Maths Sy Paper2
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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
m, n 1,21,3 2,2 2,33,3 .
(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
2 3
Complements in and .
(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 LA : n m defined by LA 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 LA : n m LA X AX . The dimension of
solution space of the system of linear equations AX 0 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 E1,E 2 ,...,E n is
1, where E j denotes the jth column of the n n identity matrix I n .
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 A det A ,det AB det Adet B .
t
(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 A 0 .
(iii) Cofactors and minors, Adjoint of an n n matrix A.
Basic results such as A. adj A det A. I n . An n n real
matrix A is invertible if and only if
1
det A 0; A1 adj A for an invertible matrix A.
det 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 nn 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 m,n 1,2 1,3 2,2 2,3 3,3 .
(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
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:
a11x1a12 x2 a13x3 0;
a21x1 a22 x2 a23x3 0;
a31x1 a32 x2 a33 x3 0.
We shall see that while solving this system of equations, we in
fact, find ourselves working only with the rectangular array of constants :
a11 a12 a13
a
21 a22 a23
a31 a32 a33
9
The unknowns x1,x2 ,x3 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.
a11x1 a12 x2 b ,
An equation of this kind is called a linear equation in the variables
x1 and x2 . More generally, we define a linear equation in the n variables
x1,x2 ,...,xn to be one that can be expressed in the form.
a11x1 a12 x2 ... a1n xn b , where a11,a12 ,...,a1n 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
a11x1 a12 x2 ... a1n xn b1
a21x1 a22 x2 ... a2n xn b2
… (*)
am1 x1 am 2 x2 ... amn xn bm
where aij , bi , 1 i m ,
1 j n .
is called a system of m linear equations in n unknown.
If b1 b2 ... bm 0
10
i.e.
a11x1 a12 x2 ...a1n xn 0
a21x1 a22 x2 ... a2n xn 0
am1x1 am2 x2 ... amn xn 0
It is called a homogeneous system of m linear equations in n unknowns.
If atleast one bi 0 , where 1 i m , it is called a non-
homogeneous system of m linear equations in n unknowns.
i.e. If
a11x1 a12 x2 ...a1n xn 3
a21x1 a22 x2 ... a2n xn 0
am1x1 am2 x2 ... amn xn 0
is called a non-homogeneous system of m linear equations.
In short,
Simple example
x 2 y 3z 0
2x 3y 4z 0 (homogeneous system)
x y 5 z 0
is called a homogeneous systems in three variables x, y and z.
If
x 2 y 3z 5
2 x 3 y z 3 (non-homogeneous system)
x y 5 z 0
is called non-homogeneous system in three variables x, y and z.
We may represent (*) in a short form as
n
aij x j bi ,1 i m .
j 1
Any n-typle x x1,..., xn which satisfies all the equations in the
system (*) is called a solution of the system. xi ,1 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. x + 2y = 4 E1
x + 2y = 6 – E2 (inconsistent)
11
2. x 2 y 4E1
2 x 3 y 7E2 (unique solution)
3. x 2 y 4E1
2 x 4 y 8E2 (infinite many solutions)
We try to solve the above systems by eliminating one variable. In
system (1). E1 E2 gives O = 2 , thus the system is inconsistent. In
system (2), 2E1 E2 gives y = 1 , substituting y = 1 in E1 , we get x = 2 .
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 AX 0 has a non-trivial solution.
s 2t 4
2s 4t 8
4 2t,t is a solution of (3) for any t . The solution set is
4 2t,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
c1,...,cn where at least one ci 0 , 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.
n
aij x j bi ; for 1 i m
j 1
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 m,n
1,2 1,3 2,2 2,3 3,3 .
i) m = 1, n = 2
a11x a12 y 0 , atleast on of a11 ,a
120 .
If a11 0 , and s,t is a solution of the system.
a11s a12t 0
S a12 / a11t ,t
The solution set is
a12
t, t t
a11
a12
i.e. ,1 t t
a11
1
If a11 0 , then a12 0 and multiplying the equations by a12 we
get equation y 0 .
The solution set is t,0 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 0
11 12
m=n=2
ii) a11x a12 y 0 E1
a21x a22 y 0 E2
a11 a12
If λ , λ λ 0 , then multiplying equation (i) by λ , we
a21 a22
get that two equations are same and the system is a11x a12 y 0 , which
a a
we discussed in (i) as 11 12 , then a11a22a12 a21 0 .
a21 a22
If s,t is a solution then a21 times E1 a11 times E2 gives
a21a12 a11a22 y 0
i.e. a11 a22 a21a12 t 0
t 0 a11a22 a21a12 0
Substituting in ( E1) and (E2 ) we get
a x a y 0
11 12
a11s 0
a21s 0
a x a y 0
s = 0 21 22
14
The system has unique solution, namely trivial solution
(Geometrically, the system represents two lines passing through origin.
iii) m = 1, n = 3
a11x a12 y a13z 0
If a11 0 and (r, s, t) is a solution of the above system
a11r a12s a13t 0 .
r a12 s a13t , s,t .
a11 a11
The solution set is
a12
a
s 13 t ,s,t s,t
a11
a11
a12
a
i.e. 0 s 13 ,0,1
,1, t s,t
a11
a11
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
a11x a12 y a13z 0 ( E1 )
a21x a22 y a23z 0 ( E2 )
a a a
Let 11 12 13 λ , then ( E1 ) and ( E2 ) are same equations.
a21 a22 a23
(multiplying E2 by λ ) and we have already discussed this in (iii).
a a a a a a
If 11 12 or 11 13 or 12 13
a21 a22 a21 a23 a22 a23
We eliminate x by a21 E1 a13 E2 . Similarly, we eliminate y by
a22 E1 a12 E2 , and z by a23E1 a13E2 . Thus we get
x y z
a12a23 a22a13 a21a13 a11a23 a11a22 a21a12
If (r, s, t) is a solution of the system.
r s t
If λ , for λ
a12a23 a22a13 a21a13 a11a23 a11a22 a21a22
The solution set is λ a11a22 a21a22 λ
= λ a12a23 a22 a13 ,a21a13 a11a23 , a11a22 a21a22 λ
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) m 3, n 3
a11x a12 y a13z 0 ( E1 )
a21x a22 y a23z 0 ( E2 )
a31x a32 y a33z 0 ( E3 )
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
a11 a12 a13
a21 a22 a23 0
a31 a32 a33
If r,s,t is a solution of the above system, we consider equation (E2)
and (E3) and get
r s t
λ , for λ
a22a33 a23a32 a31a23 a21a33 a21a32 a22a31
Substituting in E1
a11λ a22a33 a23a32 a12 λ a31a23 a21a33 a13 λ a21a32 a31
a22 0
a11 a12 a13
λ a21 a22 a23 0 and so λ 0 .
a31 a32 a33
r,s,t 0,0,0
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 aij .
n
NOTE : The system aij x j 0 , 1 i m of m homogeneous linear
j 1
equations in n-unknowns has a non-trivial solution if m < n .
Solve examples :
1) Find atleast one non-trivial solution of the system 3x y z 0 .
16
1 2
Solution : We take y 1,z 1 and get x 2 then
3 3
2
,1,1 is a non-trivial solution of the system.
3
2) Write the general solution of the system and give geometrical
interpretation.
2x 3 y 4z 0
3x y z 0
Solution : The given system is
2x 3 y 4z 0 ( E1 )
3x y z 0 ( E2 )
To eliminate y , by multiplying E2 by 3 and adding to E1 , we get
11x 7 z 0 ( E1 )
2x 3 y 4z 0 ( E2 )
7
If r,s,t is a solution of the system, then 11r 7t 0 , i.e. r t
11
Substituting in E1 , we get 2r 3s 4t 0
14t
i.e. 3s 4t 0
11
30t 10t
3s ; s
11 11
7 10t
Thus, the solution is t, ,t ,t .
11 11
7 10
The set of solution is , ,1 t t
11 11
This represents a line passing through origin and having direction
7 10 3
, ,1 in . The given system is representing two planes in
11 11
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 x 7 y 3z 0
x y 0
y 6 z 0
Solution : The given system is
4 x 7 y 3z 0 ( E1 )
x y 0 ( E2 )
y 6 z 0 ( E3 )
17
To eliminate y by performing E2 E3 we get
4 x 7 y 3z 0 ( E11 )
x y 0 ( E12 )
x6 z 0 ( E31 )
If r,s,t is a solution of the system,
E12 gives r + s = 0 i.e. r s .
r
E31 gives r + 6t = 0 i.e. t .
6
E11 gives 4r 7s 3t 0
r
i.e. 4r 7r 3 0 r 0
6
s r 0 , t r 0 . Thus the only solution of the system is
6
0, 0, 0 .
Tutorial / Assignment :
Topic :- Solving homogeneous system of m linear equations in n
unknowns by elimination for m,n
1,2 ,1,3 , 2,2 , 2,3 , 3,3 .
Theorem 1 :- Show that the system of linear equations.
n
aij x j 0,i 1,m has a non-trivial solution for
j 1
m,n
1,2 ,1,3 and 2,3 .
Theorem 2 :- Show that the system of linear equations
n
aij x j 0,1 i m has unique solution for m,n 2,2 and 3,3 .
j 1
a11 a12
If 0 (for m n 2 ) and
a21 a22
a11 a12 a13
a21 a22 a23 0 (for m n 3 )
a31 a32 a33
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) x y 2 z 0
ii) 2 x y z 0
3x 2 y 4 z 0
18
iii) 3x y z 0
x y z 0
Exercises 2 : Show that the following systems of equations have only
trivial solution.
i) 2 x 3 y 0
x y 0
ii) 4 x 5 y 0
6 x 7 y 0
iii) 3x 4 y 2 z 0
x y z 0
x 3 y 5z 0
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
n
same system of linear equations aij x j bi ,1 i m is that bi 0
j 1
for 1 i m .
System of Linear Equations :
General form of the system of linear equation is
a11x1 a12 x2 ...a1n xn b1
a21x1 a22 x2 ...a2n xn b2
(A)
am1x1 am 2 x2 ... amn xn bm
This is called as a system of m‟ equation in n-unknown variables.
Here aij ‟s and bi ‟s are real numbers and x1,x2 ,..., xn are unknowns aij ' s
are called constants.
Examples
2 x 5 y 7 z w8 7
3 y 4w 8
3x 5 z 9w 0
is a system of 4 variables. The above system (A) can be written as
19
n
aij x j bi ,1 i m
j 1
The system (A) can be written in the matrix form as
a11 a12 a1n
x1 b1
a a22
a2 n x b
21 2 2
am1 am 2
amn mn xn bm
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 c c1,c2 ,...,cn and c1 c11, c12 ,...,c1n are solutions of
n
homogeneous system AX 0 . aij x j 0,1 i m then
j 1
a) .c c1,c2 ,...,cn , is also a solution.
b) c c1 c1 c11,c2 c12 ,...,cn c1n is also a solution.
Proof (a) : Since c c1,c2 ,...,cn is the solution of the system therefore,
n
we have aij c j 0,1 i m (1)
j 1
n n n
Consider L.H.S. = aij x j = aij c j = . aij c j
j 1 j 1 j 1
= . 0 [from equation (1)]
= 0 = R.H.S.
.c c1,c2 ,...,cn is the solution of the given system.
b) Since c1 c11 ,c 1n is the solution of the given system.
12 ,...,c
We have cij c1j 0,1 i m
n
(2)
j 1
n n n n
Consider L.H.S. = aij x j = aij c j c1j = aij c j aij c1j
j 1 j 1 j 1 j 1
= 0 [from (1) and (2)]
= R.H.S.
c c1 c1 c11,c2 c12 ,...,cn c1n is the solution of the system.
20
Theorem : Consider a system of m-equation in „ n ‟ unknowns i.e. a
n
system aij x j 0,1 i m . If m n , then the system has non-
j 1
trivial solution.
Proof : We prove the result by induction method on „ m ‟.
Let m 1 , 1 n . The system of equation reduce to
a11x1 a12 x2 ... a1n xn 0 .
Let a1 p 0,1 p n .
Consider, a11x1 a12 x2 ... a1 p x p ... a1n xn 0 (I)
Let x1 1,x2 0, x3 0,...,x p 1 0
a
x p 11 ,x p 1 0,..., xn 0
a1 p
Substituting the above values of x1,x2 ,..., xn in RHS of (I). We get,
L.H.S. of (I) which is equal to zero.
x1 1,x2 0,...,x p a11 ,..,xn 0 is the non-trivial solution of (I).
a1 p
for m 1 the system has non-trivial solution.
result is true for m 1 .
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 m = k + 1 and n > k + 1 . Consider a system of
k + 1 equation.
a11x1a12 x2...a1n xn 0 E1
a21x1a22 x2 ...a2n xn 0 E2
a k 1 a k 1 x2...a k 1 xn 0
1 2 n
Ek 1
In E1 , all the coefficients are not equal to zero.
Let a1 p 0 for some p, 1 p r .
from E1
a11x1 a12 x2 ... a1 p 1x p 1 a1 p x p a1 p 1x p 1 ... a1n xn 0
Since a1 p 0, we get
a a a a1 p 1
x p 11 x1 12 x2 13 x3 ... x ...
a1 p a1 p a1 p a1 p p 1
a1 p 1 a
x p 1 ... 1n xn (II)
a1 p a1 p
21
Substituting the values of x p in remaining equation E2 ,E3 ,...,
Ek 1 , we
get, the system k equations. In variables x1 ,x2 ,...,x p 1 ,x p 1 ,...,xn we
denote there equations by E12 ,E31 ,..,
E1k & E1k 1 .
This is the system of k–equations therefore, by induction it has non-trivial
solution.
Let x1 c1 ,x2 c2 ,...,x
p 1 c p 1 ,x p 1 c p 1 ,...,xn cn be the non-
trivial solution of the above system.
