
1 , 1m n
pq p q p qp q
T a E Span E p m q n
,L U V
Now we shall show that , 1 , 1p qW p m q n is a linearly
independent set.
165
If the transformation ,1 1
m n
pq p qp q
a E
is the zero transformation for
scalars pqa then ,1 1
0m n
pq p q jp q
a E u
for each j 1 j n
1 1
0m n
pj pp q
a
But 1p p m is a linearly independent set.
0 ,pja p j
This shows that , 1 ,p qE p m q n is a basis for ,L U V .
dim ,L U V mn
Definition : Let V be a vector space over , a linear transformation.
:f V is called a linear functional on V, this means that f is a function
from V into such that 1 1f a b af bf for all vectors
1, V and scalars ,a b .
Examples :
1) Define a function : nf by 1 2 1 1 2 2, , ..., ...n n nf x x x a x a x a x
where 1 2, , ..., na a a are fixed scalars in .
2) Let n . If A is an nn matrix over , the trace of A is the scalar
tr A,
11 22 ... nntr A a a a
the trace function is a linear functional on the matrix space n n ,
because
1
n
ii iii
tr A B a b
1 1
n n
ii iii i
a b
tr A tr B
166
3) Let ,C a b be the space of continuous real valued functions on
,a b . Then : ,L C a b defined by b
a
L g g t dt is a
linear functional on ,C a b .
Definition : Let V be a vector space over the space ,L V (the set of
all linear functionals on V) is called the dual space of V denoted by *V .
* ,V L V
If V is finite dimensional vector space over then we know that
*dim dim ,V L V
dim .dimV
dimV
Let 1 2, , ..., nB be a basis for V, then by the theorem on
existence and uniqueness of a linear transformation, there is a unique
linear functional :if V such that
0 if , 1 ifi j ijf i j i j .
For 1 i n , we obtain from B in set of n distinct linear functions
1 2, , ..., nf f f on V. We show that 1, ..., nf f is a linearly independent set
in *,L V V .
Let us consider
1
, 1n
i i ii
f c f c i n
then
1 1
n n
j i i j i ij ji i
f c f c c
If f is the zero functional 0 1jc j n
1, ..., nf f is a linearly independent set in *V since *dimV n .
*1, ..., nB f f is a basis of *V . This basis of *V is called the dual
basis of B. Using above result, we shall show that each vector. V can
167
be written as a linear combinations of 1 2, , ..., n with scalars of the
form if for 1 i n .
Theorem : Let V be a finite dimensional vector space over , let
1, ..., nB be a basis for V then there is a unique dual basis
*1, ..., nB f f for *V such that i j ijf for each linear functional
:f V .
1
n
i ii
f f f
and for each vector V .
1
n
i ii
f
Proof : We have shown above that there is a unique basis 1, ..., nf f of
*V dual to the basis 1, ..., n of V.
If f is a linear functional on V, then f is some linear combination of
the if , and as we observed after the scalars jc must be given by
j jc f .
Similarly, if
1
n
i ii
U
is a vector in V, then
1 1
n n
j i j i i ij ji i
f f
So, that the unique expression for as a linear combination of the i ‟s
is,
1
n
i ii
f
Note : The expression 1
n
i ii
f
provides with a nice way of
describing what the dual basis is. If 1, ..., nB is an ordered basis
for V and *1 2, , ..., nB f f f is the dual basis, then if is precisely the
168
function, which assigns to each vector V , the co-ordinate of
relative to the ordered basis B.
When 1 1 ... n n
1 1 ... n nf f f for *f V
Example :
1) Let V be the vector space of all polynomial functions from into
, which have degree less than or equal to 2.
Let 1 2 3, ,t t t be three distinct real nos.
Let :iL V be defined by i iL p x p t for p x V
Then 1 2 3, ,L L L are linear functionals on V, *iL V . These linear
functionals are linearly independent,
For suppose, 1 2 3 0L aL bL cL
0L p x p x V
Let 21, ,p x p x x p x x
0a b c
31 2 0t a t b t c
2 2 21 2 3 0t a t b t c
1 2 3
2 2 31 2 2
1 1 1 0
0
0
a
t t t b
ct t t
But the matrix 1 2 3
2 2 21 2 3
1 1 1
t t t
t t t
is invertible, because 1 2 3, ,t t t are all distinct.
0a b c is the only solution.
1 2, ,L L L is a linearly independent set. Now *dim dim 3V V
1 2 3, ,L L L is a basis for *V .
169
We would like to investigate for the basis of V, whose dual is 1 2, ,L L L .
Let 1 2 3, ,p x p x p x be the basis of V, whose dual basis is
1 2, ,L L L .
i j ijL p x
OR
j i ijp t
These polynomials are easily seen to be
2 3
11 2 1 3
x t x tp x
t t t t
1 3
22 1 2 3
x t x tp x
t t t t
1 2
33 1 3 2
x t x tp x
t t t t
Note : If f is a non-zero linear functional, then the rank of f is 1, because
the range of f is a non-zero substance of the scalar field and must therefore
be a scalar field of reals.
If the underlying space V is finite-dimensional, the rank-nullity
theorem tells us that
nullity f = dim 1V
In a vector space of dimension n, a subspace of dimension 1n
is called a hyperspace.
Check your progress :
1) In 3 , let 1 2 31, 0, , 0,1, 2 , 1, 1, 0
a) If f is a linear functional on 3 such that
1 2 31, 1, 3f f f . Find , ,f a b c .
(Hint : Write 1 1 2 2 3 3, ,a b c x x x , find 1 2 3, ,x x x in
terms of a, b, c.
1 1 2 2 3 3, ,f a b c x f x f x f
170
1 2 33x x x )
2) Let 1 2 3, ,B be the basis of 3 defined by
1 2 31, 0, 1 , 1,1,1 , 2, 2, 0 . Find the dual basis of B.
(Hint : Let 1 2 3, ,f f f be the dual of B.
1 1 1 2 1 31; 0; 0;f f f
2 1 2 2 2 30; 1; 0;f f f
3 1 3 2 3 30; 0; 0.f f f
Write a 3 – tuple 3, ,x y z as follows :
1 2 3, ,x y z . Find , , in terms of , ,x y z .
1 2 3, ,f x y z f f f
1 2 3, ,j j j jf x y z f f f for 1 3j )
3) Let V be the vector space of all polynomial functions :p x ,
which have degree 2 or less.
20 1 2p x c c x c x
Define 3 – linear functions on V by,
1 2 1
1 2 3
0 0 0
; ;f p x p x dx f p x p x dx f p x p x dx
Show that 1 2 3, ,f f f is a basis for *V by exhibiting the basis for V
of which it is the dual.
(Hint : Assume that 1 2 3, ,p p p is a basis of V with the dual
1 2 3, ,f f f , where
21 0 1 2 2 2 0 1 2, ;p x c c x c x p x d d x d x
23 0 1 2p x c c x c x and use the fact that i j ijf p x )
171
9.4 SUMMARY :
In this chapter we have learned the following topics.
1) A mapping from one vector space to another vector space,
characterized by certain properties, called as a linear transformation or
a linear map.
2) Natural projection mappings from n to m n m are examples of
linear transformations.
3) Rotations & reflections in a plane, stretching & shearing are also linear
transformations from n to n .
4) Orthogonal projections in n as a linear transformation with extra
properties.
5) Functionals as linear transformations on n which are real-valued.
6) Every linear transformation :T U V is uniquely determined by it‟s
action on basis of U.
7) Algebra of linear transformations, ,L U V is called the space of all
linear transformations from U to V.
8) Given a vector space V, the set of all functionals in ,L U V is
denoted by *V , * ,V L V .
172
10
DETERMINANTS
Unit Structure :
10.0 Objectives
10.1 Introduction
10.2 Existence and Uniqueness of determinant function
10.3 Laplace expansion of a determinant
10.4 Summary
10.0 OBJECTIVES :
This chapter would help you understand the following terms and
topics.
To each nn matrix A over , we associate a real no. called as the
determinant of matrix A.
Determinant as an n-linear skew-symmetric function from
...n n n , which is equal to 1 on 1 2, , ..., nE E E .
Here
0
0
1
0
jE
jth
place
jE is the jth
column of the nn identity matrix nI .
Determinant of an nn matrix 1
1 2 ... n
n
RA c c c
R
as
determinant of it‟s column vectors 1 2 ... nc c c or row vectors
1
n
R
R
.
We shall see the existence and uniqueness of determinant function
using permutations.
173
Computation of matrices determinant of order 2, 3 diagonal matrices
and their determinant.
For nn matrix , det dettA A A & det det .detAB A B .
For square matrix B.
Laplace expansion of a determinant, Vandermonde determinant,
determinant of upper triangular and lower triangular matrices.
10.1 INTRODUCTION :
We wish to assign to each nn square matrix A over a scalar
(real number) to be known as the determinant of the matrix. It is always
possible to define the determinant of a square matrix A by simply writing
down a formula for this determinant in terms of the entries of A. We shall
define a determinant function on ...n n n (n times) as a function,
which assigns to each nn matrix over , the function having certain
properties. This function is linear as a function of each of the row or
columns of the matrix A.
It‟s value is 0 on any matrix having two equal rows or two equal
columns and it‟s value on the nn identity matrix is equal to 1. We shall
see that such a function exists and then that it is unique with it‟s useful
properties.
Definition : Let n be a positive integer, let D be a function which assigns
to each nn matrix A over , a scalar D A in . We say that D is n-
linear if for each , 1i i n . D is a linear function of the ith
row when
the other 1n rows are fixed.
We shall understand this definition of n-linear function D. If D is a
function from
...n times
n n n into , and if 1 2, , ..., nR R R are the rows
of the matrix A, we write 1, ..., nD A D R R , in other words D is
nothing but the function of the rows of A. To say that D is n-linear it
means that 11 2, , ..., , ...,i i nD R R R R R
11 2 1 2, , ..., , ..., , ,..., , ...,i n i nD R R R R D R R R R
Example 1 : Let 1 2, , ..., na a a be positive integers such that 1 ja n .
Let . For each nn matrix over , define
1, 2, ,1...a a n an n
D A A A A where 1 , 1ijA A i n j n
This function D is n-linear.
174
A particular n-linear function of this type is 1,1 2, 2 ,... n nD A A A A for
1 , ja j for 1 j n .
For if 1 2, , ..., nR R R are rows of the matrix A then
11 2, , ..., , ...,i i nD R R R R R
111 22 ... ...ii ii nnA A A A A i is fixed.
11 22 11 22... ... ...ii nn nnA A A A A A A
1 11 2 1 1, , ..., , ..., , ..., , ..., , ..., ...,i i n i n i nD R R R R R D R R R D R R R
Example 2 : Let us find all 2-linear functions on 2 2 matrices over .
Let 1 2,E E be the rows of the 2 2 identity matrix. By matrix theory,
11 1 12 2 21 1 22 2,D A D A E A E A E A E
Using the fact that D is 2-linear,
11 1 21 1 22 2 12 2 21 1 22 2, ,D A A D E A E A E A D E A E A E
11 21 1 1 11 22 1 2 12 21 2 1 12 22 2 2, , ,A A D E E A A D E E A A D E E A A D E E
Thus D is completely determined by the four scalars.
1 1 1 2 2 1, , , , ,D E E D E E D E E and 2 2,D E E
Example 3 : Let D be the function defined on 2 2 matrices by
11 22 12 21D A A A A A then D is a 2-linear function because
1 2D D D where 1 11 22D A A A and 2 12 21D A A A both
1 2,D D are 2-linear combination of n-linear function is n-linear.
Note : A linear combination of n-linear functions is n-linear.
This example is familiar to you, this function D was known as determinant
of 2 2 square matrix A.
11 12
21 22
A AA
A A
Let us note some of it‟s properties.
175
If I is the identity matrix, then 1D I in other words 1 2, 1D E E .
Second if the two rows of A are equal, then 11 12 12 11 0D A A A A A .
If 1A is the matrix obtained by 2 2 square matrix A by interchanging it‟s
rows, then 1 1 1 1 111 22 12 21 21 12 22 11D A A A A A A A A A
D A
Next, we shall see the meaning of n-linear skew symmetric function.
Definition : Let D be an n-linear function we say that D is skew-
symmetric if the following two conditions are satisfied.
(a) 0D A whenever two rows of A are equal.
(b) If 1A is a matrix obtained from A by interchanging two rows of A, then
1D A D A .
Any n-linear function D satisfying (a) also satisfied (b).
Definition : Let n be a positive integer. Suppose D is a function from
...n n n (nn matrices over ) into . We say that D is a
determinant function provided that D is n-linear, skew-symmetric such
that 1nD I . ( nI is the identity matrix of order n).
We shall show that ultimately there is exactly one determinant
function on nn matrices over . For 1n . If A a a , then
D A a is the only determinant function.
For 2n , the only determinant function is 2-linear function of the
following form.
11 21 1 1 11 22 1 2 12 21 2 1, ,D A A A D E E A A D E E A A D E E
12 22 2 2,A A D E E
D must be skew-symmetric.
1 1 2 20 ,D E E D E E and
2 1 1 2,D E E D E E D I
D also satisfies 1D I .
11 22 12 21D A A A A A
176
We shall prove that any n-linear function D on
...n times
n n n
satisfying 0D A whenever two adjacent rows of A are equal, must be
skew-symmetric.
Theorem : Let D be an n-linear function on nn matrices over .
Suppose D has the property that 0D A , whenever two adjacent rows
of A are equal then D is skew-symmetric.
Proof : We need to show that 0D A , whenever any two rows of A are
equal and 1D A D A , if 1A is a matrix obtained from A by
interchanging two rows of A.
First suppose that 1A is obtained by interchanging two adjacent
rows of A. say jR & 1jR then consider
1 1 1,..., , , ...,j j j j nO D R R R R R R
1 1 1 1 1, ..., , , ..., , ..., , , ...,j j j n j j j nD R R R R R D R R R R R
1 1 1, ..., , , ..., , ..., , , ...,j j n j j nD R R R R D R R R R
1 1 1 1, ..., , , ..., , ..., , , ...,j j n j j j nD R R R R D R R R R
1 1 1 1, ..., , , ..., , ..., , , ...,j j n j j nD R R R R D R R R R
1O D A D A (D is n-linear)
Now, let B be obtained by interchanging rows i & j of A, where i j .