Substituting the values of x1 ,x2 ,...,x p 1 ,..,xn in (II), we get the values
x p . Therefore, we get the values of x1 ,x2 ,...,xn which are not all zeros.
Therefore the system of k 1 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
mn.
ii) If c c1 ,c2 ,...,cn 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 x 3 y 5z 0 (i)
x yz0 (ii)
Solution :
(ii) y x z
Substituting y, in (i), we get 2 x 3 x z 5z 0 .
x 2 z
Again,
y x z 2 z z 3z
y 3z
Solution set is x, y,z / x, y,z 2z,3z,z z
z 2,3,1 z
22
EXERCISE:
i) 2x 4z 3 y 0
3x z y 0
ii) 2 x y 4 z w 0
3x 2 y 3z w 0
x y z 0
iii) x y 2 z 3w 0
2 x 4 z 4w 0
x y 2 z w 0
x 2 y 2 z 0
iv) 7 x 2 y 5z w 0
x y z 0
xz w 0
y 2 z w 0
Example 2: 2x 3 y 4z 0 (1)
3x y z 0 (2)
Solution : (2) y 3x z
Substituting y in (1), we get z 7 x
Since y 3x z 3x 7 x 10 x
y 10 x
Solution set is S x, y, z : x, y, z x,10x,7 x : x
x 1,10,
7 : x
Example 3: 2x y 4z w 0 … (1)
3x 2 y 3z w 0 … (2)
x y z 0 … (3)
Solution : (3) y x z
Substituting y x z in (2), we get w 5x 5z
4
Substituting the values of y and w in (1), we get x z
3
S z 4 ,1 ,1, 5
z
3 3 3
Example 4:
7 x 2 y 5z w 0 (1)
23
x y z 0 (2)
x zw0 (3)
y 2z w 0 (4)
(2) y xz (5)
By substituting on (1), we get
w 5x 3z (6)
(3) z 2 x x0
(5) y 0
(6) w 0
Solution set is S x, y, z, w x, y, z , w
S 0, 0, 0, 0 .
1.4 EXERCISE :
Solve the following homogeneous systems by any method :
i) 2 x1 x2 3x3 0
x1 2 x2 0
x3 x3 0
Ans : x1,x2 ,x3 : x1,x2 ,x3 i.e. 0,0,0
ii) 3x1 x2 x3 x4 0
5x1 x2 x3 x4 0
Ans : x1 s,x2 ts,x3 4s,x4 t
iii) 2 x 2 y 4 z 0
w y 3z 0
2w 3x y z 0
2w x 3 y 2 z 0
Ans : w t ,x t , y t ,z 0
iv) 2 x y 3z 0
x 2 y 3z 0
x y 4 z 0
Ans : (Only the trivial solution)
v) x1 3x2 x4 0
x1 4 x2 2 x3 0
2 x2 2 x3 x4 0
x1 2 x2 x3 x4 0
24
Ans : (Only trivial solution)
vi) 3w 2x 0
2u 4w 3x 0
2u 3 2w x 0
4u 3 5w 4x 0
Ans : u 7s 5t, 6s 4t,w 2s,x 2t
Systems of linear equations : ( points to be remembered )
1. Determine whether the equation 5x 7 y 8 yz 16 is linear.
Solution : No, since the product yz of two unknowns is of second degree.
2. Is the equation x y ez log 5 linear?
Solution : Yes, since , e,log 5 are constants.
1. Property : Consider the degenerate linear equation:
0 x1 0 x2 ... 0 xn b
i) If the equation‟s constant b 0 , then the equation has no solution.
ii) If the constant b 0 , every vector in n is a solution.
2. Property : Consider the linear equation ax b .
i) If a 0 , then x b / a is the unique solution.
ii) If a 0, b 0 , there is no solution.
iii) IF a 0 , b 0 , 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 m n matrix is a rectangular array of mn numbers arranged in m rows
and n columns. Thus,
a11 a12 a1n
a21 a22 a2n
am1 am2 amn
is an m n (real m by n) matrix, usually denoted by a single letter A. Each
of the mn numbers a11,a12,... is called an element of the matrix. The
26
element aij occurs in the ith row and jth column. The matrix is denoted by
A aij .
Transpose of a matrix : If A = aij mn , then transpose of A is the
n m matrix B b ji
nm
where b ji aij for 1 i m , 1 j n .
a11 a1n a11 am1
If A then At
a a amn
m1 amn 1n
th th
The element aij occurs in the i and j column. The matrix is denoted by
A aij .
Symmetric Matrix :
An n n matrix A over is said to be symmetric if At A .
aij a ji for 1 i n , 1 j n .
Skew – symmetric Matrix :
An n n matrix A over is said to be skew symmetric if
At A .
Exercise 1 : Let A be a n n matrix over . Show that
i) A At is symmetric.
ii) A At 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 n n matrices over , show that
A B and 2A are symmetric matrices over .
2.2 MULTIPLICATION OF MATRICES :
Let A aij be an m n matrix over and B b jk be an n p
matrix. We define the product,
AB cik ,1 i m,1 k p to be an m p matrix, where
n
cik aij b jk for 1 i m,1 k p .
j 1
27
Note 1 : The product AB may be defined but the product BA may not be
defined. However, if A, B are m n matrices over , AB and BA are both
defined. However they may not be equal.
0 1 0 0 1 0
For example, let A , B , then AB ,
0 0 1 0 0 0
0 0
BA , AB BA .
0 1
Note 2 : A, B M m n are said to be equal. If aij bij for 1 i m ,
1 j n , where mn ,B bij mn .
A aij However, we have
associated law for multiplication of matrices.
If A M mn ,B M n p ,C M pl . Then
AB C A BC and we also have distributive law.
If A M mn ,B,C M m p then A B C AB AC
proof are obvious.
We define An inductively for A M n .
An An 1. A , n N ,n 1 . 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
In is called an identity n n matrix.
0 0 1
Kronecker’s delta :
I n δij ,δij 1 if i j
0 if i j
Upper triangular matrix :
A aij is called an upper triangular matrix if aij 0 for i j
a11 a12 a1n
0 a22 a2n
A
0 0 ann
Lower triangular matrix :
nn
A aij is called a lower triangular matrix if aij 0 for j i .
28
a11 0 0
a21 a22 0
A
an1 an2 ann
Diagonal matrix :
nn
A aij is called a diagonal matrix if aij 0 for i j
a11 0 0
0 a22 0
A
0 0 ann
Scalar Matrix :
nn
A aij is called a scalar matrix if diagonal is c for some c .
c 0 0
0 c 0
A
0 0 c
Invertible Matrix :
nn
A aij is said to be an invertible matrix if B M n R such that
AB BA I n .
An invertible matrix is also called a non-singular matrix. The
matrix B is called an inverse of A.
Theorem : If A M n is a invertible matrix, the inverse of A is
unique.
Proof : If AB BA I n (i)
AC CA I n (ii)
Then BA I n ( BA )C I n .C ( post multiplying by C )
C ( In =1 )
Thus B(AC) = C BI n C (from ii)
B C ( In =1 )
We shall denote the inverse of an invertible matrix A by A1 .
Proposition 1 : If A M mn ,B M n p then AB Bt At .
t
29
Proposition 2 : If A is an n n invertible matrix over , then AT is
invertible and AT 1 A 1 T
Proof : AA1I n A 1 A
AA1 T I n T A1AT
A 1 T .AT I n AT A 1 T
AT
T
is invertible and AT 1 A1
Proposition : If A,B M n are invertible matrices, then the product
AB is invertible and AB 1 B 1A1 . More generally, if A1,..., Ak are
invertible matrices then A1 A2 ..., Ak 1 A
1 1
k ,..., A1 .
Exercises 1.2 :
1 1 1
1) Let A 0 1 1 find A2 , A3 , A4 .
0 1
0
1 a 1 b 2 3
and let A
2) Let a,b , B find AB , A , A .
0 1 0 1
Show that A is invertible and find A 1 .
1 0
3) i) Find a 2 2 matrix A such that A2 I can you find
0 1
all such matrices?
ii) Find a 2 2 non-zero matrix A such that A2 0 .
4) Let A M n of A3 0 show that I A is invertible.
5) Let A M n of A3 A I 0 show that A is invertible.
6) Let A,B,P M n . If P is invertible and B P 1AP then show
that B n P 1An P for n .
7) A,B M n of A,B are upper triangular matrices, find if AB is
upper triangular.
cosθ sin θ
8) Let R θ ,θ show that
sin θ cosθ
R θ1 R θ 2 R θ1 θ 2 for θ1,θ 2 .
30
1 2 3 4
0 2 3 4
9) Let A then find A1 ?
0 0 3 4
0 0 0 4
1 1
3 1
10) Let A 2 2 , B Is there a matrix C such that
1 0 4 4
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 :
n
aij x j bi ,1 i m . (A)
j 1
x1 b1
i.e. AX B, A aij
mn
, X x2 , B , A aij
mn
is called
x b
n m
the matrix of coefficients and the m n matrix.
a11 a1n b1
A,B the augmented matrix.
a bm
m1 amn
For example the system of linear equations:
2 x 3 y 3z 1
x 5 y z 3
x
2 3 3 1
is denoted by matrix notation as y . The
1 5 1 z 3
2 3 3 : 1
augmented matrix of the system is .
1 5 1 : 3
Examples
2 x 5 y 7 z 8w 7
3 y 4w 8 is a system of 4 variables x y, z and w.
3x 5 z 9w 0
The system (A) can be written in the matrix form as
31
a11 a12 a1n x1 b1
a a22
a2 n x b
21 2 2
am1 am 2 amn mn xn bm
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 0mn or simply 0, is the matrix
whose elements are all zero. Find x, y, z, t.
x y z 3
If 0
y 4 z w
Solution :
x y0 z30
Set :
y40 zw0
x 4, y 4, z 3, w 3
2) Show that for any matrix A , we have A A
Solution :
A aij
mn
m,n
aij
m,n
aij
x y x 6 4 x y
3) Fin x, y, z, w if 3
z w 1 2w z w 3
Solution : 3x x 4
3y x y 6
3z z w 1
3w 2w 3
x 2, y 4, z 1, w 3
Theorem : Let M be the collection of an m n matrices over a field K
of scalars. Then for any matrices A aij , B bij , C cij in M and
any scalars K1 , K2 K .
i) A B C A B C
ii) A0 A
iii) A A 0
iv) A B B A
v) K1 A B K1A K1B
32
vi) K1 K2 A K1A K2 A
vii) K1 K2 A K1 K2 A
viii) 1.A A
Problem
1. Show that 1 A A
Answer : Consider
A 1 A 1A 1 A
1 1 A
0. A
0
Thus, A 1 A 0
Or -A + A +(-1)A = 0 - A
Or 1 A A ( - A + A = 0)
2. Show that A A 2 A, A A A 3 A
2 A 1 1 A
1 A 1 A
A A
Thus 2A A A
3 A 2 1 A
2 A 1A
A A A
3A 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 :
b1 n
1
a , ..., an a b
11 ... a b
n n ak bk
b k 1
n
3
8 4 5 2 8 3 4 2 1 + (5)(-1) = 24+ (-8) – 5 = 11
1
1 3 2 0 4
Example 1 : A and B
2 1 3 2 6
33
Answer : A 2 2 & B 2 3 then AB is defined as 2 3 .
1 3 2 0 4
AB
2 1 3 2 6
1 2 3 3 1 0 3 2 1 4 3 6
11 6 14
1 3 2 0 4
Now AB
2 1 3 2 6
11 6 14
2 2 1 3 0 2 2 4 1 6
11 6 14
AB
1 2 4
2 1
1 2 5
2. Find AB if A 1 0 & B
0
.
3 4 3 4
1 8 10
Solution : AB 1 2 5
9 22 15
1 6 2
3. Find .
3 5 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
1 6 4 0 16 6 4 24
5. Let A , B than AB BA
3 5 2 1 2 5 5 7
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.
1 2 3
If A aij
m,n
, then AT = (aij)T , if A
4 5 6
then
1 4
A 2 5 .
T
3 6
Theorem : The transpose operations on matrices satisfieds.
i) A B T AT BT
AT
T
ii) A
iii) KA KAT K is scalar
T
iv) AB T BT . AT
2.5 DIAGONAL :
The diagonal of A aij consists of the elements a11,a22 ,...,ann where
A is n-square matrix.
Trace of an n-square matrix A aij is the sum of its diagonal elements
i.e. tr An a11 a22 ... ann
viz. trace of the matrix
1 2 3
A 4 5 6 is 1 5 9 15
7 8 9
Property : Suppose A aij and B bij 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
1 0
I2 is the identity matrix of order 2.
0 1
1 0 0
I3 0 1 0 is the identity matrix of order 3 etc.
0 0 1
Kronecker delta : Kronecker delta is defined as
0 if i j
ij
1 if i j
Accordingly, I ij
Note: Trace of I n n
Scalar Matrix Dk : Dk K . I
Example : Find the scalar matrices of orders 2, 3 and 4 corresponding to
the scalar k 5 .
Solution : Put 5s on the diagonal and 0s elsewhere
5
5 0 0
5 0 5
, 0 5 0 ,
0 5 0 0 5 5
5
Powers of Matrices : The non-negative integral powers of a square
matrix M may be defined recursively by
M 0 I , M 1 M , M r 1 MM r r 1,2,...
Property :
i) A p Aq A p q
ii) If A and B are commutative then A p and B q are also commutative.
1 2 1 K
Note : If A and Sk then AS K S K A S K 2 .
0 1 0 1
1 2n
Note : An
0 1
Idempotent : A matrix E is idempotent if E 2 E .
Note : i) identity matrix is idempotent I 2 I
ii) zero matrix is idempotent 02 0
2 2 4
Example : Given matrix E 1 3 4 is idempotent.
1 2 3
Nilpotent : A is a nilpotent of class p if A p 0 but A p 1 0 .
36
1 1 3
Example – Show that A 5 2 6 is nilpotent of class 3.
2 1 3
0 0 0
2
Answer : A 3 3 9
1 1 3
A3 A2. A 0
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.
1
Note : A1 A
2 5 3 5
Example 1 : Show that A and B are inverses.
1 3 1 2
1 0
Answer : AB I
0 1
1 0
BA I
0 1
a b
Example 2 : When is the general 2 2 matrix A invertible?
c d
What then is its inverse?
Answer : Take scalars x, y, z, t such that
a b x y 1 0 ax bz ay bt 1 0
c d z t 0 1 or cx dz cy dt 0 1 ,
ax bz 1 ay bt 0
cx dz 0 cy dt 1
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.
xd A, z c A ,
y b A , ta A .
d A b A 1 d b
Thus A1
c A a A A c a
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
1
iii) multiplying the matrix by
A
3 5
Example 1 : Find the inverse of A
2 3
Answer : A 1 0 , A 1 exists. Next, interchanging the diagonal
1
elements, take the negative of the other elements and multiply by .
A
5 3 5
A1 1 3
2 3 2 3
5 3
Example 2 : Find the inverse of A
4 2
Answer : A 2, A1 exists.
2 3 1 3 2
A1 1
4 5 2 5 2
2
2 3
Example 3 : Find the inverse of A
1 3
1 3 3 1 3 1 3
A1
9 1 2 1 9 2 9
Definition : A square matrix „ A ‟ of order „n‟ A 0 . Inverse of A is
Adjoint of A
denoted and defined by A 1
A
AT Transpose of A
Adjoint of A = Transpose of co-factor of A
i j
Co-factor of A = 1 A
Example :
1 1
i) Find inverse of the matrix A
2 1
Solution : A 3 0
1 3 1 3
A1
2 3 1 3
3 1
ii) If A
2 2
1 4 1 8
A1
1 4 3 8
38
1 5 2
iii) If A 2 0 1 A 0 A is not invertible.
3 1 2
1 5 2
iv) If A 2 1 1
3 1 2
Answer : A 8 0, A1 exists. Take B = matrix of cofactors called
as Adjoint of A.
111 1 11 2 1 13
1 1 1 1 1
16 8 8 16
2 1 2 2 23
B 1 8 1 8 1
131 3 13 2 5 133 11 3 5 11
1 8 3
B 1 8 5
T
1 16 11
1 8 3 1 8 1 3 8
BT 1
A 1
1 8 5 1 8 1 5 8
A 8
1 16 11 1 8 2 11 8
v) Find whether matrix is invertible and find its inverse.
3 1 4
A 2 0 1
1 1 1
Solution : A 4 0 . A 1 exists.
1 1 2
B 3 1 2
1 5 2
1 3 1
B 1 1 5
T
2 2 2
1 3 1
1 1 5
1 1
A
4
2 2 2
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:
E1 : Interchange the ith row and jth row : Ri R j .
E2 : Multiply the ith row by a non-zero scalar K; Ri KRi , K 0 .
E3 : Replace the ith row by K time the jth row plus the ith row:
Ri KR j Ri
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 K 1 , or by K 1 and then by
K , we obtain the original matrix. In other words, the operations
Ri KRi and Ri K 1Ri are inverses.
41
c) Applying the operations Ri KR j Ri and then operation
Ri KR j Ri ,
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.
F1 : Interchange the ith column and the jth column : Ci C j
F2 : Multiply the ith column by a non-zero scalar K : KCi Ci K 0
F3 : Replace the ith column by K times the jth column plus the ith column
Ci KC j Ci
We first define elementary row operations on an m n matrix over .
The following operations on A n are called elementary row
operations.
i) Interchanging i th and j th row of A denoted by Ri R j .
ii) Multiplying the i th row by a non-zero scalar λ (denoted by
Ri λRi )
iii) Adding a scalar multiple of ith row to j th row of A denoted by
R j λRi + RJ.
Two m n 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.
b1
b
AX B , A aij B 2
mn
bm
If the augmented matrix A,b
is row equivalent to A1,B1 , then the
solutions of the system AX b are same as the solutions of the system
A1X B1 .
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 m n matrix A aij is called a row echelon matrix if
A 0 or if there exists an integer r. 1 r min m,n and integers
1 K 1 K 2 K r m s.t.
i) For each i,1 i r ,aij 0 for j K i
ii) For each i,1 i r ,aik i 0 , aik i is called the pivot element of
ith row.
iii) For each i,1 i r,ask i 0 for s i .
iv) aij 0 for all i r and for all j .
Note : Sometimes we consider the pivot elements are 1.
0 2 1 0 5 6
0 0 0 5 2 1
For example, A
0 0 0 0 4 2
0 0
0 0 0 0
K 1 2,K 2 5,K 4,r 3 .
Theorem : Every matrix A, A M is row equivalent to a matrix in
row echelon form.
0 1 3 2
Example : Reduce the matrix 2 1 4 3 to row echelon form.
2 3 2 1
0 1 3 2
Answer : A 2 1 4 3
2 3 1
2
2 1 4 3
R1 R2
0 1 3 2 (bringing the left most non-zero
2 3 1
2
entry to 1st row)
2 1 4 3
R3 R3 2 R1
0 1 3 2
0 2 6 4
(making entries below 1st pivot 0)
43
2 1 4 3
R3 R3 2 R2
0 1 3 2
0 0 0 0
(making entries below 2nd pivot 0)
Exercise : Reduce the following matrices to row echelon form :
6 3 4 1 0 2
i) 4 1 6 ii) 2 1 3
1 2 5 4 1 8
1 2 3 1 1 2 1 2 1
iii) 2 1 2 2 iv) 2 4 1 2 3
3 1 2 3 3 6 2 6 5
Note : In the system of equations AX B , where A aij mn and
x1
X , the elements in the j th column are coefficients of the
xm
unknown x j (or the j th 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 R1 Ri .
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.
2 x1 3x2 7 x3 5x4 2 x5 2
x1 2 x2 4 x3 3x4 x5 2
2 x1 4 x3 2 x4 x5 7
ii) x1 2 x2 x3 2 x4 1
x1 x2 x3 x4 2
x1 7 x2 5x3 x4 3
iii) x1 x2 x3 2 x4 3
x1 x2 x3 x4 2
x1 2 x2 3x3 x4 9
x1 x2 2 x3 x4 2
iv) x1 x2 2 x3 x4 4
3x2 x3 4 x4 2
x1 2 x2 3x3 5x4 0
x1 x2 5x3 6 x4 3
v) 3x1 8x2 3x3 14 x4 1
2 x1 3x2 x3 2 x4 2
x1 2 x2 x3 10 x4 3
x1 5x2 2 x3 12 x4 1
vi) 5x y 2z 142
45
x 3y z 30
2x y 3z 5
Ans : x 39.2, y 16.7,z 18.97
vii) 10x 7y 3z 5u 6
6x8y z 4u 5
3x y 4z 11u 2
5x 9y2z 4u 7
Ans : u 1,z 7, y 4,x 5
2. Problems : Which of the following matrices are in row echelon form.
1 1 2
2 1 1 3
i) 0 0 0 ii)
0 1 0 0 0 0
0
1 0 0 3 1
1 1
iii) iv) 0 0 0 1 1
0 1 0 1
0 0 0
0 0 1
v) 0 0 1
0 0 1
3. Problems : Reduce the following matrices to row echelon form by a
sequence of elementary row operations.
0 1 2 1 2 1 1
0 1 2 2 7 2 4
i)
0 2 4 3 7 1 0
0 3 6 1 6 4 1
1 5 2 1 4 0
ii) 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 I rs denote an m n matrix which has entry 1 in the r,s
th
place and 0 elsewhere.
0 0 0
I rs 0 1 0
0 0
0
I rs are called matrix units. Let A aij mn . Then
a a1n
0 0 0
ar1 arn
I rs A 0 1 0
as1 asn
0
0 0
a amn
m1
r,s
0 0
as1 asn rth row
0 0
I rr A A
I rs .I rs 0,I rr .I rr I rr if r s
47
0 0
as1 asn
I rs I sr A rth row
ar1 arn sth row
0
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 Eij .
ii) The matrix obtained by multiplying ith row of identity matrix by
λ 0 by Ei λ .
th th
iii) The matrix obtained by adding λ times j row to i of identity
matrix by Eij λ
Theorem : Let A be an m n real matrix. Then,
i) Eij A is the matrix obtained by exchanging ith row and jth row of A.
ii) Ei λ A is the matrix obtained by multiplying ith row of A by
λ λ 0 .
iii) Eij 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.
48
Proposition : Let A be an n n 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.
B E1E2 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
A E1E2...Ek B Ek1...E11B .
A is invertible.
Theorem : An n n 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
b11 b12 b1n
0 b22 b2n
B bii 0 for 1 i n
0 0 bnn
Then multiplying ith row by bii 1 , we get a matrix C (row
equivalent to A),
1 c12 c1n
0 1 c2 n
C
0 0 1
1 c12 0
Ri Cin Rn gives 0 1 0
0 1
0 0
Similarly,
49
0
Ri Cin1Rn1 makes n 1 column 1 and so on and finally we get
th
0
identity matrix. Thus, an invertible matrix is row equivalent to identity
matrix. Further Ek ...E1 A I n .
i.e. A I n E11...Ek1 E11...Ek1 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.
1 0 3 0 0 1 1 0 0
i) E 0 1 0 ii) 0 1 0 iii) E 2 1 0
0 1 1 0 0 1
0 0 0
2. In each of the following, find an elementary matrix E such that
B EA .
2 1 2 1
i) A , B
3 1 1 2
1 2 1 2
ii) A , B
0 1 0 1
2 1 1 3
iii) A , B
1 3 2 1
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
1 0 1 0
i) ii)
5 1 0 3
50
0 1 0
5 1
iii) iv) 0 1
0
0
1
1 0 0
2 0 0 2
1 0 0 0
1
v) 0 1
1 0
0 vi)
0 0 1 0
0 0 0
0 0 0 1
5. Express A as a product of elementary matrices
2 0 0 0 0 0 0 1
0 4 0 0 0 0 5 0
i) A ii) A
0 0 5 0 0 2 0 0
0 0 0 2 0 0 0 1
1 0
A , find elementary matrices E1 ,E2 such that
2
6. Let
5
E2 E1A I .
Gauss – Elimination Method :
1) Solve the given system by Gauss – Elimination Method.
2 x1 3 x2 7 x3 5 x4 2 x5 2
x1 2 x2 4 x3 3x4 x5 2
2 x1 4 x3 2 x4 x5 3
x1 5 x2 7 x3 6 x4 2 x5 7
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 3 7 5 2 2
1
R2 R2 R1 1 1 1
2 0 0 1
R3 R3 R2 2 2 2
1 0 3 3 3 1 5
R4 R4 R1
7 7 7
2 0 1 6
2 2 2
51
2 3 7 5 2 2
1 1 1
R3 R3 6 R2 0 0 1
2 2 2
R4 R4 7 R2
0 0 0 0 1 1
0 0 0 0 0 1
2 3 7 5 2 2
1 1 1
0 0 1
R4 R4 R3 2 2 2
0 0 0 0 1 1
0 0 0 0 0 0
STOP
The unknown corresponding to columns without pivots are x3 ,x4
. The solution set is
2s t1,2 s t,s,t,
1 t 1,1,0,2,0 s,t
i.e. S 1,2,0,0,1 S 2, t 1,1,0,1,0 s,t
3.5 LINEAR ALGEBRA SYSTEM OF LINEAR
EQUATIONS :
General form of the system of linear equation is
a11x1 a12 x2 ...a1n xn b1
a21x1 a22 x2 ...a2n xn b2
(A)
am1x1 am 2 x2 ... amn xn bm
This is called as a system of m‟ equation in n-unknown variables.
Here aij ‟s and bi ‟s are real number and x1 ,x2 ,...
,xn are unknowns aij‟s
are called constants.
Row Echelon Form : A matrix in row echelon form has zeros below each
leading 1.
Example :
52
1 4 3 7 1 1 0
i) 0 1 6 2 ii) 0 1 0
0 0 1 5 0 0 0
1 1 3 4
0 1 2 6 0
0 1 3 4
iii) 0 0 1 1 0 iv)
0 0 1 5
0 0 0 0 1
0 0 0 2
Reduced row echelon form : A matrix in reduced row echelon form has
zeros below and above each leading 1.
Examples :
1 0 0 4 1 0 0
i) 0 1 0 7 ii) 0 1 0
0 0 1 1 0 0 1
0 1 2 0 1
0 0 0 1 3 0 0
iii) iv)
0 0 0 0 0 0 0
0 0 0 0 0
1 0 2 3 1 0 0 0
0 1 1 2 0 1 0 0
v) vi)
0 0 0 0 0 0 1 0
0 0 0 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.
1 0 0 5 1 0 0 4 1
a) 0 1 0 2 b) 0 1 0 2 6
0 0 1 4 0 0 1 3 2
53
1 6 0 0 4 2
1 0 0 0
0 0 1 0 3 1
c) d) 0 1 2 0
0 0 0 1 5 2
0 0 0 1
0 0 0 0 0 0
Solution :
(a) The corresponding system of equations is
x15
x2 2 By inspection
x3 4
(b) The corresponding system of equation is
x1 4x4 1
x2 2x4 6
x3 3x4 2
Since x1,x2 and x3 correspond to leading 1‟s in the augmented
matrix, we call them leading variables. Solving for the leading variables
in terms of x4 gives
x1 1 4x4
x2 6 2x4
x3 2 3x4
Since x4 can be assigned an arbitrary value, say t, we have
infinitely many solutions.
The solution set is given by the formulas.
x1 1 4t,x2 6 2t,
x3 2 3t,x4 t
(c) x1 2 4x5 6x2 , x3 1 3x5 , x4 2 5x5
x1 2 4t 6s,x2 s,x3 1 3t,x4 2 5t,x5 t
(d) 0x1 0x2 0x3 1 , 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.