Obtain B from A by a succession of interchanges of pairs of
adjacent rows, this requires j i interchanges of adjacent rows until the
rows are in the order 1 1 1 1, ..., , ..., , , , ...,i i j i j nR R R R R R R .
Now we move jR to thi position using 1j i interchanges of
adjacent rows. Thus there are 1 2 2 1j i j i j
interchanges of adjacent rows.
2 11 ( ) ( )
j iD B D A D A
177
Definition : Let A be an n n matrix over . Let A i j denote the
1 1n n matrix obtained by deleting the thi row and thj column
of A.
If D is an 1n linear function and A is an n n matrix then put
ijD A D A i j .
We have already seen that the determinant function on 2 2
and exist. Further we show by Induction that the determinant function
exists on n times
....n n n for any positive integer n.
Theorem : Let 1n D be an 1n - linear skew symmetric function on
1 1n n matrices over , for each ,j j n the function.
(n times)
: ... .n n njE defined by
1
( 1)n
i jj ij ij
i
E A A D A
is an n-linear skew symmetric
function on n n matrices over . If D is a determinant function, so is
jE .
Proof :
( )ijD A D A i j and D is an 1n - linear function.
ijD is linear as function of any row except thi row.
ij ijA D A is an n-linear function of A. We know that A linear
combination of n-linear functions is n-linear.
jE being a linear combination of n-linear functions, is an n-linear
function. To show that jE is skew-symmetric, it is enough to show that
0jE A , whenever A has two equal and adjacent rows.
Suppose 1k kR R , if , 1i k i k , the matrix A i j has two
equal rows, thus 0ijD A .
Therefore, 1
11 1k j k j
j kj kj k jE A A D A A A
178
Since 1 1k k kj k jR R A A and 1A k j A k j
0jE A
Now, assume that D is a determinant function. If nI is the nn identity
matrix then nI j j is the 1 1n n identity matrix 1nI .
Since 1,1n ij j n ni j
I E I D I
1j nE I , and therefore jE is a determinant function.
Corollary : Let n be a positive integer, then there exists atleast one
determinant function on ...n n n (n times).
Proof : The previous theorem gives us a way to construct a determinant
function on nn matrices, if such a determinant function is given on
1 1n n matrices . The corollary follows by induction.
Example 1 : If B is a 22 matrix over . Let 11 22 12 21B B B B B .
Then B D B , where D is the determinant function on 2 2 matrices.
We showed that this function on 2 2 is unique.
Example 2 : Let A be 3 3 matrix
0 1 0
0 0 1
1 0 0
A
Then 1
1 01
0 1E A
2
0 11
1 0E A
3
0 11
1 0E A
Check your progress :
1) Given that a map 3 3 3:f is 3-linear skew-symmetric.
If 3 1f I , prove that detf A A where A is any 33 matrix.
(Hint : Write a 33 matrix
11 12 13
21 22 23
31 32 33
A A A
A A A A
A A A
as
179
11 1 12 2 13 3 21 1 22 2 23 3 31 1, ,A A E A E A E A E A E A E A E
32 2 33 3A E A E
11 1 12 2 13 3 21 1 22 2 23 3, ,f A f A E A E A E A E A E A E
31 1 32 2 33 3A E A E A E
Use the fact that f is skew-symmetric.
1 1 2 2 3 3, , , 0f E E f E E f E E
Also 1 2 3, , 1f E E E .)
2) Define a function D on 33 matrices over by the rule,
22 23 21 23 21 2211 12 13
32 33 31 33 31 32
A A A A A AD A A A A
A A A A A A
Show that D is alternating (skew-symmetric) 3-linear as a function of
the columns of A.
3) Let D be an alternating (skew-symmetric) n-linear function on nn
matrices over , show that
(a) 0D A , if A has a null row.
(b) D B D A , if B is obtained from A by adding a scalar
multiple of one row to another row.
(Hint : (a) write 0 = 0 + 0, use n-linearity & skew-symmetricness of a
function D.
(b) Let 1 2 2, , ..., nB R R R R where 1 2, , ..., nA R R R .)
10.2 EXISTENCE AND UNIQUENESS OF
DETERMINANT FUNCTION :
In unit we showed the existence of a determinant function on nn
matrices over as a function of row vectors (column vectors) of nn
matrix A.
To prove the uniqueness of such a determinant function D, we try
to expand this function over a sum taken over all permutations of n
symbols 1, 2, .., n .
Recall that a permutation is a bijection from 1, 2, 3, ..., n to
1, 2, ..., n . The set of all permutations on n-objects is denoted by nS .
180
To every permutation in nS , we assign a signature as follows.
If nS then sgn 1k
where k is the number of
transpositions in the decomposition of permutation , in other words.
sgn 1k
where 1 2, , ..., k . Let us denote signature of a
permutation by .
1 or 1 depending on number of transpositions in
decomposition of .
Theorem : Let A be nn matrix over , nS and 1 2, , ..., nA R R R ,
then 1 21 2, , , , .., nnD R R R D R R R
Proof : Let us consider a transposition i j then
1 2, , ..., , ..., , ...,i j nD R R R R R
1 2, , ..., , ..., , ...,j i nD R R R R R
1 2, , ..., , ..., , ...,i j nD R R R R R
1 2, , .., nD R R R
Let , be two transpositions. Let D be a matrix defined by
1 , ..., nD R R .
1, ..., nC C say
Then 1 21 2, , ..., , , ..., nnD C C C D C C C
1 2, , ..., nD R R R
In general, 0 0... 1 0 0... 2 0 0...1 2 1 2 1 2, , ..., nk k k
D R R R
1 2 1 2... , , ...,k nD R R R
If is any permutation, which is written as a product of transpositions,
say 1 2 ... r
1 , 2 0 0...0 1 0 0...1 2 1 2, ..., , ...,n nr r
D R R R D R R
1 2 1 2... , , ...,r nD R R R
181
Suppose D is an n-linear skew-symmetric function on nn
matrices over .
Let A be an nn matrix over with rows. 1 2, , ..., nA A A . Let
1 2, , ..., nE E E be the rows of the nn identify matrix nI then we can
write.
1
n
i ij jj
A A E
, For 1 i n
1 2 31
, , ,n
j j nj
D A D A E A A A
1 2 31
, , , ...,n
j j nj
A D E A A A
Now replace 2A by 21
n
k kk
A E
2 11
, , ..., , , ...,n
j n k j k nk
D E A A A D E E A
1 21 1
, , ...,n n
j k j k nj k
D A A A D E E A
Continuing this procedure for 3 4, , ..., nA A A . We get,
1 21 2 1 21 1 11 2
... , , ...,n n n
k k nk k k kn nk k kn
D A A A A D E E E
(Using n-linearity of D)
Also, ,1 2, ..., 0k k kn
D E E E whenever two indices ik are equal.
1 21 2 1 2, , ...,1 2
... , , ...,k k nk k k kn nk k kn
D A A A A D E E E
where the
sum on R.H.S. is extended over all sequences 1 2, , ..., nR R R of positive
integers not exceeding n since a finite sequence or n-tuple is a function
defined on the first n-positive integers such a function corresponds to
the n-tuple 1 , 2 , ..., n .
182
If D is on n-linear skew symmetric function and A is any nn
matrix over , we then have
1 1 2 2 1... , ...,n n nSn
D A A A A D E E
But we know that 1 21 , ..., , , nnD E E D E E E
1 21 1 2 2 ... , , ..., nn nSn
D A A A A D E E E
1 21 1 2 2 ... , , ..., nn nSn
A A A D E E E
(*)
Denote 1 1 2 2 ... n nSn
D A A A A
Since we know that 1 2, , ..., 1nD E E E D I
detD A A D I
Remarks :
(1) The expression 1 1 2 2 .... n nSn
A A A
above depends only
on the matrix A and thus uniquely determines D A .
(2) The expression 1 1 2 2 ... n nSn
A A A
gives explicit
formula for expansion of a determinant.
For example,
11 12 13
21 22 23 1 1 2 2 3 3 1 1 2 2
31 32 33
3 3 1 1 2 2 3 3 1 1
2 2 3 3
A A A
D A A A A A A A A
A A A
A A A A A
A A
1 1 2 2 3 3
1 1 2 2 3 3
A A A
A A A
183
Where 1 2 3 1 2 3 1 2 3
, ,1 2 3 2 3 1 3 1 2
Id
1 2 3 1 2 3 1 2 3
, ,1 3 2 3 2 1 2 1 3
One can see that 1
1
The above expression is 11 22 33 12 23 31 13 21 32A A A A A A A A A
11 23 32 13 22 31 12 21 33A A A A A A A A A is the same as the determinant
of a 3 3 matrix.
From above we see that there is precisely one determinant function
on nn matrices over .
Denote this function by det,
1 1 2 2det ... n nSn
A A A A
Now, we shall see some properties of determinant function.
Theorem : Let A, B be two nn matrices, then det det .detAB A B
Proof : Let B be a fixed nn matrix over , for each nn matrix A
define detD A AB
Let 1 2, , .., nA A A be the rows of A, then
1 2, , ..., det , , ...,nD A A A A B An B
Here 1jA B n matrix since 1i i i iCA A B CA B A B & det is n-
linear D is n-linear.
If i jA A then i jA B A B and & since det is skew-symmetric.
1, ..., 0nD A A whenever i jA A i j .
Now, D is an n-linear alternating function.
detD A A D I (by *)
184
But det detD I IB B
det det detAB D A A B
Theorem : For an nn matrix A over det dettA A where tA
denotes the transpose of A.
Proof : IF is a permutation in nS ,
,
ti ii i
A A
We know that 1 1det ... n nSn
A A A
1 ,1 ,det ...tn n
Sn
A A A
Assume that i j for some ,i j
1
1, ,i i j ji j A A
Therefore, 1 11 ,1 , 1, 1 ,... ...n n n n
A A A A
Since, 1 is the identity permutation
1sgn sgn 1 or 1sgn sgn
1 as varies over nS , therefore 1 also varies
over nS .
1
1 11 1 ,det sgn ...t
n nSn
A A A
det A
Theorem : Let A be an invertible nn matrix, then 11det detA A
.
Proof : Since 1 1, det det 1AA I AA I
185
1det det 1A A
11det detA A
Check your progress :
1) If A is the matrix over given by
0
0
0
a b
A a c
b c
then show that
det 0A .
(Hint : Use the expression 1 1 3 32 2det sgn
Sn
A A A A
)
2) Prove that the determinant of the Vandermonde matrix
2
2
2
1
1
1
a a
A b b
c c
is b a c a c b .
(Hint : Replace a by b, b by c & c by a in the given matrix A, implying
that ,b a c a c b are factors of det A .)
3) Recall that An nn matrix A is called triangular if 0ijA for i j or
0ijA for i j , A is said to be upper-triangular if 0ijA for i j
and A is said to be lower triangular if 0ijA for i j . Prove that the
determinant of a triangular matrix is the product of it‟s diagonal
entries.
(Use the expression 1 1 2 2det sgn n nSn
A A A A
)
4) Prove that
2 11 1 1
2 12 2 2
2 1
1
1det
1
n
n
j i
nn n n
x x x
x x xx x
x x x
1 i j n
(Use induction on n.)
10.3 LAPLACE EXPANSION OF A DETERMINANT :
The matrix involved here is referred to as the Vandermonde matrix.
186
Now we shall see one Laplace expansion for the determinant of an
nn matrix over .
Fix 1, 2, ...,r n . Let be a permutation in nS .
Define ... 1 21 21 sgnj j j r r
rJe
and ik r i
where is a permutation which permutes the sets 1, 2, ..., r and
1, 2, ...,r r n within themselves.
Then
, , , 1 ,
1 1 1 1
...1 2 ,1 , , 1 ,
det
j r j r k r k n
Jj j i
r j j r k r k nr r r r
A A A A
A e
A A A A
Here 1 21 , 2 , ..., rj j r j
1 21 , 2 , ..., ir k r k r i k
This is one Laplace expansion for the determinant others may be
obtained by replacing the sets 1, 2, ..., r and 1, ...,r n by two
different complementary sets of indices.
Check your progress :
1) Show that the equation of the line through distinct vectors
1 1 2 2, , ,x y x y is 1 1
2 2
1
1 0
1
x y
x y
x y
.
2) Prove that the area of the triangle in the plane with vertices
1 2 1 2 1 2, , , , ,x x y y z z is the absolute value of
1 2
1 2
1 2
11
12
1
x x
y y
z z
.
3) Suppose we have an nn matrix A of the block form P Q
O R
where P
is r r matrix, R is ss matrix, Q is r s matrix & O denotes the s r
null matrix.
Then det det detA P R
187
(Hint : Define , , detP Q
D P Q RO R
, consider D as an s-linear
function of the rows of R.)
10.4 SUMMARY :
In this chapter we have learned the following topics.
1) Determinant as an n-linear skew symmetric function defined on the set
of nn matrices over as a function of rows or columns.
det : ...n times
n n n
1 1det sgn ... n nSn
A A A
where the sum on R.H.S. is taken
over all permutations in nS .
2) Determinant function exists on the set of nn matrices over and it
is unique.
3) det det detAB A B for nn matrices A, B
1
det det .det dettA A A A .
4) The Vandermonde matrix
2 11 1 1
2 12 2 2
2 1
1
1
1
n
n
nn n n
x x x
x x x
x x x
and it‟s
determinant 1j ix x i j n .
5) A laplace expansion for the determinant of nn matrix A as follows.
,1 , , 1 ,1 1 1 1
...1 2 ,1 , , 1 ,
det
j j r k r k n
Jj j j
r j j r k r k nr r r r
A A A A
A e
A A A A
where the sum is taken over all r-tuples 1 2, , ..., rj j j J such that
1 2...
rj j j
1 21 , 2 , ...,
rj j r j .
188
6) If D is an n-linear skew-symmetric function on nn matrices over ,
then detD A A D I .
7) If A is nn triangular matrix then 11 22det ... nnA A A A .
189
11
PROPERTIES OF DETERMINANTS AND
APPLICATIONS
Unit Structure :
11.0 Objectives
11.1 Introduction
11.2 Properties of determinants and Cramer‟s Rule
11.3 Determinant as area and volume
11.4 Summary
11.0 OBJECTIVES :
This chapter would help you understand the following concepts.