1 0 0 0
0 1 0 0
i.e. I
0 0 1 0
0 0 0 1
then each of the following is an elementary matrix obtained from I by the
elementary row operations indicated :
0 0 1 0
0 1 0 0
i) , by interchanges of R1 and R3 .
1 0 0 0
0 0 0 1
1 0 0 0
0 1 0 0
ii) , by 5R3 .
0 0 5 0
0 0 0 1
1 0 0 0
0 1 3 0
iii) , by R2 3R3 .
0 0 1 0
0 0 0 1
Exercises 1.3 :
Find rank and inverses of the matrices if exists :
55
1 4 3
i) A 1 2 0
2 2 3
Ans : A1 does not exists.
1 2 3 40 16 9
ii) A 2 5 3 , A1 13 5 3
1 0 8 5 2 1
1 6 4
iii) A 2 4 1 , A1 does not exists.
1 2 5
1 1 1
5 5 5
1 0 2
1 1 4
iv) A , A1 3 1 2
5 5 5
1 1 0
2 1 1
5 10 10
cos θ sin θ 0 cos θ sin θ 0
v) A sin θ cos θ 0 , A1 sin θ cos θ 0
0 0 1 0 0 1
Inverse of a matrix A :
a11 a12 a13
Let A a21 a22 a23 be an 3 3 matrix. Then inverse of A, denoted
a31 a32 a33
and defined by
Adjoint of A Adj A
A1
A A
Adj A Cofactor of A ; T indicates transpose
T
56
Co-factor of A = Co-factor of each element of A
i j a22
Co-factor of a11 1
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.
a a 2 a3
So, V = { 1 / a1 , a 2 ,…, a 6 are real numbers}. Addition and
a 4 a5 a 6
scalar multiplication in V is usual addition and scalar multiplication of
matrices.
a a2 a3 b1 b2 b3 a1 b1 a 2 b2 a3 b3
i.e. 1
a 4 a5 a6 + b4 b5 b6 = a 4 b4 a5 b5 a6 b6
a1 a2 a3 a1 a 2 a3
and a a6 = a 4 a5 a6
4 a5
Here the vectors are matrices.
0 0 0
Zero vector is
0 0 0
a1 a 2 a3 a1 a 2 a3
Negative of a a a is a a a .
4 5 6 4 5 6
Oher axioms can be easily verified. Hence V is a vector space.
Example 3: Let V = = Set of all real sequences
63
= { ( xn ) / xn are real numbers for all natural numbers n }
Define addition as ( xn ) + ( y n ) = ( xn y n ) and
scalar multiplication as ( xn ) = ( xn ) .
Under these operations V is a vector space. Vectors are sequences. Zero
vector is a zero sequence ( xn ) , where xn = 0 for all n and negative of
( xn ) is ( xn ) .
Example 4: Let V = [x] = Set of all polynomials with coefficient from
= { p(x) = a0 a1 x a2 x 2 ...... an x n / a0 , a1 , a2 , ......an are real
numbers}
Define addition and scalar multiplication as follows:
If q(x) = b0 b1 x b2 x 2 ...... bn x n then
p(x) + q(x) = a0 b0 (a1 b1 ) x (a2 b2 ) x 2 ...... ( an bn ) x n and
p(x) = a0 a1 x a2 x 2 ...... an x n
V is a vector space. Polynomials are vectors. Zero vector is a zero
polynomial with all coefficient equal to zero. Negative of
p(x) = a0 a1 x a2 x 2 ...... an x n is a0 a1 x a2 x 2 ...... an x n .
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)
a1 a 2
2) M2 2 = { / a1 , a 2 , a 3 and a 4 are real numbers}
a3 a 4
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, , a b L
Let a = (x1, y1), b = (x2, y2) y1 = mx1 and y2 = mx2 …..(*)
a b = (x1, y1) + (x2, y2) = ( x1 + x2, y1 + y2)
y1 + y2 = mx1 + mx2 = m ( x1 + x2) (from (*) )
Hence a b 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 0,0,0 0 x0 ,0 y0 ,0 z0 .
Now TST a, b L’, , a b L’
Let a = ( kx0, ky0, kz0), b = ( k‟x0, k‟y0, k‟z0), where k, k‟
a b = ( 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 a b 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, , a b P
Let a = (x1, y1, z1), b = (x2, y2, z2) 2x1 + 3y1 + z1 = 0, 2x2 + 3y2 + z2 = 0
a b = (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 a b 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 = { ( xn ) / ( xn ) is convergent sequence of real
numbers}
Since zero sequence is convergent. So W is nonempty.
If ( xn ) , ( y n ) are convergent sequences then ( xn y n ) is
convergent.
Hence ( xn ) + ( y n ) W for , W is a subspace of .
a1 0
Example 11: Let W = = { 0 a / a1 , a 2 are real numbers}. W is a set
2
of all diagonal matrices of order 2.
0 0
W, W is nonempty.
0 0
Let A, B W. and ,
a 0 b 0
A 1 B 1
0
a2 0 b2
,
a 0 b 0 a b1 0
A B 1 1 1 W
0 a2 0 b2 0 a 2 b2
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 p(x) , q(x) Pn[x] and ,
p(x) = a0 a1 x a2 x 2 ...... an x n and
q x = b0 b1 x b2 x 2 ...... bn x n
p(x) + q(x) =
a0 b0 ( a1 b1 ) x ( a2 b2 ) x 2 ...... ( an bn ) x n
Hence p(x) + q(x) 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
0
0
O = S
0
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
3
subspace of .
3) W3 = Set of upper triangular matrices of order 2
70
a1 a 2
= { 0 a / a1 , a 2 , a3 are real numbers }.Show that W3 is a
3
subspace of M2 2.
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
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 ; , then a+ b
W1 W2
Let a, b W1 W2 and ,
a+ b W1 and a + b W2 ( W1 and W2 are subspaces and
by thm 3)
a+ b W1 W2 and , . Hence W1 W2 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 . W1 W2 = {(x, 0), (0, y)/ x, y are real
numbers}.
If W1 W2 is a subspace of 2 then addition must be closed in W1 W2.
(2, 0) W1, (0, 3) W2 but (2, 0) + (0, 3) = (2, 3) W1 W2. Hence
W1 W2 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
W1 W2 is a subspace of V if and only if either W1 W2 or W2 W1.
Proof: Suppose W1 W2 or W2 W1.
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If W1 W2 then W1 W2 = W2. If W2 W1 then W1 W2 = W1. Since
W1 and W2 are subspaces of V, W1 W2 is a subspace of V.
Now let W1 W2 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 W1 W2, y W2 y W1 W2
X y
W1 W2
W1 W2 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)
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 W1 W2 …. Wn is a subspace of V. (Hint: Use
induction)
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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.
2
Example 2: u = (-1, 2) and v = (4, -6) are vectors in .
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.
2
Example 3: u = (2, 1, 0) and v = (-3, -15) are vectors in
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 3x2 – 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 (x2 + 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 2 2 matrices.
78
1 0 0 0
Let S = { , }
0 0 0 1
1 0 0 0
L(S) = { , β / , β }
0 0 0 1
0 0 0
L(S) = { 0 / , β }
0
,
0
0
L(S) = { / , β }
0
L(S) is a subspace of all diagonal matrices in M2 2 .
3
Example 7: Let V = 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
2 3
We have defined lines in and . 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 t 1 }
Definition : A parallelogram spanned by two non zero vectors u and v in a
vector space V is defined as {t1u + t2v/ 0 t1, t2 1 }
If v = u then parallelogram is degenerated (does on exist)
v+u
A parallelogram spanned by u and v
t1 v
u
t2 u
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 t 1
I.e. The line segment between P and Q is entirely contained in S
PpPPppP
S1 S2
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S1 is convex. S 2 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
S t1u t2v / 0 t1, t2 1
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
S1 S2 is convex if S1 S2
Proof: S1 S2 if S1 S2 is singleton set then it is convex.
If not, then let P,Q S1 S2
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 S1 S2
S1 S2 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/ x S} is convex.
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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.
3
Let S = {(1, 1, 0), (2, 1, 1), (1, 0, 1)} be a subset of . 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 1 0 = 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 +
0 v2 +……..0 vn = 0
2
Example 9: Let S = {(1, 0), (-1, 2), (2, -4)} be a subset of .
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.
2
Example 10: A subset S = {(1, 0), (0, 1)} of 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-1a)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) = e2t is linearly independent in V
Now. S = { et, e2t }
Let a et + b e2t = 0 for all t , where a, b ……(i)
Differentiating equation (i) with respect to t, we get
aet + 2be2t = 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 e2t = 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.
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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.
2
Note : Let v1 and v2 be two non zero vectors in . 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
3
Note : Let v1, v2, v3 be three non zero vectors in . 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
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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 :
2
1) Show that {(1, 2), (3, 4)} is linearly independent in .
4
2) Show that {(1, 1, 2, 0), (0, 1, 4, 9)} is linearly independent in .
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 + 0 vk+1 + 0 vk+2 + ….+ 0 vn = 0, where
a1, a2,.. ak are not all zero.
S is linearly dependent set.
Contradiction since S is linearly independent.
T is linearly independent.
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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)
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x = (b-1a1)v1 + (b-1a2)v2 + ……..+ (b-1an)vn, where b-1a1, b-1a2,…
b-1an
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 + 0 vi+1 + 0 vi+2 +…..+ 0 vn = 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/ x S} 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)
1 2 4 2 4 8
iv) u= ,v=
3 0 1 6 0 2
v) u = 2 – 5t + 6t2 – t3, v = 3 + 2t – 4t2 + 5t3
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
1 1 1 0 1 1
iii) , in M 22
1 1 0 1 0 0
iv) {(a, b), (c, d)} in where ad – bc 0
2
v) {t3 – 4t2 + 2t + 3, t3 + 2t2 + 4t – 1, 2t3 – t2 – 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 x, y,z in3
L S 3
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
3
Example 1: LetV
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 M 22 , Set of all 2 2 matrices.
1 0 0 1 0 0 0 0
Let B = { , , , }
0 0 0 0 1 0 0 1
1 0 0 1 0 0 0 0 0 0
Consider a1 0 0 + a2 0 0 + a3 1 0 + a4 0 1 = 0 0
93
a1 0 0 a 2 0 0 0 0 0 0
0 0 0 a3 0 0 a 4 0 0
+ + + =
0
a1 a 2 0 0
=
a 3 a 4 0 0
a1 = 0, a2 = 0 , a3 = 0, a4 = 0
B is linearly independent.
x y
Now let M 22 .
z w
x y 1 0 0 1 0 0 0 0
Since = x + y + z + w
z w 0 0 0 0 1 0 0 1
L(B) = M 22
B is a basis of M 22
Example 4: Consider a vector space Pn[x], of all polynomials of degree
≤ n with real coefficients. Let B = { 1, x, x2, ……, xn} 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 + …….oxn
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
1 1 1 1 1 1 1 0
3) Prove that 1 1
{ , 1 0 , 0 0
, 0 0 } is a basis of
M 22
4) Find the coordinates of (x, y, z) with respect to a basis
3
{(1, 1, 1), (1, 1, 0), (1, 0, 0)} of .
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
x L(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
n m
= ( a
j 1 i 1
ij xj )vj
n m
x1w1 + x2w2 + ….+ xmwm = ( a
j 1 i 1
ij x j ) v j …………….(i)
97
Hence if x1w1 + x2w2 + ….+ xmwm = 0,
n m
then ( a
j 1 i 1
ij xj )vj = 0
B = { v1, v2,….vn} islinearly independent
m
a
i 1
ij xj = 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
a i 1
ij c j = 0, where at least one cj is non zero.
n m
c1w1 + c2w2 + …..+ cnwn = ( a
j 1 i 1
ij c j ) v j (from 1)
n
= (0) v
j 1
j =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.
1 0 0 1 0 0 0 0
Example 7: Since { , , , } is a basis of
0 0 0 0 1 0 0 1
M 22 , dim M 22 = 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
ab .
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
3
Example 8: Let W be subspace of 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.
4
Example 9: Let U and W be subspaces of 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 (U W) 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) U W if and only if (x, y, z, w) U and (x, y, z,
w) W
U W = {(x, y, z, w)/ x + y + z = 0 and x + w = 0, y = 2z}
U W = {(x, y, z, w)/ x + 2z + z = 0 and x + w = 0, y = 2z}
U W = {(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 U W
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 (U W) = 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
W = {(x, y) / x = 0}
2
(i)
W = {(x, y) / y = 0}
2
(ii)
W = {(x, y) / x + y = 0}
2
(iii)
W = {(x, y, z) / x = 0}
3
(iv)
W = {(x, y, z) / y = 0}
3
(v)
W = {(x, y, z) / z = 0}
3
(vi)
W = {(x, y, z) / x + y = 0}
3
(vii)
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
2) {(1, -1), (1, 0)} or a set containing (1, -1) and any element of
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
3
(0, 1, 2) and any element of which is not a linear combination
of {(1, 0. 2), (0, 1, 2)}
6.4 RANK OF A MATRIX
Every matrix of order m n has m rows and n columns.
a11 a12 a1n
a a a 2 n
21 22
then (a11, a12, …,a1n) ,
n
Let A =
a m1 a m 2 a mn
(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.
a1 j
a2 j
Let A =
j
1 j n denote the jth column of A. Then L({A1,
a mj
A2,…Am}) 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, A2,…Am}) 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
a11 ...............a1n
a j1 ..................a jn
B= The Columns of B are B1, B2, …Bn.
a ..................a
i1 in
a ..................a
m1 mn
1B1 + 2B2 + ….+ nBn = 0
1ak1 + 2ak2 + ….+ nakn = 0 for all k = 1, 2, …m
1A1 + 2A2 + ….+ nAn = 0, where A1, A2, …An 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
a11 ...............a1n
B= ai1 .......... ....... .a
in , 0
am1 ..................amn
108
1B1 + 2B2 + ….+ nBn = 0
a12
a11 a 1n
a .