In Euclidean n-dimensional vector space n ; linear dependence and
linear independence of vectors via determinant as a function of
vectors.
The existence and uniqueness of the system AX B where A is an
nn matrix 1 ,ija i j n with det 0A . X is n1 column
vector in n-unknowns 1 2, , ..., nx x x
1
n
x
X
x
. B is given column
vector
1
n
b
B
b
. The system AX B can be represented in the
following form as a set of n-equations in n-knowns 1 2, , ..., nx x x .
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
...
...
...
n n
n n
n n nn n n
a x a x a x b
a x a x a x b
a x a x a x b
Cofactors and minors Adjoint of an nn matrix A. basic results such
as . det . nA adj A A I
190
An nn real matrix A is invertible, if and only if det 0A ; for nn
invertible matrix A.
1 1
detA adj A
A
Cromer‟s rule as a method for solving a system AX B whenever
det 0A .
Uses of determinant as area and volume.
11.1 INTRODUCTION :
Determinants also allow us to determine when vectors are linearly
independent.
For example, if you consider two vectors 1
2
and 2
4
in the plane of
2 , then the determinant of the matrix 1 2A C C where 1
1
2C
&
2
2
4C
gives you the information, whether 1C & 2C are linearly
independent vectors or linearly dependent vectors in terms of the det A ,
which is equal to 1 2
1 4 2 22 4
= 0.
This implies that 1
1
2C
& 2
2
4C
are linearly dependent,
which can be clearly because 1 22C C . In other words 1 22 0C C .
We try to generalize this result in the following theorem.
Theorem : Let 1 2, , ..., nC C C be column vectors in n and 1 2, , ..., nC C C
are linearly dependent, then 1 2, , ..., 0nD C C C .
Proof : Suppose that given n-column vectors 1 2, , ..., nC C C in n are
linearly dependent.
scalars 1 2, , ..., n in such that
1 1 2 2 ... 0n nc c c 0 n
and some scalars are non-zero.
Assume that 0j for some j 1 j n
191
1
nk
j kjk j
k
C C
1 2, , ..., , ...,j nD C C C C 1 2
1
, , ..., , ...,n
kk n
jk j
k
D C C C C
1 2
1
, , .., , ..,n
kk n
jk j
k
D C C C C
11 21 1 1 2 2
11 2 1 1 1 1 1
1
, ..., , ..., , , ..., , ..., ....
, , ..., , , ..., , ..., , , ...,
... , ..., , ...,
jn n
j j j
jj j n j j n
j
nn n
j
D C C C D C C C C
D C C C C C D C C C C
D C C C
= 0
Since each determinant has a column equal to the jth
column.
Colollary : If 1 2, , ..., nC C C are column vectors of n such that
1 2, , ..., 0nD C C C then there exist scalars 1 2, , ..., n such that
1 1 2 2 ... n nB c c c for given column vector B in n .
Proof : Since 1 2 1 2, , ..., 0 , , ...,n nD C C C C C C are linearly
independent n-vectors in n (by above theorem).
1 2, , ..., nC C C being a linearly independent set of n-vectors in n , it
forms a basis of n .
Any vector of n can be written as a linear combination of
1 2, , ..., nC C C . Thus for nB there are scalars 1 2, , ..., n such that
1 1 2 2 ... n nB c c c
Note : The above corollary states that the system of n-equations in n-
unknowns 1 2, , ..., nx x x namely,
192
AX B OR
11 1 12 2 1 1
1 1 2 2
...
...
n n
n n nn n n
a x a x a x b
a x a x a x b
has a solution
provided that det 0A .
We shall also see that the converse of the above theorem is also
true, in other words the determinant.
1 2 1 2, , ..., 0 , , ...,n nD C C C C C C are linearly dependent.
Before providing this result we prove certain results about on nn matrix
A.
Definition : An nn matrix A is said to have rank k, 1 k n , if A has k
linearly independent rows or columns (k is largest such number).
For example,
1 0 1
2 1 2
3 1 3
A
has rank equal to 2.
1 1
2 1 2
3 3
.
1 0 0
0 1 0
0 0 1
B
has rank equal to 3. ( 1 2 3, ,E E E are linearly independent.)
1 0 1 0
2 1 0 2
3 1 1 2
0 1 2 1
C
has rank equal to 3,
3, 1, 1, 2 1, 0, 1, 0 2, 1, 0, 2
We have seen that an nn matrix A is invertible iff rank A n ,
because if 1 2 ... nA A A A is an nn matrix then A is invertible iff
1 2, , ..., nA A A are linearly independent, which has same meaning that rank
A n .
To compute determinants recall that we use mainly the following
two column operations.
1) Add a scalar multiple of one column to another.
2) Interchange two columns.
These are called as elementary column operations.
193
Note : There is a term called as elementary row operations, the difference
is that we add a scalar multiple of one row of another and we interchange
any two rows.
Definition : We say that two matrices A, B are row equivalent or (column
equivalent) if either can be obtained from the other by elementary row
operations or (Column operations).
For example,
1)
1 0 1 0
2 1 0 2
3 1 1 2
0 1 2 1
and
1 0 1 0
2 1 0 2
0 0 0 0
0 1 2 1
are row equivalent.
2)
2 0 1 1 2 0
1 1 0 and 0 1 1
0 1 3 3 0 1
are column equivalent.
Next we shall state a result for an nn matrix A, which is column
equivalent to a triangular matrix.
Theorem : Let A be an nn matrix, then A is column equivalent to a
triangular matrix B of the form
11
21 22
1 2
0 0 0
0 0
n n nn
b
b bB
b b b
.
Proof : To prove this theorem use induction on n and consider two cases.
1) All elements in the first row of A are 0.
2) Some elements in the first row of A are not 0.
For 1n there is nothing to prove.
Assume that 1n , let , 1, 2, ..., , 1, 2, ...,ijA a i n j n
Case (1) 21 22 2
1 2
0 0 0
n
n n nn
a a aA
a a a
194
Consider
22 2*
2 1 1
n
n nn n n
a a
A
a a
.
By Induction hypothesis *A is equivalent to a triangular matrix, say
22
32 33
2 3
0 0
0
n n nn
C
C CC
C C C
.
Now, the column operations performed on *A to obtain C do not
affect the first column of the matrix A, hence A changes to a matrix.
21 22
31 32 33
1 2 3
0 0 0
0 0
0 0
n n n nn
a C
a C C A
a C C C
is equivalent to a linear triangular
matrix.
Case (2) : Assume that 11 0a .
Performing column operations again, we can reduce each of the
elements 1 ja to 0 for 2, 3, ...,j n .
A is column equivalent to
11
21 22 23 2
1 2 3
0 0 0
n
n n n nn
a
a p p pM
a p p p
Now, consider the 1 1n n matrix
22 2
2
n
n nn
p p
p
p p
.
By induction hypothesis, the matrix p is row equivalent to
22
32 33
2 3
0 0
0
n n nn
q
q q
q q q
.
195
Again, the column operations on p do not affect the first column of
M nor the zero elements of first row of M. Hence A is column equivalent
to
11
21 22
1 2 3
0 0 0
0 0
n n n nn
a
a q
a q q q
which is a lower triangular matrix.
This completes the proof.
We have seen that if the column vectors 1 2, , ..., nC C C are linearly
dependent then 1 2, , ..., 0nD C C C . Here we prove it‟s converse.
Theorem : If 1 2, , ..., 0nD C C C then the column vectors 1 2, , ..., nC C C
are linearly dependent.
Proof : Let 1 2, , ..., nA C C C , jC is jth
column of the matrix A. By the
previous theorem, A is column equivalent to say
11
21 22
1 2
0 0
0 0
n n nn
b
b bB
b b b
and det detA B
1 2det , , ..., 0nB D C C C .
Since, B is a triangular matrix.
11 22det ... nnB b b b .
Assume that 0kkb for some k, 1 k n .
11
21 22
1, 1
1,
1 2
0 0
0 0
kkbk k
k k
n n nn
b
b b
bB
b
b b b
Reduce 1, 1k kb to 0 by multiplying kth
column by 1, 1
1,
k k
k k
b
b
and
subtracting from 1k th column continuing in this manner of reducing
196
kkb to 0 upto k n , using such column operations will lead to a matrix
column equivalent to B,
Say
Since D is column equivalent to B and B is column equivalent to A.
rank = rankA D .
However, the set 1 2, , .., nD D D of column of D is linearly
dependent set, as the last column is 0.
rank 1D n
rank 1A n 1 2dim , , ..., 1nC C C n
1 2, , ..., nC C C are linearly dependent.
Corollary : A square nn matrix A is invertible iff det 0A .
Now, we shall see the existence and uniqueness of the system
AX B , where A is an nn matrix with det 0A .
Given a system of n-equations in n-unknowns
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
...
...
...
n n
n n
n n nn n n
a x a x a x b
a x a x a x b
a x a x a x b
can be represented in the following form AX B where A is the nn
matrix.
1 2 ...t
nX x x x is a column vector in n .
1 2 ...t
nB b b b is a column vector in n .
nB is fixed column vector.
d11000
n1 n2d d
dK-1,K-1
n-1d
00
0
D =
197
If A is invertible 1A exists.
By premultiplying the system AX B by 1A we get
1 1A AX A B .
1 1A A X A B
1IX A B
1X A B
This implies that the system has a unique solution namely 1X A B provided that A is invertible matrix.
We know that if det 0A A is invertible hence det 0A
implies that the system AX B has a solution, which is unique.
Now, we shall prove basic results about the determinants, before
that let us understand few more concepts of cofactors and minors.
Definition : Let A be an nn matrix over A i j denotes the
1 1n n matrix obtained from A by deleting ith
row and jth
column
of A.
1 ,ijA a i j n , for 1 ,i j n
The cofactor of ija is the scalar given by 1 deti j
A i j
.
Example :
1) Consider a 3 3 matrix
1 1 0
0 2 2
1 3 1
A
.
The cofactor of 1 1
11 1 det 1 1a A
2 2
2 6 43 1
Here 2 2
1 13 1
A
.
198
The cofactor of 2 1
21 1 det 2 1a A
1 0
1 0 10 1
Here 1 0
2 13 1
A
.
2) Consider a 4 4 matrix
1 1 2 0
0 1 1 2
3 2 1 0
1 0 0 3
A
The cofactor of 4 2a
4 2
1 det 4 2A
1 2 01 2 0 2
0 1 2 1 2 01 0 3 0
3 1 0
2 2 0 6 2 12 10
Definition : Let A be an nn matrix then the 1 1n n matrix.
A i j obtained by deleting ith
row and jth
column of the matrix A is
called as the ,i j minor of the matrix A.
Example :
1) Consider a 3 3 matrix
3 1 0
1 2 5
1 2 0
A
the 1, 1 minor of the matrix A
2 5
1 12 0
A
2) Consider a 4 4 matrix
1 1 0 2
2 1 3 0
0 2 2 1
3 4 1 3
A
199
The 3, 4 minor of the matrix A =
1 1 0
2 1 3
3 4 1
Definition : Let A be an nn matrix. 1 ,ijA a j j n . Let ijC
denote the cofactor of ija for all ,i j ,a i j n .
1 deti j
ijC A i j
for 1 ,i j n write 1 ,ijC C i j n the
C is an nn matrix called as the matrix of cofactors of A.
(Note : ijC = the cofactor of ija )
Example :
1) For a 3 3 matrix
2 1 3
0 1 1
1 2 0
A
1 1
11
1 11 det 1 1 2
2 0C A
1 2
12
0 11 det 1 2 1
1 0C A
1 3
13
0 11 det 1 3
1 2C A
21 22 23 31 32 336, 3, 5, 4, 2, 2C C C C C C
The matrix of cofactors of A
2 1 1
6 3 5
4 2 2
C
Definition : Let A be an nn matrix over then the transpose of the
matrix of cofactors of A is called as adjoint of A denoted by adj A.
Example :
1) In above example
2 6 4
1 3 2
1 5 2
Tadj A C
our aim behind
finding out adjoint of the matrix A is to simply steps in order to obtain the
inverse of the matrix A, whenever det 0A .
200
Check your progress :
1) Find adj A for the following matrices
i)
1 1 0
2 1 3
1 2 0
ii)
1 2 3
2 1 2
4 5 3
iii)
1 1 2 0
0 3 5 2
2 2 3 1
1 0 5 4
2) For the 4 4 matrix A given by
1 1 2 0
2 3 3 1
4 5 0 3
2 1 3 2
A
, find
i) ,i j minor of A for all ,i j , 1 , 4i j
ii) the cofactor of ija for all ,i j . 1 , 4i j
11.2 PROPERTIES OF DETERMINANTS AND
CRAMER’S RULE :
Now, we have already proved that a determinant function defined
on nn matrices is given by 1
1 detn
i jj ij
i
E A a A i j
whenever
det is a determinant function.
But the uniqueness of a determinant function tells us that, if we fix
any column index j.
1
det 1 detn
i jij
i
A a A i j
(the expansion by minors of the jth
column)
1
detn
ij iji
A a c
Let B be the matrix obtained by replacing jth
column of A by kth
column of A (where j k )
B has two equal columns det 0B & B i j A i j .
1
det 0 1 detn
i jij
i
B b B i j
201
1
1 detn
i jik
i
a A i j
1
n
ik iji
a C
1 det & 1i j
ij ij ikC A i j b a i n
If
1
0n
ik iji
j k a C
This means that
1
detn
ik ij iki
a C A
0 if , 1 ifjk j k j k
1
det .n
ij ik jki
C a A
1
det .n
ik jkjii
adj A a A
. det .adj A A A I
Theorem : For on nn matrix A over .
det nA adj A A I
Proof : We already know that . det nadj A A A I .
For an nn matrix A.
Applying this result for tA .
. det . dett t tadj A A A I A I
. dett tadj A A A I
Taking transpose on both the sides,
. det .t
t tadj A A A I
det .t t
t tA adj A A I
202
. det .tA adj A A I
But ttadj A adj A
det .t
tA adj A A I
. det .A adj A A I
Theorem : Let A be an nn matrix over , then 1A exists if and only if
det 0A whenever A is invertible, the unique inverse for A is
11 1
detdet
A A adj A adj AA
Proof : For an nn matrix A over , we know that
. det .A adj A A I .
If A is invertible 1A exists premultiplying by 1A on both the sides,
we get,
1 1 detA A adj A A A I
1 1. detA A adj A A A
1 1
detA adj A
A
Conversely, we know that for an nn matrix over ,
. det nA adj A A I and . det nadj A A A I
If 1
det 0det
nA A adj A IA
Also 1
detnadj A A I
A
1A exists and 1 1
detA adj A
A
We have use of the following theorem in proving further properties
of determinant.
203
Theorem : (Cramer’s Rule) : Let A be the matrix with column vectors
1 2, , ..., nA A A such that 1 2, , ..., 0nD A A A .
Let B be a column vector
1
2
n
b
b
b
such that AX B where
1
2
n
x
xX
x
is a
column vector consisting of n-unknowns, then for each j.
det
det
jj
Mx
A
Where jM is the nn matrix obtained from A by replacing the jth
column of A by B.
1 2
th
, , ..., , ...,
j column
j nM A A B A
Proof : For 1 j n , consider 1 2det , , ..., , ...,j nM D A A B A
We know that 1 1 2 2 ... n nx A x A x A B .
1 1 1 2 2det , ..., ... , ...,j n n nM D A x A x A x A A
1 1 1 2 1 2 2
1 1 2
, ..., , ..., , , ..., , ..., ...
, ..., , ..., ... , , ..., , ...,n n
j j n n n n
x D A A A x D A A A A
x D A A A x D A A A A
1 2, , ...,j nx D A A A
Since every term in this sum except jth
them is 0, because two
columns are same.
det
det
jj
Mx
A
Note : The above theorem gives us the way of obtaining the unique
solution of the system AX B where A is an nn matrix.
1 2 1 21 , , , , ..., , , , ...,t t
ij n nA a i j n X x x x B b b b
OR
204
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
...
...
...
n n
n n
n n nn n n
a x a x a x b
a x a x a x b
a x a x a x b
which is a system of n-equations in n-unknowns Cramer‟s rule
gives the method of solving the system, when det 0A .
Example :
1) Consider the following system of linear equations
3 2 4 1
2 0
2 3 1
x y z
x y z
x y z
Solution : Let
3 2 4
2 1 1 det 5
1 2 3
A A
1 2 41 1
0 1 15 5
1 2 3
x
3 1 41
2 0 1 05
1 1 3
y
3 2 11 2
2 1 05 5
1 2 1
z
Check your progress :
1) For an nn matrix A, show that A adj adj A , whenever det 1A .
2) Use adjoint to find the inverse of
i) a b
c d
, given that 0ad bc
ii)
1 2 0
1 1 1
1 2 1
205
3) Compute
5 0 0 0
7 2 0 0
9 4 1 0
9 2 3 1
4) Solve the following systems using Cramer‟s rule.
i) 3 0x y z ii) 2 1x y z
0x y z 3 2 0x y z
iii) 2 1x y z iv) 2 3 4x y z a
2x y z 5 6 7x y z b
2 5x y z 8 9 9x y z c
11.3 DETERMINANT AS AREA AND VOLUME :
Now we shall see how a 2 2 determinant represents the area of a
parallelogram and a 3 3 determinant represents the volume of a
parallelepiped.
Consider a parallelogram spanned by the two vectors u & where
1 2 1 2, , ,u u u both in 2 such parallelogram is denoted by
,P u .
u+v
o u
, 0 , 1P u u
Let ,A u denote the area of ,P u
, 0A u
Here we introduce the concept of oriented area denoted by
, ,A u A u if 1 1
2 2
0u
u
206
,A u if 1 1
2 2
0u
u
,A u has same sign as 1 1
2 2
u
u
We shall show that ,A u is actually the determinant
1 1
2 2
,u
D uu
. Consider the following axioms about area, which
are accepted to be true always.
1) The area of a line segment is zero.
2) If G is some region in a plane then the area of G is same as the area of
translation of G by any vector .
A G A G
G g g G
3) If 1G and 2G are two regions which are non-intersecting or their
intersection has area 0, then 1 2 1 2A G G A G A G
The main good in this section is to show that the oriented area of a
parallelepiped spanned by vectors ,u in a plane is same as ,D u .
Theorem : , ,A u D u
Proof : In order to show this result, we shall show that ,A u is a 2-
linear skew-symmetric function from 2 2 such that
1 1, 1A e e where 1
1
0e
, 2
0
1e
, then by the uniqueness theorem
of the determinant function.
, ,A u D u
Therefore we need to check the following properties of 2 2:A .
1) A is linear in each variable ,u
2) , 0, , ,A u u A u A u
3) 1 2, 1A e e
207
1) Firstly show that following.
i) , , ,A nu n A u n
ii) 1 1
, , .A u A u nn n
iii) For , 0C C , ,A cu CA u
iv) For any , , ,C A cu CA u & 0 , ,A u c CA u
v) , ,A u A u
vi) , ,A u A u
vii) , ,A cu d CA u
viii) 1 2 1 2, , ,A u u A u A u
2) 1 0A u u is clear from the definition of oriented area, let
, 0D u .
, , ,A u A u A u
, , ,A u A u A u
, ,A u A u
The case , 0D u is same.
(Here ,u are linearly independent vectors.)
3) 1 2,A e e is the area of a unit square which is equal to
1 21 , 1A e e
(1), (2) and (3) together show that , ,A u D u
Now, we turn to showing that if , ,u are vectors in 3 then the
oriented volume of the parallelepiped spanned by , ,u is , ,D u .
Denote the volume of the parallelepiped , ,P u by , ,V u and the
corresponding oriented volume by , ,V u .
208
u
We extend the postulates we made about areas to volumes.
1) The volume of a plane area or a line segment is 0.
2) If G is a certain region in 3 then the volume of G is the same as the
volume of the translation of G by any vector thus for a vector
3Z V G Z V G .
3) If 1G and 2G are two regions, which are disjoint or such that their
intersection has volume zero, then 1 2 1 2V G G V G V G .
Theorem : , , , ,V u D u
Proof : We have to show that , ,V u is a 3-linear skew-symmetric
function on 3 3 3 such that 1 2 3, , 1V e e e where
1 2 3
1 0 0
0 , 1 , 0
0 0 1
e e e
Thus we need to show that
1) V is linear in each variable , ,u
2) , , 0V u if u or , , , , ,u u
, , , ,u u
, , , ,u u
3) 1 2 3, , 1V e e e
1) Firstly show the following.
i) , , , ,V nu nV u for n
ii) For 3n , 1
, , , ,m
V u V un n
209
iii) , , , ,V u V u
iv) For any real no. C, , , , ,V cu CV u
v) , , , ,V u V u
vi) For any real no. C
, , , , , , , ,V cu V u c V u c CV u
vii) , , , ,V u V u
viii) , , , ,V u V u
Use that , , , ,D u D u
ix) , , , ,V au b c aV u
x) 1 2 1 2, , , , , ,V u u V u V u
1 2 1 2, , , , , ,V u V u V u
1 2 1 2, , , , , ,V u V u V u
Consider linearly independent vectors , ,u .
There are numbers, , , , ..., , ,a b c such that
1u au b c
2u u .
1 2, , , ,V u u V a u b c
, ,a V u
, , , ,aV u V u
, , , ,V au b c V u
1 2, , , ,V u V u
This above theorems can be interpreted in terms of linear maps as
follows. Let ,u be two linearly independent vectors in the plane. Then
we know that there is a unique linear transformation 2 2:T such
that 1u Te and 2Te .
Let 1 1 2 2u e e & 1 1 2 2e e .
210
Then m T w.r.t. the standard basis on both sides is 1 1
2 2
, denote
m T by detT .
,A u = absolute value of det ,u
= absolute value of det m T
detT
If C is a unit square, then
1 1 2 2 1 20 , 1C t e t e t t
1 1 2 2 1 20 , 1T C T t e t e t t
1 1 2 2 1 20 , 1t Te t Te t t
1 2 ,t u t P u
Theorem : Let 2 2:S be a linear map, then Area of
, detS P u S (Area of ,P u ).
Proof : Since ,P u is spanned by two vectors ,u . ,S P u is
spanned by ,S u S .
u
uS
S
There is a linear map 2 2:T such that 1Te u and
2Te , then as seen above ,P u T C where C is the unit square.
211
Hence ,S P u S T C S T C .
, detA S P u S T
det detS T
det ,S A u
We shall repeat the above arguments for the parallelepiped in 3 .
Let , ,u be three linearly independent vectors in 3 . Then
there is a unique linear transformation 3 3 3: such that
1 2,Te Te and 3Te .
Write 1 1 2 2 3 3u u e u e u e
1 1 2 2 3 3e e e
1 1 2 2 3 3e e e
1 1 1
2 2 2
3 3 3
u
m T u
u
, , det , ,V u u
det detm T T
If C is a unit cube, then
1 1 2 2 3 3 0 1jC e e e
1 1 2 2 3 3T C T e e e
1 1 2 2 3 3 0 1 , ,ie e e P u
Theorem : Let 3 3:S be a linear map, then volume of
, , detS P u S volume of , ,P u .
212
Proof : , ,S P u is spanned by , ,S u S S .
Now, , ,P u T C
, ,S P u S T C S T C
, , det det . detV S P u S T S T
det , ,S V P u
11.4 SUMMARY :
In this chapter we have learned the following topics.
1) Determinant as a function of vectors in n .
2) A set of n-vectors in n is linearly independent iff their determinant is
non vanishing.
3) Given a system of n-equations in n-unknowns
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
...
...
...
n n
n n
n n nn n n
a x a x a x b
a x a x a x b
a x a x a x b
OR AX B
The system has a unique solution namely 1X A B provided that the
co-efficient matrix. A has non-vanishing determinant.
4) Basic results like det . nA adj A A I , for an nn matrix A over
.
5) use of adj A to find inverse of an invertible nn matrix A over as
1 1
detA adj A
A
.
6) Cromer‟s rule as a method for solving the system AX B , whenever
det 0A , in this case the unique solution is given in terms of the
determinant function as follows.
det
det
jj
Mx
A for 1 j n
7) Determinants can be used to find the area and volume of regions in n .
213
12
RELATION BETWEEN MATRICES AND
LINEAR TRANSFORMATIONS
Unit Structure:
12.0 Objectives
12.1 Introduction
12.2 Representation of a linear transformation by matrix
12.3 The matrices associated with composite, inverse, sum of linear
transformation.
12.4 The connection between the matrices of a linear transformation
with respect to different bases.
12.5 Summary
12.0 OBJECTIVES :
This chapter would help you understand the following concepts and
topics.
Representation of linear transformation from U to V. Where U and
V are finite dimensional real vector spaces by matrices with
respect to the given ordered bases of U and V.
The relation between the matrices of linear transformation from U
to V with respect to different bases of U and V.
Matrix of sum of linear transformations and scalar multiple of a
linear transformation.
Matrices of composite linear transformation and inverse of a linear
transformation.
12.1 INTRODUCTION :
In this chapter we shall investigate for the relation between the
matrices and linear transformations. Given a linear trams formation.
2 3:T defined by
214
, , ,T x y x y x y we shall see the images of standard basis
element 1
1
0e
. 2
0
1e
of 2 under this linear transformation.
1
11,0,1
0Te T
2
00,1, 1
1Te T
Since 31 2,Te Te and 1,0,0 , 0,1,0 , 0,0,1 is a basis for 3 .
1 11 21 311,0,0 0,1,0 0,0,1Te a a a
2 12 22 321,0,0 0,1,0 0,0,1Te a a a
For some scalars 11 21 31 12 22, 32, , , ,a a a a a a which are uniquely
determined.
We found that
11 1a 12 0a
21 0a and 22 1a
31 1a 32 1a
If we form a matrix of order 3 2 as follows:
1 2
1 3ijj
A a i
3 2
1 0
0 1
1 1
A
If we consider AX, where 1
2 2 1
xX
x
then we get 11 2 1 2
2
1 0
0 1 , ,
1 1
xx x x x
x
1 2, ;T x x AX TX AX
In this way we associated a 3 2 matrix A to a given linear
transformation, not only this such matrix A is unique.
215
Recall that the set 1,0 , 0,1 forms a basis for the vector space
2. As a set it is not different from the set 0,1 , 1,0 . So we
afterwards try to distinguish between these sets by using a term called
ordered basis.
12.2 REPRESENTATION OF A LINEAR
TRANSFORMATION BY MATRIX :
Assume that V is an n-dimensional vector space with ordered basis
1 2, ,..., nB and let V be an m-dimensional vector space with
order basis 1 2, ,..., mB .
Let :T V V be a linear transformation. Since 1jT j n is
an element in V , it can be expressed as a linear combination of elements
of B . Therefore for each j, 1 .j n
We introduce m numbers 1 2,j j mja a a
we produce mn numbers in all
Write
1
m
j ij ii
T a u
This gives rise to an m n matrix. 1
1ijj n
A a i m
A has thj column 22, ,...,
ij
tjij j mj
mj
a
aa a a
a
11
21
1m
a
a
a
is the first column of A.
12
22
2m
a
a
a
is the second columns of A.
216
1
2
n
n
mn
a
a
a
is the thn column of A.
Call this matrix A to be the matrix of T with respect to the ordered basis
,B B of ,V V respectively, denote this matrix by B
Bm T
.
Note: B
Bm T
is uniquely determined for given ordered bases ,B B of V
and V respectively.
We shall use this procedure of finding B
Bm T
to establish a one-
to-one correspondence between ,L V V and m nM .
Theorem: Let V be an n-dimensional vector space with ordered basis
1 2, ,..., n and let V be an m-dimensional vector space with basis
1 2, ,..., m . Then there is a one-to-one correspondence between
,L V V and m nM .
Proof:
Let , , :T L V V T V V is a linear transformation.
1
m
j ij ii
T a
for each j, 1 j n .
This gives rise to the matrix B
Bm T
.
Define a function : , m nL V V M by
B
BT m T
We show that for each m nA M these is a unique linear map from V to
V , whose matrix is precisely A.
Let ,1 ; 1ijA a i m j n .
Let 1 1 11 11 1 21 2 1... m my a a a
1 1 12 12 1 22 2 1... m my a a a
217
1 1 11 2 2 ...n m n mn my a a a
Here 1 2, ,..., ny y y are vectors on V . We know that a unique linear
transformation :T V V such that j jy T for j =1, 2 …, n
B
Bm T A
.
: , m nL V V M is onto map.