1 ai1 . + 2 i 2 +… + n a in . = 0
am1 a mn
a
m2
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 + 2A2 + ….+ nAn = 0, where A1, A2, …An 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
a11 ...... .a1n
ai1 ........ ain
B=
a a . ...........a .a
ji i1 jn in
am1 ......... amn
1B1 + 2B2 + ….+ nBn = 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 + 2A2 + ….+ nAn = 0, where A1, A2, …An 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 At 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 At.
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)
1 0 ...0.........0
0 1 ....0.........0
0 0 ....1......... 0 (r 0)
0 0 0 0
0 0 0 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 Aj,
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.
a11 0 0 ..... 0
0 a a .....a
22 23 2n
Hence we get A equivalent to 0
0 a m 2 a m 3 .....a mn
110
a 22 a 23 .....a 2 n
Let A* = A* is (m -1) (n -1) matrix
a m 2 a m3 .....a mn
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 1st
column of A.
Hence Performing elementary row and column operations on A* as we
a 22 0 0 ..... 0
0 a a .....a
32 33 3n
performed on A, we get A* i.e. A equivalent to 0
0 a m 2 a m 3 .....a mn
a11 0 ...0.........0
0 a ....0.........0
22
By repeating this process we get A equivalent to 0 0 ....a rr ......... 0
0 0 0 0
0 0 0 0
By multiplying rows A1, A2,…Ar by (1/a11), (1/a22),…(1/arr) respectively
we get
1 0 ...0.........0
0 1 ....0.........0
A= 0 0 ....1......... 0 where rows Ar+1, Ar+2,…Am are zero.
0 0 0 0
0 0 0 0
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.
1 2 1 3
Example 11: Let A =
3 6 3 2
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
1 1 0 1
Example 12: We reduce A = 3 2 1 1 to echelon form
1 0 1 3
A1,A2, A3 and A4 are columns of A
We perform following elementary column operations
By A2 + (-1)A1, A4 + A1 on A , we get
112
1 0 0 0
A ~ 3 1 1 4
1 1 1 4
By (-1)A2, we get
1 0 0 0
A ~ 3 1 1 4
1 1 1 4
By A1 + 3A2, A3 + A2, A4 + 4A2 we get
1 0 0 0
0 0
A~ 0 1
2 1 0 0
Now we perform following elementary row operations
By A3 + 2A1 and A3 + A2 we get
1 0 0 0
0 0
A~ 0 1
0 0 0 0
Hence rank A = 2
Check your progress
1) Find the rank of the matrix A where
2 1
7
(i) A= 3
6 1
1 3
0 2
(ii) A =
5 1
2 3
113
1 2 3
2 1 0
(iii) A=
2 1 3
1 4 2
2) Determine which of the following matrices are invertible by
finding rank of the matrix.
1 1 2
0 1 5
(i)
2 1 5
1 1 3
2 1 10
(ii)
3 51 1
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
3
{(1, 1, 1), (1, 1, 0), (1, 0, 0)} of 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
1 1 0 1 1 1 1 0 2 3
{ 1 1 , 1 , , } where A = 4 7
0 0 0 0 0 .
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
a b
(v) W = { c d / 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, W1 W2,
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 3 1 1 0 1 1 1
2) = (-7) + 11 + 14 + (-5)
4 7 1 1 1 0 0 0
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
3
7) A basis of containing (1, 0, 2) is {(1, 0, 0), (0, 1, 0), (1, 0, 2)}
116
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
n
x and y as x . y = x1y1 + x2y2 + …….+ xnyn = xy
i 1
i i
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 + x22 + …..+ xn2 0
x . x = 0 x12 + x22 + …..+ xn2 = 0
x12 = x22 = …..xn2 = 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) x x1 , x2 ,....., xn
x . y x1 y1 x2 y2 ..... xn yn
x1 y1 x2 y2 ....... xn yn
x1 y1 x2 y2 ..... xn yn
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 V V 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> = <0 0, 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
2
Example 1: Let V =
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 + 5x22 0
<x, x> = 0 2x12 + 5x22 = 0
x12 = 0, x22 = 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> =
a
f (t ) g (t )d t
For f, g, h C[a, b]
(i) <f, f> =
a
f (t ) f (t )d t
120
( f (t )) d t 0 for all t [a, b] (Since (f(t))2 = f2(t) 0)
2
=
a
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
f ( t ) f ( t ) d t > 0
b t
0 = <f, f> = a
f (t ) f (t )d t =
a
f (t ) f (t )d t +
t b
t
f (t ) f (t )d t +
t
f (t ) f (t )d t > 0
Contradiction
Hence <f, f> = 0 f = 0
<f, f> = 0 f = 0
(ii) <f, g> =
a
f (t ) g (t )d t
= g( t ) f ( t ) d t
a
= <g, f>
b
(iii) For , < f, g> = ( f )(t ) g (t ) d t
a
b
=
a
f (t ) g (t )d t
=
a
f (t ) g (t )d t
= <f, g>
(iv) <f+ g, h> = ( f g ) ( t ) h ( t ) d t
a
b
= ( f ( t ) g ( t ) ) h( x ) d t
a
121
b b
=
a
f (t )h(t )d t + g( t ) h ( t ) d t
a
= <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
zw)
For z , w , t C
(i) If z = a + i b , z z = a2 + b2
< z , z > = Re( z z ) = a2 + b2 0
< z , z > = 0 Re( z z ) = 0
a2 + b2 = 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
a1 a 2
Let A = then tr(A) = Trace of A = a1 + a4.
a3 a 4
If A, B M2 , define <A, B> = tr(ABt)
122
b1 b2 t b1 b3
Let B b b then B
= = b b and ABt =
3 4 2 4
a1b1 a 2 b2 a1b3 a 2 b4
a b a b a b a b
3 1 4 2 3 3 4 4
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 + a22 + a32 + a42 0
<A, A> = 0 a1 + a2 + a3 + a4 = 0
2 2 2 2
a12 = a22 = a32 = a42 = 0
a1 = a2 = a3 = a4 = 0
a1 a 2
=O
a3 a 4
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 + BCt)
= tr(ACt) + tr(BCt)
= <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 + b2x2 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 + b2x2
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 = v, v
Example 6: We know that IRn is an inner product space with usual dot
product.
For x ( x1 , x 2 ,......, x n ) x = x, x = x1 2 x 2 2 .... x n 2
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.
2
(i) The usual inner product in is <x, y> = x1y1 + x2y2 , for x =
(x1, x2),
y = (y1, y2)
x = x, x = x1 x 2
2 2
(3 , 4 ) = 32 4 2 =5
124
(ii) The inner product given in example 1 is <x, y> = 2x1y1 + 5x2y2
x = x, x = 2 x1 5 x 2
2 2
(3 , 4 ) = 2( 3 2 ) 5 ( 4 2 ) = 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 = x, x 0 (< x , x > 0)
x =0 x, x = 0 x, x = 0 x = 0
(ii) x = x, x = x, x = 2 x, x = x, x
= x
Definition : Let x ( 0) be in an inner product space V.
x x x
Then since = = 1, is said to be a unit vector in the
x x x
direction of x .
Theorem 2: (Cauchy – Schwarz inequality) : Let V be an inner product
space.
If x , y V then x, y x y
The equality holds if and only if x yor y x for some in .
Proof: Let x, y V.
If y 0 then x, y x,0 0 and x y = x 0 =0
equality holds.
Suppose y 0
Define f : by f( t ) = x t y
f t 0 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 t 0 is an extremum (maximum or minimum) the f „( t 0) = 0
2< x , y > + 2 t 0< y , y > = 0
x, y
t0 = ,<y, y> 0 ……………….(ii)
y, y
f t = 2< y , y > > 0 for all t
f is minimum at t = t 0
0 f( t 0) f( t ) for all t
Putting the value of t 0 from (ii) in (i), we get
x, y x, y 2
<x, x> + 2 <x, y> + ( ) <y, y> 0
y, y y, y
x , y 2 x , y 2
x -2 + 0
y, y y, y
x , y 2
x
2
- 2 0
y
x , y 2 x 2 y 2
x, y x y
The equality holds if and only if f( t 0) = 0
i.e. x t0 y, x t0 y 0
i.e x t0 y 0
i.e x t0 y
Theorem 3: (Triangle inequality): Let V be an inner product space.
If x, y V then x y x + y
Proof: Consider x y = x y, x y
2
= x, x + y, x + x, y + y, y
+ 2 | x, y | + y
2 2
= x
126
( x
2 2
+ 2 x y + y )2 (By Cauchy
Schwarz inequality)
x y 2 x 2
( + y )2
By taking non negative square roots of both sides
x y x + y .
Note : x y = x + y if and only if x = y or y = x for some
in .
+ 2 | x, y | + y
2 2
Since x y
2
= x
By Cauchy Schwarz inequality x, y = x y if and only if x =
y or y = x for some in .
Corollary : For x , y V, x y x y
Proof: For x, y V , x x y y
By Triangle inequality
x = (x y) y x y + y
x - y x y ……………(i)
Now y y x x
y = ( yx) x y x + x
y - x y x
- ( x - y ) ( y x = ( x y) = x y ) ………….(ii)
x y x y
Note : x y denotes the distance between vectors x and y in the inner
product space V.
Check your progress
1 2
1) Find the norm of a matrix 1 1 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)
b
(iii) (C[a, b], < , >), where <f, g> =
a
f ( t ) g ( t ) w(t ) d t
Where w(t) 0 for all t in [a, b] and w c[a, b]
e
(iv) (C[1 , e], < , >), <f, g> = log
1
t f (t ) g (t )d t
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
x, y
x y 1
x, y
-1 x y 1
x, y
There exist unique in [0, ] such that Cos
x 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 IR
3
with usual inner product (dot product) is given by
(1, 1, 1) . (0, 1, 1)
Cos =
(1, 1, 1) (0, 1, 1)
1 0 1 (1) (1) (1)
= =0
12 12 (1) 2 0 2 (1) 2 (1) 2
= /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
x, y
i.e. if and only if =0
x y
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.
3
Note: We can verify that vectors (1, 1, -1) and (0, -1, -1) in 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 x y = x 2 +
2
2
y
Proof: x y = x y, x y
2
= x , x + y, x + x , y + y, y
+ 2 x, y + y
2 2
= x
if and only if x , y = 0
2
Thus x y
2 2
= x + y
2
Hence x y
2 2
= x + 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 + AC2 = 2OA2 + 2OC2
132
The length of the side OA is x and that of side OC is y and diagonal
OB is x y . The end points of diagonal AC are x and y. So length of
AC is x y
To show that x y 2
+ x y
2
=2 x
2
+2 y
2
+ 2 x, y + y
2
Now, x y
2 2
= x
- x, y + y
2
x y 2
= x
2
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
2
While studying vectors in plane , 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. v Pu (v ) v u for all .
The equality holds if and only if Pu(v) = u.
Proof: Consider < v Pu (v ) , u > = < v , u > - < Pu (v ) , 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
< v Pu (v ) , u > = 0
v Pu (v ) is orthogonal to u
v Pu (v ) is orthogonal to u for all in
Since Pu (v ) is along u Pu (v ) - u is also along u
< v Pu (v ) , Pu (v ) - u > = 0 …………….(i)
v u 2 = v Pu ( v ) Pu ( v ) u 2
= v Pu ( v )
2
+ Pu ( v ) u 2
(From (i) and Pythagoras
theorem)
v u v Pu ( v ) 2 for all in .
2
v Pu (v ) v u for all in .
v, w
Note: For any vector w in V Pw (v ) = w
w, w
Check your progress
1) Find the orthogonal projection of (1, 1) along (1, -2) with respect
to the usual inner product. Ans :
1 ,2
5
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),
3
(0, 1, 0), (0, 0, 1)} is a basis of .
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
n
2 2
|| x || = <x, v1> + <x, v2> + …..+ <x, vn> = 2 2
x,v v
i 1
i i
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
n
Thus, x = <x, v1>v1 + <x, v2>v2 + ……+ <x, vn> vn = x,v v
i 1
i i
|| x ||2 = <x , x>
n n
=< x,v v , x,v v
i 1
i i
j 1
j j >
n n
= ( x , vi x , v ) vi, vj j
i 1 j1
n n
= ( x , vi x , vi ) vi , vi ( vi, vj = 0 if i j )
i 1 i 1
n
= ( x , vi )2 ( vi , vi = 1)
i 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 n
a v
i 1
i i ,v j a v ,v
i 1
i i j ai
(Since i j vi , v j 0 )
But a
i 1
i vi 0
0, v j 0
v j 0 for each j
S is linearly independent.
136
u
Note : We know that If u V then is the unit vector along u.
u
Let { u1 , u 2 , ......, u k } be an orthogonal set in V.Then
u1 u2 uk
= =……..= =1
u1 u2 uk
ui uj ui , u j
Also , 0 if i j
ui uj ui u j
ui , ui
= 1 if i j
ui ui
u1 u u
{ , 2 , ........, k } is an orthonormal set.
u1 u2 uk
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
v 2 , u1
For k = 2, we take u 2 v 2 u1
u1 , u1
v 2 , u1
Then u1 , u 2 u1 , v2 u1
u1 , u1
v2 , u1
u1 , v 2 u1 , u1
= u1 , u1
u1 , v2 v2 , u1
=
=0
v 2 , u1
u 2 v2 u1
Now if u2 = 0 then u1 , u1 =0
v 2 , u1
v2 u1
u1 , u1 = 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})
k 1 vk , ui
Now we define u k v k ui
i 1 ui , ui
Then for 1 j k - 1, (j fixed)
k 1 vk , ui
u k , u j vk ui , u j
i 1 ui , ui
k 1 vk , ui
vk , u j ui , u j
= i 1 ui , ui
vk , u j
vk , u j uj , uj
= u j , u j
138
Since ui ,u j 0if i j
= vk , u j vk , u j
= 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
k 1 vk , ui
u k vk u i for k = 2, 3, …..n
i 1 ui , ui
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.
u1 u u
From Note 8.4.1 { , 2 , ........, n } is orthonormal basis
u1 u2 un
of V.