Let BX denotes co-ordinate vector in n of a vector V .
BX T denotes co-ordinate vector in m of a vector T V .
We show that
B
B BBX T m T X
Let
1
2,
n
j j Bj i
n
x
xV x X
x
Let ijT A a
Let iA the thi row of the matrix A.
1 2, ,...,i i ina a a .
1
21 2, ,...,i B i i in
n
x
xA X a a a
x
1
n
ij jj
a x
11 12 1 1
21 22 2 2
1 2 3
n
nB
m m mn
a a a x
a a a xX
a a a x
218
1
21
1
n
ij jj
n
j jj
n
mj jj
a x
a x
a x
Also,
1 1
n n
j j j jj j
T T x x T
1 1 1 1
n m m n
j ij i ij j ij j i j
x a a x
1
1
n
ij jj
B B
n
mj j Bm T X Bj B
a x
X T X
a x
B
B BBX T m T X
We use this result to prove injectivity of .
Let 1 2B B
B BT m T
1 1B
B BBX T m T X
2 2B
B BBm T X X T
1 2T T for all V
1 2T T
is an injective map
There is a on to one correspondence between ,L V V and m nM .
219
12.3 THE MATRICES ASSOCIATED WITH
COMPOSITE, INVERSE, SUM OF LINEAR
TRANSFORMATION :
Given linear transformations S, T, we saw that S T is again a
linear transformation in the next theorem, we understand the relation
between ,m S T m S and m T .
Theorem: Let , ,V V V be vector spaces with bases , ,B B B
respectively.
Let :T V V and :S V V be linear transformations. Then
B B B
B B Bm S T m S m T
Proof: Consider,
B
BBm S T X
BX S T
BX S T
B
BBm S X T
B B
BB Bm S m T X
Using comparison B B B
B B Bm S T m S m T
e.g.
Consider a linear transformation
:T Id V V T (V is finite dim vector space)
Then, for 1 j n
1
n
j j ij ii
id a
Since 1 2, ,..., n is a basis of V.
Therefore, 1jja and 0ija if i j for 11
i nj n
11
ij nm Id i n Ij n
(the n n identity matrix)
Corollary:
220
Let V be an n-dimensional vector space and :T V V be a linear
transformation. Let B
BA m T , where B is a fixed ordered basis
1 2 3, ,..., of V. Then T is non-singular iff A is non-singular, and in
that case 1 1B
Bm T A .
Proof:
Suppose that T is non-singular. Then 1 1T T Id T T .
1 1B BB
BB Bm T T m Id m T T
1 1m T m T I m T m T
1 1A m T I m T A
Hence m T A is invertible and 1 1A m T conversely,
Suppose that A is invertible. Then 1A exists and also there exists a
transformation.
:S V V such that 1B
Bm S A
1 ( )I AA m T m S m T S
Similarly 1( )m S T m S m T A A I
m T S I m S T
However I m Id
By one-to-one correspondence between ,L V V and n nM
T S Id S T
1T S
Thus T is invertible and by uniqueness of inverse 1T S .
Corollary: Let V be a vector space with ordered bases B and B .
Then B B
B Bm Id m Id I
B B
B Bm Id m Id
In the previous theorem take V V V and S T Id .
Replace B by B . Then
B B B
B B Bm Id m Id m Id
and
221
B B B
B B Bm Id m Id m Id
Since B
Bm Id I for all bases.
B B B B
B B B Bm Id m Id I m Id m Id
Theorem: Let :T V V be a linear transformation and let ,B B be bases
of V. Then there is an invertible matrix N such that
1B B
B Bm T N m T N
Proof:
Write :T V V as follows:
T Id T Id
B B
B Bm T m Id T Id
B B
B Bm Id T m Id
B B B
B B Bm Id m T m Id
Put B
BN m Id
Since B B B B
B B B Bm Id m Id T m Id m Id
N is invertible matrix and 1 B
BN m Id
1B B
B Bm T N m T N
Examples:
1) Let ,V V be vector spaces. Let B be a basis of V and B be a basis
of V .
Let S,T be two linear maps from V to V .
Then show that
B B B
B B Bm s T m S m T
and for any scalar ,
B B
B Bm T C m T
Solution:
Let 1 2 3, ,...,B and 1 2, ,..., mB be bases for ,V V
respectively.
222
Let 1 11 1
ij ijm S a i m m T b i mj n j n
and 11
ijm S T C i mj n
Then 1
m
j ij ii
S T C
for each j = 1, 2, n
also j j jS T S T
1 1
m m
ij i ij ii i
a b
1
m
ij ij ii
a b
ij ij ija a b for all ,i j
1 ; 1i m j n
Hence m S T m S m T
Also, 1 1
m m
j j ij i ij ii i
T T b b
ij ijm T b b m T
2) Let 3 5:T P P the linear transformation given by
21 2T p x x x p x . Find the matrix of T relative to the
standard bases of 3P and 3P .
Solution : We know that 3P has standard basis 2 31, , ,B x x x and
5P has standard basis 1 2 3 4 51, , , , ,B x x x x x .
21 1 2T x x
2 32T x x x x
2 2 3 42T x x x x
3 3 4 52T x x x x
223
1
1 0 0 0
2 1 0 0
1 2 1 0
0 1 2 1
0 0 1 2
0 0 0 1
1 2 3 4
B
Bm T
T T T T
3) Calculate the matrix of the linear transformation 41:T P given
by 1 2 3 4 1 3 2 4, , ,T x x x x x x x x x relative to the basis.
1 2 3 41,1,1,1 , 1,1,1, 0 , 1,1, 0, 0 , 1, 0, 0, 0B
of 4 and basis 1 1 11 21 , 1B x x of 1P .
Solution : 1 11 1 21,1,1,1 2 2 2 0T T x
1 12 1 2
3 11,1,1, 0 2
2 2T T x
1 13 1 21,1, 0, 0 1 0T T x
1 14 1 2
1 11, 0, 0, 0 1
2 2T T
1
3 12 1
2 2
1 10 0
2 2
B
BM T
4) Let 3 3:T be a linear transformation defined by
, , , , 0T x y z x x y . Find matrix of ,T m T
i) with respect to a natural basis B on both sides.
ii) with respect to basis 1,1, 0 , 0,1,1 , 1,1,1B on both
sides.
iii) with respect to basis 1 2 3, ,B e e e for the domain and basis
1 1,1, 0 , 0,1,1 , 1,1,1B for the co domain.
Solution :
224
i) Let 1 2 3, ,e e e be a natural basis for 3 on both the sides,
1 1, 0, 0e , 2 30,1, 0 , 0, 0,1e e
then 1 1 2 31, 0, 0 1,1, 0 1. 1. 0.Te T e e e
2 1 2 30,1, 0 0,1, 0 0. 1. 0.Te T e e e
3 1 2 30, 0,1 0, , 0 0. 0. 0.Te T e e e
1 0 0
1 1 0
0 0 0
m T
w.r.t natural basis for 3 on both the sides.
ii) Let 1 2 31,1, 0 ; 0,1,1 ; 1,1,1u u u
1 1 1 2 2 3 31,1, 0 1, 2, 0Tu T u u u
2 1 1 2 2 3 30,1,1 0,1, 0Tu T u u u
3 1 1 2 2 3 31,1, 1, 2, 0Tu T u u u
, ,i i i are scalars.
1 1 2 2 3 3 31, 2, 0 , , 0 0, , , ,
1 3 1 2 3 2 3, ,
1 2 32, 1, 1
1 1 2 32 1 1Tu u u u
Similarly,
2 1 2 30,1, 0 1 1 1Tu u u u ,
3 1 2 31, 2, 0 2 1 1Tu u u u
2 1 2
1 1 1
1 1
m T
w.r.t. basis 1 2 3, ,B u u u on both the
sides.
iii) 1 1 1 2 2 3 31,1, 0Te u u u
2 1 1 2 2 3 30,1,Te u u u
3 1 1 2 2 3 30, , 0Te u u u
225
Where , ,i i i are all scalars.
1 1 2 31,1, 0 1,1, 0 0,1,1 1,Te
1 3 1 2 3 2 31,1, 0 , ,
1 2 31, 0, 0
1 1 2 31,1, 0 1. 0. 0.Te u u u
Similarly,
2 1 2 30,1, 1. 1. 1Te u u u
3 1 2 30, , 0 0. 0. 0.Te u u u
1 1 0
0 1 0
0 1 0
m T
w.r.t. basis 1 2 3, ,e e e of 3 and basis
1 2 3, ,B u u u of 3 (co domain).
5) Let 2 2V M (set of 2 2 matrices over ). Let :T V V
be defined by 1 1
1 1T A A
. Find the matrix of T with respect to
basis 1 2 3 4
1 0 0 1 0 0 0 0, , ,
0 0 0 0 1 0 0 1E E E E
of
2 2M .
Solution :
1 1 2 3 4
1 1 1 0 1 01. 0. 1. 0.
1 1 0 0 1 0T E E E E E
2 1 2 3 4
1 1 0 1 0 10. 1. 0. 1.
1 1 0 0 0 1T E E E E E
3 1 2 3 4
1 1 0 0 1 01. 0. 1. 0.
1 1 1 0 1 0T E E E E E
226
4 1 2 3 4
1 1 0 0 0 10. 1. 0. 1.
1 1 0 1 0 1T E E E E E
The matrix of the transformation is
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
. Now it‟s
your turn to solve certain exercises.
Check your progress :
1) 2 2:T is defined by 1 1 2T and 2 1 2T , where
1 1 22e e and 2 1 2e e
i) write matrix of T, relative to basis 1 2, on both the sides.
ii) write matrix of T, relative to basis 1 2,e e on both the sides.
2) Let 1 2
3 4M
and T be the linear operator defined on the set of
22 matrices having basis 1 2 3
1 0 0 1 0 0, , ,
0 0 0 0 1 0E E E
4
0 0
0 1E
by .T A M A , find the matrix of T.
3) Let 1 1 2 31,1,1 , 1,1, 0 , 1, 0, 0B u u u be a basis of 3
and 2 1 21, 3 , 2, 5B be a basis of 2 . Let 3 2:T
be defined by , , 3 2 4 , 5 3T x y z x y z x y z . Find 2
1
B
Bm T .
4) Consider the set 3 3 2 3, ,t t te t e t e a basis of a vector space V of
functions ::f . Let D be the differential operator on V.
1D f t f t . Find the matrix of D in the given basis.
(Hint : Find 3 3 2 3, ,t t tD e D te D t e , write all these as a linear
combination of 3 3 2 3, ,t t te t e t e .)
227
12.4 THE CONNECTION BETWEEN THE MATRICES
OF A LINEAR TRANSFORMATION WITH RESPECT
TO DIFFERENT BASES :
Suppose V is an n-dimensional vector space and 1 2,B B be it‟s two
bases, the rule for assigning a matrix of T depends not only on T alone, but
also on the choice of basis for V.
Given on n-dimensional vector space with bases 1,B B , what can
be the relation between B
Bm T and
1
1
B
Bm T . To find the answer we state
the following theorem.
Theorem : Let V be a vector space over with basis
1 2, , ..., nB . Let ,T L V V be such that
1 ,B
ijBm T A a i j n . If B is another nn matrix such that
1B N AN for some invertible matrix N nij , then another basis
11 2, , ..., nB of V such that
1
1
B
Bm T B .
Proof : Since B
BA m T
1
n
j ij ii
T a
Define a transformation S corresponding to the matrix
1 ,ijN n i j n with respect to basis B, in other words
1
n
j ij ii
S n
.
Since 1N exists 1S exists.
Let j jS , since 1, ..., n is a basis of V and S is non-singular
transformation with the fact that 11 2, , .., nB is also a basis of V.
We claim that 1
1
B
Bm T B
Let 1
11 ,
Bij
Bm T D i j n
228
1 1
n n
j ij i ij ii i
T S
1 1
n n
ij ki ki k
n
1 1
n n
ki ij ki k
n
1 1
n n
ki ij kk i
n
(i)
Also
1
n
j ij ii
T T n
1 1 1
n n n
ij j ij ki ki i i
n T n a
1 1
n n
ki ij ki i
a n
1 1
n n
ki ij ki i
a n
(ii)
From (i) and (ii), we conclude that
ND AN , in other 1D N A N B
1
1
B
BB m T
In the above theorem, we saw that, how one can connect the two
matrices A, B relative to bases B and 1B of V respectively, in terms of on
invertible matrix N.
Definition : Two matrices , nA B M are said to be similar if there
exists a non-singular matrix C in nM such that 1B C AC .
Here we can observe that 1
1 1 1 1B C AC A CBC C B C
.
229
It is senseful to talk about two matrices as being similar to each
other, without specifying a particular order.
The above theorem can also be rephrased as follows. Two
matrices , nA B M represent the same linear mapping ,T L U V
relative to different bases for V if and only if A and B are similar.
N is called the transition matrix or change of basis matrix, from the
basis 1, ..., n to the basis 1 2, , ..., n . Also the equation
1
n
j ij ji
n
gives you the columns of N as the co-ordinates of -basis
vectors expressed in terms of the -basis, in other words expressing each
j in terms of a linear combination of vectors i gives columns of the
transition matrix N.
e.g. (1) Consider the polynomial space 3P with basis
2 31 2 3 41, , ,B x x x
and 1 2 2 31 2 3 41, 1 , ,B x x x x x
Find 1
1,
B B
B BA m D B m D , the transition matrix N and verify that
1B B A N where 3 3:D P P is a differential operator.
Solution :
0 1 0 0
0 0 2 0
0 0 0 3
0 0 0 0
BBm D A
2 3
2 3
2 2 3
3 2 2 3
1 0 0.1 0. 0. 0.
1 1.1 0. 0. 0.
2 0.1 2. 0. 0.
3 0.1 0. 3. 0.
D x x x
D x x x x
D x x x x x
D x x x x x
230
and 1
1
0 1 1 1
0 0 2 1
0 0 0 3
0 0 0 0
B
Bm D B
2 2 3
2 2 3
2 2 2 3
2 3
1 0 0.1 0. 1 0. 0.
1 1 1.1 0. 1 0. 0.
2 1.1 2. 1 0. 0.
2 3
D x x x x x
D x x x x x x
D x x x x x x x x
D x x x x
2 2 2 31.1 1 . 1 3. 0.x x x x x
Also the transition matrix
1 1 0 0
0 1 1 0
0 0 1 1
0 0 0 1
N
2 31
2 32
2 2 33
2 3 2 34
1.1 0. 0. 0.