Example 4: We now apply Gram Schmidt process to convert linearly
3
independent set {(0, 1, -1), (1, 2, 1), (1, 0, 1) } of 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
v 2 , u1
u 2 v2 u1
u1 , u1
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)
v2 , u1 v2 , u 2
u 3 v3 u1 u2
u1 , u1 u2 , u2
(1, 0 , 1) , (0, 1, 1) (1, 0 , 1) , (1, 1, 3 / 2 )
u3 (1, 0, 1) (0, 1, 1) (1, 1, 3 / 2 )
(0, 1, 1) , ( 0 ,1, 1) (1, 1, 3 / 2 ) , (1, 1, 3 / 2)
1 10
u3 (1, 0, 1) (0, 1, 1) (1, 1, 3 / 2 )
2 17
1 1 10 10 15
u3 (1, 0, 1) (0, , ) ( , , )
2 2 17 17 17
10 3 32
u3 ( , , )
17 34 34
10 3 32
{u1, u2, u3 } = {(0, 1, -1), (1, 1, 3/2), ( , , ) } is an
17 34 34
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)
v 2 , u1
u 2 v2 u1
u1 , u1
(0 ,1, 2) , (1, 0, 3)
( 0, 1, 2 ) (1, 0, 3)
= (1, 0, 3) , (1, 0 , 3 )
6
= ( 0, 1, 2 ) (1, 0, 3)
10
3 1
= ( , 1, )
5 5
3 1
{(1, 0, -3), ( , 1, ) } is an orthogonal basis of W
5 5
3 1 3 1
( , 1, ) ( , 1, )
(1, 0, 3) 5 5 (1, 0, 3) 5 5
{ , } = { , }
(1, 0, 3) 3 1 10 35
( , 1, )
5 5 5
1 3 3 5 1
= { ( ,0, ), ( , , ) } is an orthonormal basis
10 10 35 35 35
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
Hence W 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
xW
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
k
Let w v , ei ei
i 1
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 >
k
=<v- v,e e
i 1
i i , ej >
= < v, ej > - v,e e ,e
i 1
i i j
= < 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.
Ans : , 62 75 ,65 75 ,58 75
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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.
Ans : 1,1,0 , 1 2 ,1 2 ,1
, 13 ,2 3 ,2 3
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)}
Ans : 0, 1
2
, 1
2 , 2 1 1
3 1, 2 , 2 ,
1
2 3 3
1 , 1 , 4
3
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.
1 1 1 0 0 1 1 0
i) {
, , 0 1 , 0 1 }
0 0 1 0
1 0 1 2 1 2 13 1 1 2 1 2
Ans : , , 1
3
, 1
1 0 1 0 3 1 2 1 2
1 1 1 0 1 0 1 0
ii) { , , , }
0 1 1 1 0 1 0 0
1 0 13 2 3 15 2 5 1 2 0
Ans : , 1 , 2 1 , 1
1 1 1
3 5 5 0 2
1 2 0 1
5) Find the projection of along
2
in the inner product
1 3 1
0 1
space given in problem 4) Ans :
1 2
6) Find the cosine of angle between (1, -3, 2) and (2, 1, 5) in 3 with
9
respect to usual inner product. Ans :
14 30
2 1 1 1
7) Find the cosine of angle between and in M2
3 1 2 3
with respect to inner product given in problem 4). Ans :
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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
1
Say we have the vector in n , and we rotate it through 90
0
0
degrees (anticlockwise), to obtain the vector . We can also stretch a
1
2 4
vector U to make it 2U, for example becomes or, it we look at
3 6
the projection of one vector onto the x-axis, extracting it‟s x-component.
2 2
E.g. to , these all are examples of mapping between two vectors,
3 0
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-
x x
axis. T : 2 2 , and T maps to clearly this is linear map.
y 0
Y
(x, y)
X
(x, 0)
We shall check for both the properties of preserving scalar multiplication
and vector addition.
i) T x, y x1, y1 T x x1, y y1
x x1, 0
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x, 0 x1, 0
T x, y T x1, y1
ii) T x, y T x, y x, 0
. x, 0 . T x, y
Clearly, T preserves the null vectors.
T 0, 0 0, 0 .
(2) Let V be the space of all continuous functions from into .
Define a function T : V V by
x
T f x f t dt
0
then T satisfies the properties of linear transformation.
For, continuous functions f, g :
x
T f g f g t dt
0
x
f t g t dt (By properties of Riemann Integration)
0
x x
f t dt g t dt
0 0
T f T g
For any scalar ,
x x
T f f t dt . f t dt
0 0
x
f t dt
0
.T f (By properties of Riemann Integration)
So that T is a linear transformation.
(3) Let v1, v2, ..., vn be a basis of a finite n – dimensional vector
space V over . Define a map T : V n by associating to each
148
element V , it‟s unique co-ordinate vector with relative to this basis of
V.
If 11 22 ... nn, j
then T 1, 2, ..., n .
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 :
T Ou Ou T Ou T Ou
(By addition property)
T Ou Ov T Ou T Ou
Ou Ou Ou
T Ou Ov (By cancellation on both the sides)
Exercises :
Check whether the following transformations satisfy the properties of a
linear transformation or not.
i) T : 3 2, T x1, x2, x3 x1 x2, 0
ii) T : 2 2
T x, y x cos y sin , x sin y cos
iii) T : 3 2, T x, y, z x2, 0, 0
iv) Let A be m n matrix over . Define a function LA : n m by
x1
LA X AX where X x2 a column vector in n .
x
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.
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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. T u u1 . T u . T u1 for all , and
vectors u,u1 U .
Proof : Assume that T : U V is a linear transformation, then for
, , u,u1 U .
T u u1 T u T u1 (By addition property)
T u T u1 (By Scalar multiplication property)
Conversely, assume that T u u1 T u T u1 .
Put 0 T 0 u1 O . T u . T u1
T u1 . T u1
Similarly for 1 ,
T 1. u 1. u1 1. T u 1. T u1
T u u1 T u T u1
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) T Ou Ov
(ii) T – u – T u
(iii) T u – u1 T u – T u1
Proof :
(i) Ou Ou Ou
T Ou T Ou T Ou Ov
T Ou Ov (By adding T Ou on both the sides)
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(ii) Ou u – u
T Ou T u T – u
Ov T u T – u
T u – T – u OR T – u T u
(iii) T u – u1 T u – u1 T u T – u1
T u – T u1 [By (ii)]
Given a linear transformation T :U V . We saw that T 0 0 , 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 u U T u 0 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 :
x
(1) The projection of a vector in a plane on the x-axis given by
y
T x, y x, 0 .
In this example all the vectors on y-axis constitute the kernel of T.
KerT x, y 2 T x, y 0, 0
x, y 2 x, 0 0, 0
x, y 2 x 0
0, y 2 y
= Y-axis
Clearly ImT consists of all vectors on Real line (x-axis) in a plane.
(2) Consider the linear map.
T : 2 2 defined by T x, y x – y, – x y
In this example,
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kerT is the line passing through the origin in a plane having the
equation y x .
Ker T
O
X
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 u, u1 ker T then we need to show that
u u1 ker T .
For that consider T u u1 . T u . T u1
.0 .0 0 0 0
u u1 ker T
kerT is a subspace of U.
(ii) Let , 1 ImT there exist u, u1 U such that T u and
T w 1 .
Let , be scalars we need to show that d 1 Im T .
Consider T u u1 T u T u1 T u T u1
1
152
1 T u u1
1 ImT u u1 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 T 0 .
Proof : Assume that T is injective map.
Let u ker T T u 0 .
T u T 0 and injectivity of T implies that u 0 .
ker T 0
Conversely, Assume that ker T 0 . We need to show that T is one-to-
one, for assume that
T u T u1 for u, u1 U
T u T u1 0 T u, u1 0
u – u1 ker T but ker T 0
u – u1 0
u u1 , 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 B u1, u2, ..., um is a basis for U and we know the
values of T on u1, u2, ..., um , then we can determine the action of a linear
transformation T naturally on any vector u U .
For example consider a linear transformation T : 3 3 , that
maps a natural standard basis 1, 0, 0 0, 1, 0 0, 0, 1 of 3 as follows.
T 1, 0, 0 0, 0, 1
T 0, 1, 0 1, 0, 0
T 0, 0, 1 0, 1, 0
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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 x, y, z x 1, 0, 0 y 0, 1, 0 z 0, 0, 1
T x, y, z xT 1, 0, 0 yT 0, 1, 0 zT 0, 0, 1
T x, y, z x 0, 0, 1 y 1, 0, 0 z 0, 1, 0
0, 0, x y, 0, 0 0, z, 0 y, z, x
Therefore, T can be defined uniquely as follows.
T x, y, z y, z, x for x, y, z 3 .
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 u1, u2, ..., un 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 T ui i for all I, 1 i n .
Proof : Let u U be any vector since u1, u2, ..., un is a basis of U.
There are unique scalars 1, 2, ..., n in such that
u 1 u1 d2 u2 ... n un .
Define a mapping T : U V as follows.
T u 1 1 2 2 ... n n
We shall show that T is a linear map.
For c, d & u, u1 U .
Suppose u 1 u1 2 u2 ... n un and u1 11 u1 a12 u2 ... a1n un
for scalars 11, ..., 1n ,
then cu du1 c 1 u1 2 u2 ... n dn d 1n u1 ... 1n un
154
c1 d 11 u1 ... cn d 1n un
T cu du1 T c1 d11 u1 ... c n d 1n un
c1 d 11 1 ... cn d 1n n
c 1 1 ... n n d 11 1 ... 1n n
T cu du1 cT u dT u1
T is a linear transformation
Also, we can write ui 0 . u1 0 . u2 ... 1. ui ... 0 . un
T ui 0 . T u1 ... 1. T ui ... 0 . T un
0 . 1 ... 1. i ... 0 . n i
T ui i 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 ui to i for all i, 1 i n , then
M u M 1 u1 2 u2 ... nun 1 1 2 2 ... n n
T u
M T on U
Now, it‟s your turn to solve few problems.
Check your progress :
1. Check whether F : 2 , defined by F x, y xy is a linear
transformation or not.
(Hint : Take x 1, 2 , y 3, 4 in 2 . Find T x y &
T x T y )
2. Let T : 3 3 be a linear transformation defined by
T x, y, z x – y 2z, 2x y, – x – 2y 2z . Find ker T, ImT .
Find a basis for each of them and their dimensions.
155
Hint : To find kerT , solve the system of equations.
x y 2z 0
2x y 0
– x – 2y 2z 0
1 –1 2
Using matrix form, after reducing the matrix 2 1 0 to row
–1 – 2 2
1 –1 2
reduced echelon form 0 1 – 4 .
3
0 0 0
4
y z and x – 2z .
3 3
ker T –2, 4, 3
To find ImT see that
3x
T x, y, z 2x y –1, 1, – 2 z 2, 0, 2
2
3
Because 1, 2, – 1 2 –1, 1, – 2 2, 0, 2
2
ImT – 1, 1, – 2 2, 0, 2 ,
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) T : 3 3 , T x, y, z x, y, 0
ii) M : 3 , M x, y, z 2x – 3y 5z
iii) I : V V , I for all V
iv) O : V V , O 0 for all V
4. Show that the following maps are not linear
i) Q : 2 , Q x, y y x
ii) T : 3 2 , T x, y, z x 2, 0
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5. Let T : 4 3 be defined by
T x, y, z, w x – y z w, x 2z – 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 n n matrices over . Let B be a fixed
n n 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 u1, u2, ..., um are linear by independent vectors in U then show that
Tu1, Tu2, ..., Tum are linearly independent vectors in V.
m
(Hint : Assume that j Tu j 0 and show that all scalars
j 1
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 T : 2 defined by
T x, y x y then dimkerT nullity = 1 and dimImT 1 .
(Here ker T 1, – 1 )
Theorem : Let U, V be vector spaces and T : U V be a linear
transformation then
dimU dim ker T dim ImT
= rank T + nullity T
157
Proof : If ImT 0 then kerT U and dimU dim ker T dim ImT .
Assume that 1, 2, ..., m be a basis of ImT . Let
u1, u2, ..., um be a set of vectors in U such that Tui i for all i,
1 i m.
Let k1, k2, ..., k p be a basis of kerT. In order to prove the result we need
to show that k1, k2, ..., k p, 1, 2, ..., m forms of basis for U.
Let u U T u ImT , hence there are scalars 1, 2, ..., m such that
T u 1 1 2 2 ... m m
1 Tu1 2 Tu2 ... m Tum
T 1 u1 2 u2 ... m um
u – 1 u1 2 u2 ... m um ker T
This means that there are scalars 1, 2, ..., p such that
u – 1 u1 2 u2 ... m um 1 k1 2 k2 ... p k p .
In other words,
u = 1 u1 2 u2 ... m um 1 k1 2 k2 ... p k p
Here we have proved that U is generated by the set of vectors
u1, u2, ..., um, k1, k2, ..., k p .
We have to show that this is a linearly independent set. So consider
scalars 1, 2, ..., m, 1, 2, ..., p such that
1 u1 2 u2 ... m um 1 k1 2 k2 ... p k p 0
T 1 u1 2 u2 ... m um 1 k1 2 k2 ... p k p T 0 0
1 Tu1 2 Tu2 ... m Tum 1 Tk1 2 Tk2 ... p Tk p 0
Here k1, ..., k p ker T Tk j 0 for all j, 1 j p
1 1 ... m m 0
158
Since 1, ..., m are linearly independent vectors.
1 2 ... m 0
Similarly, 1, 2, ..., p are all zero.
Thus dim U p m dim ker T dim Im T
= nullity T + rank T
Corollary : If U, V are vectors spaces such that dim U dim 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 T 0 .
dim ker T 0 dim U dim Im T dim V
dim Im T dim V T is onto Im T V
Conversely, assume that T is onto.
ImT V
dim U dim ker T dim Im T
dim ker T dim V
But dim U dim V dim ker T 0
ker T 0
Therefore T is injective map.
Note : If dim U dim 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 L U, V T : U V T is a linear transformation , 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.
S T u u1
S u u1 T u u1
S u T u S u1 T u1
S T u S T u1 for any u, u1 U
Similarly, S T u S T u u U,
s u u1 S u u1 S u S u1
S u S u1
S u S u1
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 u, u1 U and , then
SOT u u1 S T u u1
S T u T u1 (T is linear)
S T u S T u1 (S is linear)
S T u S T u1
160
SOT u SOT u1
This proves that SOT is a linear transformation.
Example :
1) Let T : 3 3 be defined by T x, y, z x, y, 0 and S : 3 3
be defined by S x, y, z x, z, 0 , then SOT : 3 3 is defined by
SOT x, y, z S x, y, 0 x, 0, 0
whereas TOS x, y, z T x, z, 0 x, z, 0 .
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 T 2 in general the
TOTO
operator ...OT by T n .
n times
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 TOS IV and SOT IU
where IV and IU 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 T 1 by S.
We shall show that S : V U is a linear map.
Let V1, V2 V , then u1, u2 U such that v1 Tu1 and v2 Tu2
or u1 S 1 and u2 S 2 .
S a 1 b 2 S a.Tu1 b.Tu2
S T au1 bu2
T 1T au1 bu2
au1 bu2 aS 1 bS 2 for a,b
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S is a linear transformation..
By definition of inverse function, it follows that T S IV and
S T IU .
Definition : A linear map T : U V which has inverse is called
invertible or non singular transformation or an isomorphism.
Example : F : 3 3 defined by
F x, y,z x y 2 z,x 2 y z,2 x 2 y 3z is invertible linear
transformation.
Here, x, y,z ker F
x y 2 z 0
x 2 y z 0
2 x 2 y 3z 0
1 1 2 x 0
This system can be written as 1 2 1 y 0 .
2 2 3 z 0
1 1 2
The coefficient matrix 1 2 1 has it‟s row reduced echelon form
2 2 3
1 0 0
as 0 1 0 .
0 0 1
1 0 0 x 0
The system is equivalent to 0 1 0 y 0
0 0 1 z 0
x y z 0 is the only solution of this system.
ker F 0
F is invertible.
Check your progress :
162
1) Define T : 3 3 by T x, y,z x y z,x 2 y z,3x 5 y z .
Find if T is non-singular. If not, find u 0 in 3 such that Tu 0 .
x y z 0
(Hint : Solve the system x 2 y z 0 x, y,z ker T
3x 5 y z 0
x 3z, y 2 z
the solution is z 3,2,1 ,z .
ker T 0 , if u 3,2,1 then Tu 0 )
2) Let U, V, W be vector spaces over . Let T : U V be a linear
transformation and let S1,S2 be two linear transformations of V into
W . Then show that S1 S2 T S1 T S2 T .
(Hint : For u U , find S1 S2 T u )
3) In above example prove that if , then S1 T S1 T .
(Hint : For u U , S1 T u )
4) Find a map F : 3 4 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 wi of v1 a unique linear map T : V V 1 s.t.
1 i n.
T i wi
Here Im F Span1,2,0,4
Let w1 1,2,0,4 ,w2 2,0,1,3
We want w3 such that spanw1,w2 ,w3 spanw1,w2
w3 0,0,0,0 is an obvious choice.
F x, y,z F xe1 ye2 ze3
xw1 yw2 zw3 x 1,2,0, 4 y 2,0,1,3
x 2 y,2 x, y,4 x3 y .)
163
5) Show that T : 2 2 defined by T x, y x y,x y is
invertible.
6) Let L : V V be a linear map s.t. L2 2L I 0 . Show that L is
invertible.
7) Let V be a vector space and T : V V be a linear map such that
T 2 T show that V ker T ImT .
(Hint : V can be written as T T )
8) Let V and V 1 be two vector spaces over . Show that there is a
bijective linear transformation T : V V 1 iff dimV dimV 1 .
9) Let A,B be linear maps of V into itself. If ker A 0 ker B , show
that ker AOB 0 .
10) Let A be a linear map of V to itself such that A2 A I 0 . Show
that A is invertible, and A1 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 B u1,u2 ,...,un and B1 1,2 ,...,m 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
E p,q : U V as follows.
164
E p,q u j p if jq
0 if jq
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 E p, q
1 p m,1 q n forms a basis for
L U ,V .
Let T be any linear transformation from U to V. Suppose that
Tu1 a111 a212 ... am1m
Tu2 a121 a222 am2m
Tun a1n 1 a2n 2 ... amn m
m n
We show that T a pq E p,q
p 1 q 1
m
Consider a pq E p,q u j
n
p 1 q 1
m n
a pq E p,q u j
p 1 q 1
m n
a pq jq p
p 1 q 1
m
a pj Tu j
p 1
m n
T a pq E p,q SpanE p,q
1 p m ,1 q n
p 1 q 1
L U ,V
Now we shall show that Wp,q 1 p m,1 q n is a linearly
independent set.
165
m n
If the transformation a pq E p,q is the zero transformation for
p 1 q 1
m n
p 1 q 1
scalars a pq then a pq E p,q u j 0 for each j 1 j n
m n
a pj p 0
p 1 q 1
But p 1 p m is a linearly independent set.
a pj 0 p, j
This shows that E p,q 1 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 f a b1 af bf 1 for all vectors
,1 V and scalars a,b .
Examples :
1) Define a function f : n by f x1,x2 ,..., xn a1x1 a2 x2 ... an xn
where a1,a2 ,...,an are fixed scalars in .
2) Let n . If A is an n n matrix over , the trace of A is the scalar
tr A,
tr A a11 a22 ... ann
the trace function is a linear functional on the matrix space nn ,
because
n
tr A B aii bii
i 1
n n
aii bii
i 1 i 1
tr A trB
166
3) Let C a,b be the space of continuous real valued functions on
b
a,b . Then L : C a,b defined by L g g t dt is a
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
dimV * dim L V ,
dimV . dim
dimV
Let B 1,2 ,...,n be a basis for V, then by the theorem on
existence and uniqueness of a linear transformation, there is a unique
linear functional fi : V such that
fi j ij 0 if i j, 1if i j .
For 1 i n , we obtain from B in set of n distinct linear functions
f1, f2 ,..., f n on V. We show that f1,..., f n is a linearly independent set
in L V , V * .
n
Let us consider f ci fi ,ci 1 i n then
i 1
n n
f j ci fi j ci ij c j
i 1 i 1
If f is the zero functional c j 01 j n
f1,..., f n is a linearly independent set in V * since dimV
* n.
B* f1,..., f n 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 fi for 1 i n .
Theorem : Let V be a finite dimensional vector space over , let
B 1,...,n be a basis for V then there is a unique dual basis
B* f1,..., f n for V * such that fi j ij for each linear functional
f :V .
n
f f i fi
i 1
and for each vector V .
n
fi i
i1
Proof : We have shown above that there is a unique basis f1,..., fn 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 fi , and as we observed after the scalars c j must be given by
cj f j .
n
Similarly, if U i i is a vector in V, then
i 1
n n
f j i f j i i ij j
i 1 i 1
So, that the unique expression for as a linear combination of the i ‟s
is,
n
fi i
i 1
n
Note : The expression fi i provides with a nice way of
i 1
describing what the dual basis is. If B 1,...,n is an ordered basis
for V and B* f1, f 2 ,..., f n is the dual basis, then fi is precisely the
168
function, which assigns to each vector V , the co-ordinate of
relative to the ordered basis B.
When 11 ... nn
f f 1 1 ... f n n 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 t1,t2 ,t3 be three distinct real nos.
Let Li : V be defined by Li p x p ti for p x V
Then L1,L2 ,L3 are linear functionals on V, Li V * . These linear
functionals are linearly independent,
For suppose, L aL1 bL2 cL3 0
L p x 0 p x V
Let p x 1, p x x, p x x 2
abc 0
t1a t2b t 3c 0
t12a t22b t32c 0
1 1 1 a 0
t1 t2 t3 b 0
2 2 3 c 0
t1 t2 t2
1 1 1
But the matrix t1 t2 t3 is invertible, because t1,t2 ,t3 are all distinct.
2 2 2
t1 t2 t3
a b c 0 is the only solution.
L1,L2 ,L is a linearly independent set. Now dimV * dimV 3
L1,L2 ,L3 is a basis for V * .
169
We would like to investigate for the basis of V, whose dual is L1,L2 ,L .
Let p1 x , p2 x , p3 x be the basis of V, whose dual basis is
L1,L2 ,L .
Li p j x ij
OR
p j ti ij
These polynomials are easily seen to be
p1 x
x t2 x t3
t1 t2 t1 t3
p2 x
x t1 x t3
t2 t1 t2 t3
p3 x
x t1 x t2
t3 t1 t3 t2
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 = dimV 1
In a vector space of dimension n, a subspace of dimension n 1
is called a hyperspace.
Check your progress :
1) In 3 , let 1 1,0, ,2 0,1,
2 ,3 1, 1,0
a) If f is a linear functional on 3 such that
f 1 1, f 2 1, f 3 3 . Find f a,b,c .
(Hint : Write a,b,c x11 x22 x33 , find x1,x2 ,x3 in
terms of a, b, c.
f a,b,c x1 f 1 x2 f 2 x3 f 3
170
x1 x2 3x3 )
2) Let B 1,2 ,3 be the basis of 3 defined by
1 1,0,1 ,2 1,1,1
,3 2,2,0 . Find the dual basis of B.
(Hint : Let f1, f2 , f3 be the dual of B.
f1 1 1; f1 2 0; f1 3 0;
f2 1 0; f2 2 1; f2 3 0;
f3 1 0; f3 2 0; f3 3 0.
Write a 3 – tuple x, y,z 3 as follows :
x, y,z 1 2 3 . Find ,, in terms of x, y,z .
f x, y,z f 1 f 2 f 3
f j x, y,z f j 1 f j 2 f j 3 for 1 j 3 )
3) Let V be the vector space of all polynomial functions p x : ,
which have degree 2 or less.
p x c0 c1x c2x 2
Define 3 – linear functions on V by,
1 2 1
f1 p x p x dx; f 2 p x p x dx; f3 p x p x dx
0 0 0
Show that f1, f 2 , f3 is a basis for V by exhibiting the basis for V
*
of which it is the dual.
(Hint : Assume that p1, p2 , p3 is a basis of V with the dual
f1, f2 , f3 , where
p1 x c0 c1x c2x2 , p2 x d0 d1x d2x 2 ;
p3 x c0 c1x c2x 2 and use the fact that fi p j x ij )
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 n n 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 E ,E ,...,E 1 2 n
.
0
j 0
Here E jth place
1
0
E j is the jth column of the n n identity matrix I n .
R1
Determinant of an n n matrix A c1 c 2 ... c n as
Rn
determinant of it‟s column vectors c1 c 2 ... c n or row vectors
R1
.
Rn
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 n n matrix A,det At det A & det AB det A. det 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 n n 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 n n 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 n n 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 n n matrix A over , a scalar D A in . We say that D is n-
linear if for each i,1 i n . D is a linear function of the ith row when
the other n 1 rows are fixed.
We shall understand this definition of n-linear function D. If D is a
function from n n ... n into , and if R1,R2 ,...,Rn are the rows
n times
of the matrix A, we write D A D R1,...,Rn , in other words D is
nothing but the function of the rows of A. To say that D is n-linear it
means that D R1,R2 ,...,Ri Ri1,...,Rn
D R1,R2 ,...,Ri ,...,Rn D R1,R2 ,...,Ri1,...,Rn
Example 1 : Let a1,a2 ,...,an be positive integers such that 1 a j n .
Let . For each nn matrix over , define
D A A1,a1 A2,an ... An,an where A Aij 1 i n,1 j n
This function D is n-linear.
174
A particular n-linear function of this type is D A A1,1
A2,2 ... An,n for
1 , a j j for 1 j n .
For if R1,R2 ,...,Rn are rows of the matrix A then
D R1,R2 ,...,Ri Ri1,...,Rn
A11 A22 ... Aii Aii1 ... Ann i is fixed.
A11 A22... Aii ... Ann A11 A22... Ann
D R1,R2 ,...,Ri Ri1,...,Rn D R1,...,Ri ,...,Rn D R1,...,Ri1...,Rn
Example 2 : Let us find all 2-linear functions on 2 2 matrices over .
Let E1,E2 be the rows of the 2 2 identity matrix. By matrix theory,
D A D A11E1 A12E2 , A21E1 A22E2
Using the fact that D is 2-linear,
D A A11D E1, A21E1 A22E2 A12D E2 ,A21E1 A22E2
A11 A21D E1,E1 A11A22D E1,E2 A12 A21D E2E1 A12 A22D E2 ,E2
Thus D is completely determined by the four scalars.
D E1,E1 ,D E1,E2 ,D E2 ,E1 and D E2 ,E2
Example 3 : Let D be the function defined on 2 2 matrices by
D A A11 A22 A12 A21 then D is a 2-linear function because
D D1 D2 where D1 A A11 A22 and D2 A A12 A21 both
D1,D2 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.
A A12
A 11
A21 A22
Let us note some of it‟s properties.
175
If I is the identity matrix, then D I 1 in other words D E1,E2 1 .