1 1.1 1. 0. 0.
0.1 1. 1. 0.
0.1 0. 1. 1.
x x x
x x x x
x x x x x
x x x x x
It can be easily seen that 1N A N B .
2) Let D be the differentiation operator on 3P . Find the matrix B
representing D with respect to basis 21, ,B x x and find the matrix A
representing D w.r.t. 1 21, 2 , 4 2B x x .
Solution :
2
2
22
1 0 0 1 0 0
1 1 1 0 0
2 0 1 2 0
xD x
D x x x
x x xD x
231
0 1 0
0 0 2
0 0 0
B
Now,
2
2
2 2
1 0.1 0.2 0. 4 2
2 2.1 0.2 0. 4 2
4 2 0.1 4.2 0. 4 2
D x x
D x x x
D x x x
1
1
0 2 0
0 0 4
0 0 0
B
BA m D
To obtain the transition matrix N, we write
2
2
2 2
1 1.1 0. 0.
2 0.1 2. 0.
4 2 2 .1 0. 4
x x
x x x
x x x
1
11 0
21 2 21
0 2 0 0 02
0 0 41
0 04
N N
1N B N A
Check your progress :
1) Let 3 3:S be the linear transaction
, , , 0,S x y z y x z x .
a) Write the standard and matrix representation for operator T.
b) Let 1 2 31, 0,1 , 0, 1, 1 , 1,1, 0B u u u be another
basis of 3 . Find the transition matrix N from the basis B to the
standard basis 1 2 3, ,e e e of 3 .
c) Find the matrix representation of T with respect to the basis B.
232
d) Find the matrix representation of T with respect to the basis
1 2 31,1, 1 , 1,1, 1 , 0,1,1G of 3 .
2) Suppose that the linear mapping 3 3:T is represented by the
matrix
1 2 3
1 1 2
2 2 2
A
relative to the standard basis of 3 .
Find the matrix representation of T in the basis
3,1, 3 , 2, 0, 0 , 1, 1, 0 .
3) Find the matrix for the rotation in the plane through an angle
(anticlockwise) 2 2:R .
, cos sin , sin cosR x y x y x y
4) Find the matrix for the reflection in the plane along a line e making an
angle with positive x-direction given by,
, cos sin , sin cosT x y x y x y
12.5 SUMMARY :
In the chapter we learned the following concepts.
1) There exists a one-to-one correspondence between the set of mn real
matrices and the set 1,L V V where dimV n and 1dimV m
assuming given ordered bases 1, .., nB and 1 1 11, ..., mB of
V and 1V respectively.
2) For a linear operator :T V V on a finite dimensional vector space
V, if 1&B B are given bases of V then B
Bm T and
1
1
B
Bm T are
similar matrices, in other words, there exists an invertible matrix N s.t.
1
11
.B B
BBm T N m T N .
3) The matrix of the sum of linear transfer motions S and T is the sum of
matrices of linear transformations S and T, with respect to given bases
on both the sides,
1 1 1B B B
B B Bm STT m S m T
233
Similarly, 1 1
.B B
B Bm T m T where T is a linear
transformation defined by .T T for all V .
4) Given two similar matrices , nA B M , where dimV n , such
that B
BA m T and
1
1
B
BB m T for bases B and 1B of V.
234
13
LINEAR EQUATIONS AND MATRICES
Unit Structure:
13.0 Objectives
13.1 Introduction
13.2 Solutions of the homogeneous system AX= 0
13.3 Solution of the non-homogeneous system AX = B, where 0B
13.4 Summary
13.0 OBJECTIVES :
This chapter would help you understand the following concepts:
Rank of the linear transformation n mAL : defined by
AL X X and it‟s connection with the rank of on mn
matrix A.
The dimension of the solution space of the system of linear
equations 0AX where A is an mn matrix and X is n1
column sector or a system of m linear equations in n unknowns
consisting of a set of equations:
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x .... a x b
a x a x a x b
a x a x .... a x b
Where ija and ib are real numbers and 1 2 nx ,x ,...,x are unknowns
The dimension of the solution space of the system 0AX is equal
to n – rank A.
The solutions of non-homogeneous system of linear equations
represented by AX B or Existence of a solution when rank
A = rank (A,B). The general solution of the system AX = B is the
sum of a particular solution of the system AX = B and the solution
of the associated homogeneous system.
235
13.1 INTRODUCTION :
We would use linear algebra to methods of solving a system of
linear equations. Recall that by a system of m –linear equations in n-
unknowns, we mean a set of equations:
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
n n
n n
m m mn n m
a x a x .... a x b
a x a x a x b
a x a x .... a x b
Where ija and ib are real numbers and 1 nx ,...,x are n-unknowns
The mn matrix
11 12 1
21 22 211
1 2
n
ni mijj n
m m mn m n
a a ...a
a a ...aA a
a a ...a
is called the co-efficient matrix of the system or the matrix associated
with the system Let
1 1
2 2
n m
x b
x bX and B
x b
then the system
can be written as a single matrix equation AX = B.
A solution to the system is a vector
1
n
q
Q
q
in n such that
AQ = B
The set of all solutions to the system is called the solution set of the
system. When B = 0 the system is called homogeneous.
The homogeneous system corresponding to AX = B is AX = 0.
Any homogenous system AX = 0 has atleast one solution, namely
1
0
00
0n
such solution is called the trivial solution.
236
13.2 SOLUTIONS OF THE HOMOGENEOUS SYSTEM
AX= 0 :
Given a system AX = 0, the solution set is nonempty subset of n , not
only this but the solution set is a vector subspace of n .
Theorem: The solution set of the system AX = 0 is a subspace of n .
Proof: Let S be the solution set of the system AX = 0 Let P, Q be any two
solutions of the system AX = 0. Let a,b , then A (aP + bQ) =
0 0 0aAP bAQ a b .
aP bQ is a solution of AX = 0.
Hence aP bQ S, thus S is a subspace of n .
Now, we turn to the most important result in this chapter that gives us the
connection between the dimension of the solutions of the system AX = 0
and the rank of the co-efficient matrix A.
Theorem: The dimension of the solution of the system AX = 0 is n-rank A.
Proof: Let n mT : be defined by,
11 1
1
n
ij jj
nmn
mj jj
a qq A Q
T( Q ) T
q A Qa q
Where
1
m
A
A
A
iA ' s are rows of the matrix A.
Then
1 1 1
n n m
ap bq A aP bQ
T aP bQ T
ap bq A aP bQ
237
1 1
m m
aA P bA Q
aA P bA Q
1 1
m m
A P A Q
a b aT P bT Q
A P A Q
T is a linear transformation.
Also, Q kerT if 0T(Q ) , in other words.
1
1
0
0
n
ij jj
n
mj jj
a q
a q
iff
11 1 12 2 1
21 1 22 2 2
1 1 2 2
n n
n n
m m mn n
a q a q a q
a q a q a q
a q a q a q
Q S (the solution set of the system AX = 0)
ker dim ker dimT S T S
We know that ImT under the standard basis 1,..., me e of m is the
matrix A, that is m T A .
By Rank – nullity theorem,
dim ker dim Imn T T
dim dimIm dimS T S rank A
dimS n r
Corollary: If m n , the system 0AX has a non-trivial solution.
Proof: Suppose m n we know that rank A m and rank A n rank
.A m n
dim 0.S n rank A Thus 0S
a vector 0nX s.t. 0X S , that is 0AX has a
nontrivial solution 0X .
Note: Equivalently, this corollary states that, for a homogeneous system of
equations in which number of variables is more than the number of
equations. We always have a non-trivial solution. e.g. consider the system
of 2-equations in 3-variables.
238
0x y
0x y z
Here m =2 & n = 3,
2 3
1 1 0
1 1 1A
1 1 0, , is a non-trivial solution of this system.
Definition:
If 1
1ijj n
A a i m
is the matrix associated with the system of
equations. AX B the 1m n matrix
11 1 1
1 2
n
m m mn m
a a a b
a a a b
is
called the Augmented matrix of the system ,AX B denoted by
,A B or A B .
Theorem: The system of non-homogeneous linear equations AX B has
a solution iff rank A = rank (A, B).
Proof: Suppose nQ is a solution of the system AX = B.
11 12 1
1 2
n
m m mn
a a a
a a a
1
n
q
q
=
1
m
b
b
1 11 1 12 2 1
1 1 2 2
n n
m m m mn n
b a q q q a q
b a q a q a q
11 12 1
21 221 2
1 2
n
n
m m mn
a a a
a aq q q
a a a
B being a linear combination of columns of A, rank A = rank (A, B).
Conversely, assume that rank A = rank (A, B) = r say.
239
Let 1 2, , nA A A be columns of the matrix A. There are r of them, which
are linearly independent. Assume that 1
,...,i irA A be linearly independent.
Let if possible, suppose that B is not a linear combination of 1
,...,i irA A .
The columns 1
,...,i irA A B are linearly independent
, 1rank A B r
This is a contradiction
B must be a linear combination of 1 2, ,..., ,nA A A hence there are
numbers 1 2, ,... n such that
1 1 2 2 ... n nB A A A
Let
1
n
P B AP
The system AX B has a solution namely P.
13.3 SOLUTION OF THE NON-HOMOGENEOUS
SYSTEM AX = B, WHERE 0B :
Given an non homogeneous system of the form AX = B where
0,B there is a method that describes all solutions of this system.
Theorem:
Suppose the system AX = B has a particular solution 0X . Then a
vector X in n is also a solution iff 0 ,X X Y where Y is a solution
of the corresponding homogeneous system AX = 0.
Proof: Assume that Y is any solution of the homogenous system AX = 0.
Let 0 ,X X Y
Where 0X is a particular solution of the non-homogenous system
AX = B.
Then 0 0 0AX A X Y AX AY B B , which implies that X is
a solution of the non-homogeneous system AX = B.
Conversely, assume that X is any solution of AX = B since 0X is also a
solution,
0 0 0A X X AX AX B B
240
0X X is a solution of the corresponding homogeneous system, call it
Y. Then 0 ,X X Y
Here Y satisfies AX = 0
Note: When we are asked to find all solutions of the non-homogeneous
system AX = B. 0,B we need to find one solution of the corresponding
non-homogeneous system and secondly to solve the homogenous system.
Given a system of n non-homogeneous linear equations ni n
unknowns, the system possesses a unique solution provided that the
associated n n matrix is invertible. We prove this result in the following
theorem.
Proof: Let AX = B be the system of n-non-homogenous linear equations
in n- unknowns, where 0.B Suppose 0X is the unique solution of this
system. Let Y be any solution of the corresponding homogeneous system
AX = 0 AY = 0.
0 0A X Y AX AY B indicating that 0X Y is also a solution of
AX = B.
The uniqueness implies that 0 0 0.X Y X Y
Thus the homogeneous system AX = 0 has only trivial solution.
dim 0 0S n rank A n rank A
(S is the solution space of AX = 0)
A must be invertible matrix conversely, suppose that A is invertible
rank A n .
Since A B contains one more column, namely B.
rank A B rank A n ……… (1)
However A B has n-rows & (n + 1) columns.
rank A B minimum of , 1n n
rank A B n ………..(2)
(1) and (2) rank A B n
rank A rank A B The system AX = B has a solution by
previous theorems.
All the solutions of the system AX = B are given by 0 ,X Y where Y is
any solution of AX = 0.
241
But dim 0rank A n S
0S
0X Y can take only one value, namely
0 00X X .
Thus the solution of the system AX = B is unique.
We would make use of certain symbols, which are interpreted as follows:
:i jR R Exchange thi row and thj row
( )iR k : Multiply thi row by some non zero k .
i jR k R : Multiply thi row by a scalar k and add it to thj row.
These operations are called as the elementary row operations.
Examples:
1) Check whether the following system of equations possess a non-trivial
solution or not
2 3 0x y z
3 4 4 0x y z
7 10 12 0x y z
Solution:
The given system can be represented in the following form.
1 2 3 0
3 4 4 0
7 10 12 0
x
y
z
We try to calculate the rank of the matrix
23 3 22 173 1
1 2 3 1 2 3 1 2 3
3 4 4 0 2 5 0 2 5
7 10 12 0 4 9 0 0 1
R RR R
R R
the rank of
1 2 3
0 2 5
0 0 1
is 3
Since A is row equivalent to
1 2 3
0 2 5
0 0 1
242
rank (A) = 3. Hence the only solution is the trivial solution, namely 0.
2) Solve
2 2 5 3 0x y z w
4 0x y z w
3 2 3 4 0x y z w
3 7 6 0x y z w
Solution:
The system can be written as follows:
2 2 5 3 0
4 1 1 1 0
3 2 3 4 0
1 3 7 6 0
x
y
z
w
4 , 32 1 3 11 424 1
1 3 7 6
4 1 1 1
3 2 3 4
2 2 5 3
R R R RR R
R R
1 3 7 6 0
0 11 27 23 0
0 7 18 14 0
0 4 9 9 0
x
y
z
w
2 4 3 4( 3) ( 2)R R and R R gives
1 3 7 6 0
0 1 0 4 0
0 1 0 4 0
0 4 9 9 0
x
y
z
w
3 2 4 2( 1) 4R R and R R gives
1 3 7 6 0
0 1 0 4 0
0 0 0 0 0
0 0 9 7 0
x
y
z
w
The rank of the associated matrix is 3.
243
dim 4 3 1S
All the solutions of the system are scalar multiples of one single non
zero vector.
The last matrix equation gives
3 7 6 0x y z w
4 0y w
9 7 0z w
74 & , 3 7 6
9y w z w x y z w
49 5
12 69 9
w w w w
5 7
, , , , 4 , ,9 9
x y z w w w w w
5,36,7,99
w
Thus the solution space is 5,36,7,9 /S
3) Show that the only real value for which the following equations
have nonzero solution is 6.
2 3 ;x y z x
3 2 ;x y z y
2 3x y z z .
Solution: The system can be rewritten as.
1 2 3 0
3 1 2 0
2 3 1 0
x
y
z
1 2R R gives
4 3 5 0
3 1 2 0
2 3 1 0
x
y
z
1 3R R gives
244
6 6 6 0
3 1 2 0
2 3 1 0
x
y
z
If 6 , the coefficient matrix
6 6 6
3 1 2
2 3 1
becomes
0 0 0
3 5 2 ,
2 3 5
whose rank is 2.
dim 3 2 1S
the system has non-trivial solutions
If the associated matrix becomes, after operation
1
1 1 11
, 3 1 26
2 3 1
R
2 1 3 13 , 2R R R R gives
1 1 1
0 2 1
0 1 1
If 2 , this matrix is
1 1 1
0 0 1
0 1 1
, which has rank 3.
In this case these is only one solution namely, the trivial (zero) solution.
If 2, after performing operation 2 32R R
1 1 1
0 2 1
0 1 1
becomes
2
1 1 1
0 0 3 3
0 1 1
245
2 3 3 0 for any value of real
2 2
1
3 3R
gives
1 1 1
0 0 1
0 1 1
1 2 3 2R R and R R gives
1 1 1
0 0 1
0 1
the rank of this matrix is 3, irrespective of whether 0 0or .
the system can have trivial solution only.
The system can have a non-trivial solution if 6 .
4) Show that the equations
2 6 11 0;x y
6 20 6 3 0;x y z
6 18 1 0y z are not consistent.
Solution:
The augmented matrix A B IS
32 1
2 6 0 11 2 6 0 11
6 20 6 3 0 2 6 30
0 6 18 1 0 6 18 1
R R
33 2
2 6 0 11
0 2 6 30
0 0 0 91
R R
2 6 0
6 20 6 2
0 6 18
rank
and rank
2 6 0 11
0 2 6 30 3
0 0 0 91
Hence the system has no solution. In other words the system is
inconsistent.
5) Show that the equations
2 3;x y z
3 2 1;x y z
246
2 2 3 2;x y z
1x y z are consistent and solve them.
Solution:
The augmented matrix
1 2 1 3
3 1 2 1
2 2 3 2
1 1 1 1
A B is
, 22 4 3 4
1 2 1 3
2 0 1 2
0 0 1 4
1 1 1 1
R R R R
,2 3 4 3
1 2 1 3
2 0 0 2
0 0 1 4
1 1 0 5
R R R R
1,1 3 2
2
1 2 0 7
1 0 0 1
0 0 1 4
1 1 0 5
R R R
,1 2 4 2
0 2 0 8
1 0 0 1
0 0 1 4
0 1 0 4
R R R R
21 4
0 0 0 0
1 0 0 1
0 0 1 4
0 1 0 4
R R
Since rank A = rank 3A B (number of) variables.
247
The system has precisely one solution which is same as the solution of
the system
0 0 0 0
1 0 0 1
0 0 1 4
0 1 0 4
x
y
z
1, 4, 4x y z
6) For what values of , the simultaneous equations.
6;x y z
2 3 10;x y z
2x y z have
(i) no solution (ii) a unique solution
(iii) an infinite number of solutions?
Solution:
1 1 1 6
1 2 3 10
1 2
A B
3 2
1 1 1 1
1 2 3 10
0 0 3 10
R R
2 1
1 1 1 6
0 1 2 4
0 0 3 10
R R
(i) If 3 and 10 then rank A = 2 but rank 3,A B then the
given system has no solution.
(ii) If 3 3,Rank A rank A B in this case the system has a
unique solution.
(iii)If 3 and 10 , both A and A B are of rank 2 these are
infinitely many.
248
Solution:
The solution space 2,4,0 1, 2,1 /
7) Discuss the system
4 6;x y z
2 2 6;x y z
6x y z for different values of .
Solution:
1 1 4 6
1 2 2 6
1 1 6
A B
2 1
1 1 4 6
0 1 6 0
1 1 6
R R
1 2
1 0 10 6
0 1 6 0
1 1 6
R R
3 2
1 0 10 6
0 1 6 0
0 7 6
R R
73 1
10
1 0 10 6
0 1 6 0
7 90 0
10 5
R R
the system is consistent iff 7
10
In case 7
10 , rank A = rank A B and the system has a unique
solution.
Check your progress
1) Show that the following system of equations is not consistent.
4 7 14;x y z
3 8 2 13;x y z
249
7 8 26 5x y z
2) Solve the following systems
(i) 6;x y z
2 3 10;x y z
2 4 1x y z
(ii) 2 3 9;x y z
2 3 6;x y z
3 2 8x y z
(iii) 2 5 9;x y z
3 2 5;x y z
2 3 3;x y z
4 5 3x y z
3) Find the dimension of the solution space of the system.
2 0;x y z
3 2 0;x y
x y z
4) For the system of equations
4ax y z
3x by z
2 4x by z
Find in each of the following cases all possible real values of a and b
such that
(i) the system has no solution.
(ii) the system has a unique solution.
(iii) the system has infinitely many solutions.
13.4 SUMMARY :
1) In this chapter we learned how to solve the system of equations
AX B , provided that the system is consistent.
2) The given system of equations AX B is consistent if rank A = rank
A B .
3) The dimension of the solution space of the system 0AX is given by
n-rank A.
4) The general solution of the non homogeneous system AX B , 0B
is the sum of a particular solution of AX B and the solution of the
associated homogeneous system.
250
14
SIMILARITY OF MATRICES (Characteristic Polynomial of A Linear
Transformation)
Unit Structure :
14.0 Objectives
14.1 Similarity of Matrices
14.2 The Characteristic Polynomial of a Square Matrix
14.3 Characteristic Polynomial of a Linear Transformation
14.4 Summary
14.0 OBJECTIVES
In this chapter we introduce following two concepts related to square
matrices :
I. A relation, called “similarity”, between pairs of matrices of the
same size.
II. A polynomial Ap t in the real variable to associated with each
matrix A.
We will also combine the above two concepts and obtain a
polynomial Tp t for a linear transformation T: V → V, Tp is the
“characteristic polynomial” of the transformation T. We are interested in
the roots of the characteristic polynomial Tp because the roots carry
important information about the transformation T. We will study the
characteristic polynomial Tp in detail in the next chapter.
14.1 SIMILARITY OF MATRICES
Let A, B be two real matrices of the same size n n .
Definition 1 : A and B are similar if there is an invertible n x n matrix E
satisfying the equation : 1B EAE (1)
251
We indicate similarity between the matrices A and B by the notation :
A B (2)
Thus, for example, 4 1
2 5A
and
3 7
0 6B
are similar because there is
the invertible matrix 3 2
1 1E
which gives the equality 1B EAE .
But note that a pair of matrices A, B is not always in similarity
relation. For example, if 4 1
2 5A
and
3 7
0 6B
then A, B are not
similar matrices. To see this, first note that the matrix A is invertible, but
B is not. Now if A, B were similar, then there would exist an invertible
2 x 2 matrix E satisfying. 1B EAE
Now note that 1EAE is invertible, its inverse being 1 1EA E . But this
implies that B is invertible which in fact is not invertible. Consequently
such an invertible E does not exist, that is, A and B are not similar.
Let ,M n denote the set of all n x n real matrices. Then clearly
the similarity A B between A and , ,A B M n is a relation on the set
,M n . Following proposition lists the basic properties of this
“similarity of matrices” relation :
Proposition 1 : The relation has the following properties :
(1) is reflexive, that is, A A holds for every ,A M n
(2) is symmetric that is, if A B holds (for A, B in ,M n then
B A holds.
(3) is transitive in the sense that if A, B, C are any three matrices in
,M n satisfying A B and B C then A C.
Proof : Recall, the identity matrix
1 0
I 1
0 1
Is invertible with 1I . Consequently for any ,A M n we have.
1A I A IA I I A I IAI , This shows that A A proving (1).
252
We suppose A B. Then by the definition of , there exists invertible E
such that 1B EAE We put E-1
= F so that F is an invertible matrix, (its
inverse 1F being equal to E). now the equality 1B EAE gives
1FB FEAE FE AF IAF AF , . .i e AF FB
Multiplying on the right of each side of this equation by 1 ( )F E we
get : 1 1( )AF F FBF
i.e. 1 1( )A FF FBF
i.e. A I = FBF -1
i.e. A = FBF -1
proving A B and hence the proof of (2).
Finally consider A B. Then there exists an invertible matrix E such that 1B EAE (*)
Also, since B C, there exists and invertible F such that 1C FBF
(**)
Combining (*) and (**) we get,
1 1 1 1F (EAE ) ( ) ( )C FBF F FE A FE
Since 1 1 1( )FE E F showing that FE = D is invertible matrix
with 1 1 1D E F . Thus we have 1C DAD .
A C proving (3).
Let GL (n, ) = { E M (n, ) : E is invertible}.
Also for AM (n, ) we put :
[A] = { B M (n, ) : A B}.
Clearly, [A] is the subset of M (n, ) consisting of all BM (n, ) which
are of the form B = EAE-1
for some EGL (n, ) (i.e. E invertible). Thus
[A] = { EAE-1
: E GL (n, )}
Definition 2 : [A] is the similarity class of the matrix A.
Clearly, A[A] holds for every AM (n, ) and therefore, a
similarity class of matrices is not a vacuous concept.
We prove below that two similarity class are either the same or
disjoint subsets of M (n, )
Proposition 2 : If A, B are any two matrices in M (n, ) then either
[A] = [B] or [A] [B] = .
253
Proof : Suppose [A] [B] is not empty. We have to prove [A] = [B] now.
We accomplish this by choosing a matrix C from the non-empty set
[A] [B] and verify the equations: [A] = [C] = [B].
We prove [A] = [C] only. (the other equality, namely [C] = [B] is proved
by similar method.)
Now [A] = [C] implies C[A] and therefore, there exists GGL(n, )
such that C = GA G-1
(*)
Multiplying on the left by G-1
and on the right by G, the equality (*)
gives: G-1
C G = A (**)
Now, let X[A]. Then there exists EGL (n, ) such that X= EAE-1
(***)
Combining (**) and (***), we get,
X = EAE-1
= E.G-1
. C. G. E-1
Now, note that the matrix F = E. 1G is invertible with F
-1 = G. 1E Thus
the equality X = E.G-1
. C. G. 1X becomes X F. C. 1F and therefore we
have : X [C]
This being true for every X[A] we get [A] [C]
Next, let Y[C]. Therefore there exists EGL (n, ) such that Y = E. C.
E-1
= E. G. A. G-1
. E-1
using (*)
But again E. GGL (n, R) and therefore Y[A]. Thus every Y[C] is
contained in [A] and therefore [C] [A].
Now we have both : [A] [C] and [C] [A] and therefore the
equality [A] = [C]. As remarked above we prove [B] = [C] similarly and
therefore [A] = [B].
14.2 THE CHARACTERISTIC POLYNOMIAL OF A
SQUARE MATRIX
Now we want to associate with every n x n matrix A, a polynomial
Ap t of degree n in the real variable t in such a way that two similar
matrices will have the same polynomial associated with it. If A B, then
Ap t Bp t . We use the notion of the determinant det (A) of a square
254
matrix A to get the polynomial Ap t . Recall if A = 1 ,ij i j n
a
then
det(A) is the sum: det (A) = 1 (1) 2 (2) n (n)( ) . aa a
(*)
The sum is ranging over all the permutations
: ,n n being the signature of the permutation
.
To get the desired polynomial Ap t , we require only a few
properties of the determinant det(A) which we list below:
(I) det(I) = 1, where I is the identity matrix.
(II) det(A.B) = det (A). det B; A, B being any two matrices of size n x n.
(III) In particular, if E is an invertible matrix then
det ( 1E ) = 1
det ( )E.
Let A = ija be any n x n matrix. For any tR we consider the
n n matrix : 1 ,
I ij ij i j nA t a t
Clearly det (A – tI) =
i i i iS n
a t
is a polynomial in t, its
degree being n. We denote this polynomial by Ap t . We prove that
Ap t depends only on the equivalence class [A] of A. Thus, let
A = EAE -1
. Then for any t , we have
B – t I = EAE – t I
= EAE – E tI E-1
= E (A – t I) E-1
Therefore Bp t = det (E (A – T I) E-1
)
= det (E). det (A – t I). (det (E)-1
By the multiplicative property (II) quoted above
Bp t = det (A – t I) = Ap t
This completes the proof that if A B then PA(t) PB(t)
Definition 3 : The polynomial PA(t) is the Characteristic polynomial of
the matrix A.
For example for a 2 x 2 matrix a b
Ac d
we have :
255
t
a - t b
c d - t
a b tA tI
c d
and therefore det (A – t I) = (a – t) (d – t) – bc
= t2 – (a + d) t + ad – bc.
= t2 – tr (A). t + det (A)
Also, for a 3 x 3 matrix
a b c
A d e f
g h i
( ) detA
a t b c
p t d e t f
g h i t
= t3 – (a + e + i) t
3 + ……0
14.3 CHARACTERISTIC POLYNOMIAL OF A
LINEAR TRANSFORMATION
We consider a linear transformation T : V → V. Let 1 2{ , , }nB e e e
basis of the vector space.
Recall, the vector basis B associates with T a n x n matrix. Thus, if
T( je )= 1
1 ,n
ij ji
a e for j n
then the coefficients ija determine the n x n
matrix A = [ ija ].
If 1 2{ , , , }nC f f f is another vector basis of V then C associates (in a
similar way) another matrix 1 ,ijB b i j n with T, where
T jf = 1
(1 )n
ij ii
b f j n
We claim that the matrices A and B are similar. For, let 1 ,[ ]ij i j nC c
be the matrix given by 1
(1 )n
j kj kk
f c e j n
Clearly C is an invertible n x n matrix. We prove : 1CBC A = and
therefore A B
256
Towards this claim we apply T to the equation
1
1n
j kj kk
f c e j n
To get, 1
( ) ( ).n
j kj kk
T f c T e
1
1 1
1 1
( )
( )
n
j kj kk
n n
kj kk
n n
k kjk
NowT f b f
b c e
c b e
and 1 1 1 1 1
( ) ( )n n n n n
kj k kj k k kjk k k
c T e c a e a c e
Putting together these results, we get,
1 1 1 1
( ) ( )n n n n
k kj kjk k
kc b e a c e
For the range 1 j n . Comparing the coefficients of each e we get.,
1 1
1 ,n n
k kj k kjk k
c b a c j n
This set of equalities gives the matrix equality CB = AC or
equivalently A = CBC-1
= CBC-1
Now for any t and for any two matrices A, B with A B we
have
1
-1 1
1
t
CBC
C (B - t I) C
A tI CBC
CtIC
Consequently, we get :
det (A – t I) = det (C (B – t I) C -1
)
= (det C). det (B – t I). (det – C) -1
= det (B – t I).
257
Thus we have proved the following important result:
Proposition 3 : If A and B are similar matrices, then their characteristic
polynomials are equal is A Bp t p t
We use this fact to associate with a linear transformation T : V → V a
polynomial.
Choose a vector basis 1 2{ }ne e e and let A be the matrix of T with
respect to the basis B. Then we consider Ap t .
Thus, a choice of a vector basis of V enables to associate a
polynomial PT(t) with every linear transformation T : V → V, namely the
characteristic polynomial pA(t) of the matrix of T with respect to any
vector basis B of V. The above result guarantees that this polynomial pT(t)
indeed depends on T only and not on any vector basis of V : If B and C are
any two basis of V giving matrices A and B respectively then
pT(t) = pA(t) = pB(t).
We call pT(t) the characteristic polynomial of T.
For example :
(I) Let T : 2 → 2 be given by T (x, y) = (4x + 3y, x + 4y)
For all (x, y) 2 .
Clearly, the matrix of T with respect to the standard basis of T is
4 3
1 4A
and therefore the characteristic polynomial pT(t) of T is :
4 3( ) det
1 4T
tp t
t
= (4 – t)
2 – 3 = t
2 – 8t + 13.
14.4 SUMMARY
Two matrices A, B of size n x n are said be similar if there exists an
invertible matrix E such that B = E. A. E -1
. We indicate similarity of A and
B by the notation: A B . The Similarly A B between two matrices A, B
is a binary relation on the set M (n, ) of all matrices of size n x n. this
relation has the following properties :
I. A for every A M(n, )
II. If A B then for any A, B in M (n, )
258
III. If A B , C then A C A, B, C being arbitrary elements of
1( )M n
[A] denotes the subset of M (n, ) consisting of all B with A B . If
GL(n, ) denotes the subset of M(n, ) consisting of all invertible
elements, then [A] is given by 1[ ] { : ( , )}A EAE E GL n
Using the notion of determinant det (A) of a square matrix A, we
associate a polynomial PA(t) of degree n (in the real variable t) with every
AM(n, ), PA(t) is the characteristic polynomial of the matrix.
Characteristic polynomial has the property: If A B , then PA(t)=PB(t).
Using this property of characteristic polynomials of matrices, we associate
a polynomial PT(t) with every linear transformation T: V→V. It is the
characteristic polynomial of the linear transformation T.
We relate the roots of PT(t) with eigen values of T (to be defined in
the next chapter)
EXERCISES
1) If A is an invertible square matrix, prove that any B with A B is
also invertible.
2) Find [ I ].
3) Let T : 2 → 2
be the transformation given by T (x, y) = (3x +
4y, 2x + 3y) for all (x, y) 2. Find its characteristic polynomial.
4) Determine the characteristic polynomial of each of the matrices
1 0 3 3 1 0 5 0 0
1 2 3 1 4 0 0 5 0
0 1 4 0 0 1 0 0 5
A B C
5) Let B = A + 3I. Obtain the characteristic polynomial of B in terms
of that of A.
259
15
EIGENVALUS AND EIGENFUNCTIONS
Unit Structure
15.0 Objectives
15.1 Eigenvalues and Eigenvectors
15.2 Finding Eigenvalues and eigen-vectors of a linear transformation
15.3 Summary
15.0 OBJECTIVES
In this chapter we will study in more detail a single linear
transformation : T V V of a finite dimensional real vector space V. We
will define two important concepts related to such a T namely the
“Eigenvalues” and “eigenvectors” of T. We will explain how Eigenvalues
and eigenvectors of or given T are calculated and how their use facilitates
calculations involving T. We will also derive some simple properties of
Eigenvalues, eigenvectors, eigen spaces etc.
15.1 EIGENVALUES AND EIGENVECTORS
Let V be an n dimensional real vector space and let : T V V be a
linear transformation. (Recall the self map of V is also called a “linear
endomorphism” of V, one more name of it is an “operator” on V. This
later term, being shorter, is naturally more popular.)
Definition 1: A real number is an eigenvalue of T if there exists a non-
zero vector v V such that T v v is satisfied.
Any such non-zero v V is said to be an eigenvector of T associated with
the eigenvalue . (Often the eigen vector v of T is said to “belong” to the
eigenvalue of T.)
We consider the subset of V consisting of all the eigenvectors of T
belonging to a specific eigenvalue and the zero vector. We denote it by
E . : 0 0,E v V v or v T v v
E is called the eigenspace of T associated with the eigenvalue .
Proposition 1 Each eigenspace E is a subspace of V.
260
Proof: It is enough to verify the following statement : If v , w are in E
and if a, b are any two real numbers then av bw E
Clearly if 0av bw , then there is nothing to prove. Therefore, we
assume 0av bw and then proceed to prove that it is an eigen-vector
with eigen-value .
T av bw aT v bT w by the linearly of T
a v b w , since ,v w are eigen vectors.
av bw
This shows that the non-zero av bw is an eigen-vector with eigen-value
and hence is in E . This completes the proof that , , ,v E a b
implies av bw E .
Here is a simple example of a linear transformation with eigen-
values and eigen vectors.
Let 2V , with its usual (2-dimensional) real vector space
structure; Let 2 2:T be given by 2, (2 , 3 ); ( , ) .T x y x x y x y
Then the linear transformation T has, 2 and 3 as its eigenvalues. For, the
vector 1, 1v satisfies 1, 1 2 1, 1T while the vector 0,1w
satisfies 0,1 3 0,1T
Here is another example of a linear transformation not having any
(real) eigenvalues. Again we take 2V and the linear transformation 2 2:S is given by , ,S x y y x , for every 2,x y Clearly,
S is linear but we explain that no real number is an eigenvalue of S: Let
be arbitrary. If were an eigen value of S, then there should
correspond a non-zero ,v a b satisfying S v v
Assuming (without loss of generality), 0a under the assumption
0,0v the above equation gives , ,b a a b or equivalently
(i) b a and (ii) a b . Combining these two equations, we get
a b a 2a
Thus, 2a a and 0a implies 2 1 which is not true.
Therefore such a does not exist, that is, S has no eigenvalues (and
eigenvectors).
On the other hand on an n dimensional real vector space V, there
are linear transformation :T V V having exactly n distinct eigenvalues.
261
To see this, choose a vector basis E 1 2, ne e e of V and consider the
linear transformation :T V V
given by , 1k kT e k e k n That is if 1 1 2 2 n nv v e v e v e is any
element of V, then 1 1 2 22 k k n nT v v e v e kv e nv e . Clearly, T
is a linear transformation having the distinct eigenvalues 1, 2, 3,
….n, the respective eigen-vectors being 1 2 3, , ne e e e .
In passing, notice that the matrix T of T with respect to the
chosen basis 1 2, , ne e e which consists of eigen-vectors is the diagonal
matrix:
1
02
0
T k
n
and consequently, the matrix (with respect to the same vector basis of V)
of any power T of T is :
1 1
02 2
0
T T k
n
This goes to show that the algebra of a linear transformation is
simplified if we carry the calculations on its matrix obtained using a vector
basis consisting of eigen-vectors. Indeed eigenvalues and eigenvectors are
useful ingradiants of a linear transformation.
Now we prove the following important poperty of eigenspaces of a
linear transformation.
262
Proposition 2 Let and be two distinction eigen-values of a linear
transformation :T V V . The we have 0E E
Proof: Pick an arbitrary v E E
We prove 0v
Assuming the contrary . . 0i e v , we get that v is an eigen vector
of T with the two distinct eigen values and .
T v v v
and therefore 0v v i.e. 0v with 0 which implies
0v a contraction. Therefore 0E E .
Using this result, it can be proved that T can have at most n distinct
Eigenvalues. Below, we give an alternate proof.
15.2 FINDING EIGENVALUES AND EIGEN-VECTORS
OF A LINEAR TRANSFORMATION
We prove below that eigen-values of a liner :T V V are the roots
of the characteristic polynomial of T (Recall characteristic polynomial of
T was introduced in the proceeding chapter). Therefore we obtain all the
eigenvalues of T by solving the polynomial equation. Towards this aim,
we consider a suitable vector basis 1 2 , nE e e e of V. Now we
consider an eigenvalue of T and an eigen-vetor belonging to .
T v v
Suppose ijT a is the matrix of T with respect to the chosen vector
basis E . (Thus we have 1
n
j ij ii
T e a e
)
Therefore 11
n
j j nj
T v T x e x x
being the components of
v with respect to E.
T(v) 1
n
j jj
x T e
1 1
n n
ij ij i
jx a e
1
n n
ij j ii j l
a x e
Also, 1
n
i ii
v x e
. Therefore the equation T v v gives;
263
1 1 1
n n n
ij j i i ii j i
a x e x e
Equating the coefficients of 1 ,ie i n we get ;
1
n
ij j ij
a x x
1 i n
Thus the equation T v v gives rise to the system of simultaneous,
linear equations (in 1 nx x involving the unknown quantity );
1
n
ij j ij
a x x
1 i n
But we can write 1
n
i ij jj
x x
and therefore the above equations (*)
can be rewritten in the form :
1
0n
ij ij jj
a x
1 i n
Now this system has a “non trivial” solution 1 nx x [non-trivial in the
sense that 1 2, , , 0,0, 0nx x x remembering v should not be the zero
vector] if and only if
det 0ij ija
Note that det ij ija is a polynomial in of degree n. It is the
characteristic polynomial Tp of the linear transformation T.
Thus, we solve the polynomial equation 0Tp or equivalently
put det 0ij ija to get all the eigen-values 1 2 n of T. Next
to get an eigen vector belonging to one of the eigen values i we consider
the system of simultaneous linear equations :
11 1 12 1 1 1n n ia x a x a x x
21 1 22 2 2 2n n ia x a x a x x
1 1 2 2 1n n n n i na x a x a x x
264
A non-zero solution of these equation will give an eigen vector
1 2,i nv x x x .
We illustrate this method (of getting eigen values and eigenvectors
of a T) in the following two examples.
Example 1: Obtain eigen-values and an eigen vector for each eigen vector
of the linear transformation 2 2:T given by
2, 2 3 , 4 ,T x y x y x y x y .
Solution:
Clearly the matrix T of T with respect to the standard basis
1 21,0 , 0,1e e of 2 is : 2 3
1 4T
Therefore the characteristic polynomial Tp of T is :
2 3
det 01 4
i.e. 2 4 3 0Tp
2 6 5 0Tp
Its roots are 1 5, 1 . Thus, the eigen-values of T are 1 25, 1 .
Therefore, the simultaneous linear equations determining the eigen-vector
,v x y belonging to the eigen-value 1 5 are
2 3 5x y x
4 5x y y
It has 1 1v as a non-zero solution.
Therefore 1 1v is an eigen-vector of T belonging to the eigen-value 5
of T.
Next, we consider the simultaneous equations
2 3x y x
4x y y
Corresponding to the eigen value 2 1 . A solution of this pair gives
3, 1 , it is an eigen vector belonging to the eigen-value 2 1
265
Example 2: Let 3 3:T be the linear transformation given by
, ( , 2 , 2 5 )T x y z x y z y z . Find the eigen-values and eigen-vectors.
Solution; The matrix of T with respect to the standard basis
1 2 31,0,0 , 0,1,0 , 0,0,1e e e of 3 is;
1 0 0
0 2 1
0 2 5
T
and therefore the characteristic polynomial Tp of T is:
1 0 0
det 0 2 1
2 5
Tp
Therefore, the eigen-values are 1 2 31, 3, 4 . Now, an eigen-
vector , ,u x y z of T belonging to the eigen-value 1 1 satisfies the
equation : 0 0 1x x
0 2 1y z y
0 2 5 1y z z
Clearly a solution of this system is (1, 0, 0) and therefore, an eigen vector
of T belonging to the eigen value 1 1 is 1,0,0u .
Next, an eigen vector belonging to 2 3 is , ,v x y z which must
satisfy the equation :
0 0 3x x
0 2 3y z y
0 2 5 3y z z
A solution of this system is an eigen vector 0,1,1v belonging to the
eigen value 2 3 and finally an eigenvector , ,x y z of T belonging
to the eigen-value 3 4 must satisfy the system of simultaneous
equations : 0 0 4x x
0 2 4y z y
0 2 5 4y z z
1 3 4
266
A solution of this system is 0,1,2 which is an eigen-vector of T
belonging to the eigen value 3 4 .
15.3 SUMMARY
Two important attributes of a linear transformation T of a finite
dimensional real vector space V are (i) eigen vectors of T, and (ii) eigen
vectors of T belonging to the eigen values.
If is an eigen value of T, the subspace E of V consisting of all
the eigen vectors belonging to the eigen value together with the zero
vector is the eigen space of T. If and are distinct eigen values of T,
then 0E E . A Linear transformation may not admit any eigen
values at all, if it does, there are at most n eigen values, n being the
dimension of V.
Eigenvalues of T are the real roots of its characteristic polynomial.
Given an eigen value we obtain the eigen vectors of T belonging to be
given eigen value by solving a system of first order linear equations.
Exercises:
1) Let be an eigen value of a linear T. Prove that 2 is an eigen value
of 2T . More generally k is an eigen value of kT .
2) What way the eigen spaces 2 3,E E E
are related?
3) Let E be a subspace of V and let T be the projection map of V onto V.
Prove that E is precisely the eigen space of T associated with the eigen
value 1 .
4) Find eigen values and eigen vector of the following transformations:
(i) 2 2: , , 5 2 , 2 2T T x y x y x y
(ii) 2 2: , , 4 6T T x y x y
(iii) 2 2: , 10 4 ,18 12T T xy x y x y
(iv) 2 2: , 3 4 , 4 3T T xy x y x y
(v) 2 2: 2 , 3T T xy x y y
(vi) 3 3: , 3 , 8 ,4T T x y z x y z
(vii) 3 3: , , , 3 5 3 , 4 6 ,T T x y z x y z y z z
(viii) 3 3: , , , 7 ,0, 2T T x y z y z
267
(ix) 3 3: , , , 2 , , 42
yT T x y z x z x z
(x) 3 3: , , , , ,T T x y z ax y x ay z y az