Second if the two rows of A are equal, then D A A11 A12 A12 A11 0 .
If A1 is the matrix obtained by 2 2 square matrix A by interchanging it‟s
rows, then D A1 A11
1 1 1 1
A22 A12 A21 A21 A12 A22 A11
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) D A 0 whenever two rows of A are equal.
(b) If A1 is a matrix obtained from A by interchanging two rows of A, then
D A1 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 (n n matrices over ) into . We say that D is a
determinant function provided that D is n-linear, skew-symmetric such
that D I n 1 . ( I n is the identity matrix of order n).
We shall show that ultimately there is exactly one determinant
function on n n matrices over . For n 1 . If A a a , then
D A a is the only determinant function.
For n 2 , the only determinant function is 2-linear function of the
following form.
D A A11 A21D E1E1 A11 A22D E1,E2 A12 A21D E2 ,E1
A12 A22D E2 ,E2
D must be skew-symmetric.
D E1 E1 0 D E2 ,E2 and
D E2 E1 D E1,E2 D I
D also satisfies D I 1 .
D A A11A22 A12A21
176
We shall prove that any n-linear function D on n n ... n
n times
satisfying D A 0 whenever two adjacent rows of A are equal, must be
skew-symmetric.
Theorem : Let D be an n-linear function on n n matrices over .
Suppose D has the property that D A 0 , whenever two adjacent rows
of A are equal then D is skew-symmetric.
Proof : We need to show that D A 0 , whenever any two rows of A are
equal and D A1 D A , if A1 is a matrix obtained from A by
interchanging two rows of A.
First suppose that A1 is obtained by interchanging two adjacent
rows of A. say R j & R j 1 then consider
O D R1,..., R j R j 1,R j R j 1,..., Rn
D R1,...,R j R j 1,R j ,...,Rn D R1,...,R j R j 1,R j 1,...,Rn
D R1,...,R j ,R j ,...,Rn D R1,...,R j 1,R j ,...,Rn
D R1,...,R j ,R j 1,...,Rn D R j ,...,R j 1,R j 1,...,Rn
D R1,...,R j 1,R j ,...,Rn D R1,...,R j ,R j 1,...,Rn
O D A1 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 R1,...,Ri 1,Ri 1...,R j ,Ri ,R j 1,...,Rn .
Now we move R j to ith position using j i 1 interchanges of
adjacent rows. Thus there are j i j i 1 2 j 2 1
interchanges of adjacent rows.
D B 1 D( A) D( A)
2 j i 1
177
Definition : Let A be an n n matrix over . Let A i j denote the
n 1 n 1 matrix obtained by deleting the ith row and j th column
of A.
If D is an n 1 linear function and A is an n n matrix then put
Dij 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 n .... n for any positive integer n.
n times
Theorem : Let n 1 D be an n 1 - linear skew symmetric function on
n 1 n 1 matrices over , for each j, j n the function.
E j : n n ... n . defined by
(n times)
n
E j A (1)i j Aij Dij A is an n-linear skew symmetric
i 1
function on n n matrices over . If D is a determinant function, so is
Ej .
Proof :
Dij ( A) D A i j and D is an n 1 - linear function.
Dij is linear as function of any row except ith row.
Aij Dij A is an n-linear function of A. We know that A linear
combination of n-linear functions is n-linear.
E j being a linear combination of n-linear functions, is an n-linear
function. To show that E j is skew-symmetric, it is enough to show that
E j A 0 , whenever A has two equal and adjacent rows.
Suppose Rk Rk 1 , if i k ,i k 1, the matrix A i j has two
equal rows, thus Dij A 0 .
k j k 1 j
Therefore, E j A 1 Akj Dkj A 1 A k 1 j A
178
Since Rk Rk 1 Akj A k 1 j and A k j A k 1 j
E j A 0
Now, assume that D is a determinant function. If I n is the n n identity
matrix then I n j j is the n 1 n 1 identity matrix I n 1 .
Since I n i, j ij E j I n D I n 1 1
E j I n 1 , and therefore E j 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 n n matrices, if such a determinant function is given on
n 1 n 1 matrices . The corollary follows by induction.
Example 1 : If B is a 2 2 matrix over . Let B B11B22 B12 B21 .
Then B D B , where D is the determinant function on 2 2 matrices.
We showed that this function on 2 2 is unique.
0 1 0
Example 2 : Let A be 3 3 matrix A 0 0 1
1 0 0
1 0
Then E1 A 1
0 1
0 1
E2 A 1
1 0
0 1
E3 A 1
1 0
Check your progress :
1) Given that a map f : 3 3 3 is 3-linear skew-symmetric.
If f I3 1 , prove that f A det A where A is any 3 3 matrix.
A11 A12 A13
(Hint : Write a 3 3 matrix A A21 A22 A23 as
A31 A32 A33
179
A A11 E1 A12 E2 A13E3 , A21E1 A22 E2 A23E3 , A31E1
A32 E2 A33E3
f A f A11 E1 A12 E2 A13E3 , A21E1 A22 E2 A23E3 ,
A31E1 A32 E2 A33E3
Use the fact that f is skew-symmetric.
f E1,E1 f E2 ,E2 f E3 ,E3 0
Also f E1,E2 ,E3 1 .)
2) Define a function D on 3 3 matrices over by the rule,
A A23 A A23 A A22
D A A11 22 A12 21 A13 21
A32 A33 A31 A33 A31 A32
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 n n
matrices over , show that
(a) D A 0 , 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 B R1 R2 ,R2 ,...,
Rn where A R1,R2 ,...,Rn .)
10.2 EXISTENCE AND UNIQUENESS OF
DETERMINANT FUNCTION :
In unit we showed the existence of a determinant function on n n
matrices over as a function of row vectors (column vectors) of n n
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 S n .
180
To every permutation in S n , we assign a signature as follows.
then sgn 1
k
If Sn where k is the number of
transpositions in the decomposition of permutation , in other words.
sgn 1
k
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 n n matrix over , Sn and A R1,R2 ,...,Rn ,
then D R1 ,R 2 ,R n D R1,R2 ,..,Rn
Proof : Let us consider a transposition i j then
D R1 ,R 2 ,...,R i ,...,R j ,...,R n
D R1,R2 ,...,R j ,...,Ri ,...,Rn
D R1,R2 ,...,Ri ,...,R j ,...,Rn
D R1,R2 ,..,Rn
Let , be two transpositions. Let D be a matrix defined by
D R1 ,...,R n .
C1,...,Cn say
Then D C1 ,C 2 ,...,C n D C1,C2 ,...,Cn
D R1,R2 ,...,Rn
In general, D R 0 0... 1 ,R 0 0... 2 ,...,R 0 0... n
1 2 k 1 2 k 1 2 k
1 2 ... k D R1,R2 ,...,Rn
If is any permutation, which is written as a product of transpositions,
say 12 ...r
D R1, R 2 ,...,R n D R102 0...0r 1 ,...,R102 0...r n
1 2 ... r D R1,R2 ,...,Rn
181
Suppose D is an n-linear skew-symmetric function on n n
matrices over .
Let A be an n n matrix over with rows. A1, A2 ,..., An . Let
E1,E2 ,...,En be the rows of the n n identify matrix I n then we can
write.
n
Ai Aij E j , For 1 i n
j 1
n
D A D A1 j E j , A2 , A3 , An
j 1
n
A1 j D E j , A2 , A3 ,..., An
j 1
n
Now replace A2 by A2k Ek
k 1
n
D E j , A2 ,..., An A1k D E j ,Ek ,..., An
k 1
D A A1 j A2k D E j ,Ek ,..., An
n n
j 1 k 1
Continuing this procedure for A3 , A4 ,..., An . We get,
n n n
D A
A1k1 A2k2 ... Ankn D Ek1 ,Ek2 ,...,Ekn
k1 1 k2 1 kn 1
(Using n-linearity of D)
Also, D Ek1, Ek2 ,..., Ekn 0 whenever two indices ki are equal.
D A
A1k 1 A2k2 ... Ankn D Ek1 ,Ek2 ,...,Ekn where the
k1,k2 ,...,kn
sum on R.H.S. is extended over all sequences R1,R2 ,...,Rn 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 n n
matrix over , we then have
D A
Sn
A11 A2 2 ... An n D E1 ,...,E n
But we know that D E1 ,...,E n D E1,E2 ,En
D A A11 A2 2 ... An n D E1,E2 ,...,En
Sn
A11 A2 2 ... An n D E1,E2 ,...,En (*)
S
n
Denote D A A11 A2 2 ... An n
Sn
Since we know that D E1,E2 ,...,En D I 1
D A det A D I
Remarks :
(1) The expression A11 A2 2.... An n above depends only
Sn
on the matrix A and thus uniquely determines D A .
(2) The expression A11 A2 2... An n gives explicit
Sn
formula for expansion of a determinant.
For example,
A11 A12 A13
D A21 A22 A23 A1 1 A2 2 A3 3 A11 A2 2
A
31 A32 A33
A3 3 A1 1 A2 2 A3 3 A11
A2 2 A3 3 A11 A2 2 A3 3
A11 A2 2 A3 3
183
1 2 3 1 2 3 1 2 3
Where Id , ,
1 2 3 2 3 1 3 1 2
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 A11 A22 A33 A12 A23 A31 A13 A21 A32
A11 A23 A32 A13 A22 A31 A12 A21 A33 is the same as the determinant
of a 3 3 matrix.
From above we see that there is precisely one determinant function
on n n matrices over .
Denote this function by det,
det A A11 A2 2... An n
Sn
Now, we shall see some properties of determinant function.
Theorem : Let A, B be two n n matrices, then det AB det A. det B
Proof : Let B be a fixed n n matrix over , for each n n matrix A
define D A det AB
Let A1, A2 ,.., An be the rows of A, then
D A1, A2 ,..., An det A,B,..., AnB
Here A j B 1 n matrix since CAi Ai B CAi B Ai1B & det is n-
linear D is n-linear.
If Ai A j then Ai B A j B and & since det is skew-symmetric.
D A1,..., An 0 whenever Ai A j i j .
Now, D is an n-linear alternating function.
D A det A D I (by *)
184
But D I det IB det B
det AB D A det A det B
Theorem : For an n n matrix A over det At det A where At
denotes the transpose of A.
Proof : IF is a permutation in S n ,
Ait, i A i i
We know that det A A11... An n
Sn
det At A1,1... A n ,n
Sn
Assume that i j for some i, j
i 1 j A i ,i A
j ,1 j
Therefore, A1,1... A n ,n A 1 ... A 1
1, 1 n, n
Since, 1 is the identity permutation
sgn sgn 1 1 or sgn sgn 1
1 as varies over S n , therefore 1 also varies
over S n .
det At
Sn
sgn 1 A111... An,1n
det A
Theorem : Let A be an invertible n n matrix, then det A1 det A 1
.
Proof : Since AA1 I ,det AA1 det I 1
185
det A det A1 1
det A1 det A
1
Check your progress :
0 a b
1) If A is the matrix over given by A a 0 c then show that
b c 0
det A 0 .
(Hint : Use the expression det A sgn A11 A
2 2 3 3
A )
Sn
2) Prove that the determinant of the Vandermonde matrix
1 a a2
A 1 2
b b is b a c a c b .
1 c c2
(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 n n matrix A is called triangular if Aij 0 for i j or
Aij 0 for i j , A is said to be upper-triangular if Aij 0 for i j
and A is said to be lower triangular if Aij 0 for i j . Prove that the
determinant of a triangular matrix is the product of it‟s diagonal
entries.
(Use the expression det A sgn A11 A2 2 An n )
Sn
1 x x12 x1n 1
1
1 x2
4) Prove that det
x22 x2n 1
x x
j i 1 i j n
1 xn xn2 xnn 1
(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
n n matrix over .
Fix r 1,2,...,n . Let be a permutation in S n .
j j ... j r r 1 2
Define eJ 1 1 2 r sgn and ki r i
where is a permutation which permutes the sets 1,2,...,r and
r 1,r 2,...,n within themselves.
Then
A j ,r A j ,r Ak , r 1 Ak , n
1 1 1 1
det A eJ
j j ...i A j ,1 A j , r Ak , r 1 Ak , n
1 2 r
r r r r
Here 1 j1, 2 j2 ,..., r jr
r 1 k1, r 2 k2 ,..., r i ki
This is one Laplace expansion for the determinant others may be
obtained by replacing the sets 1,2,...,r and r 1,...,n by two
different complementary sets of indices.
Check your progress :
1) Show that the equation of the line through distinct vectors
x y 1
x1, y1 , x2 , y2 is x1 y1 1 0 .
x2 y2 1
2) Prove that the area of the triangle in the plane with vertices
x1 x2 1
1
x1,x2 , y1, y2 , z1,z2 is the absolute value of y1 y2 1 .
2
z1 z2 1
P Q
3) Suppose we have an n n matrix A of the block form where P
O R
is r r matrix, R is s s matrix, Q is r s matrix & O denotes the s r
null matrix.
Then det A det P det R
187
P Q
(Hint : Define D P,Q,R det , consider D as an s-linear
O R
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 n n matrices over as a function of rows or columns.
det : n n ... n
ntimes
det A sgn A11 ... An n where the sum on R.H.S. is taken
Sn
over all permutations in S n .
2) Determinant function exists on the set of n n matrices over and it
is unique.
3) det AB det A det B for n n matrices A, B
det At det A. det A det A
1
.
1 x x12 x1n 1
1
1 x2 2 n 1
x2 x2
4) The Vandermonde matrix and it‟s
1 xn xn2 xnn 1
determinant x j xi 1 i j n .
5) A laplace expansion for the determinant of n n matrix A as follows.
A j ,1 A j ,r Ak , r 1 Ak , n
1 1 1 1
det A eJ
j j ... j A j ,1 A j , r Ak , r 1 Ak , n
1 2 r
r r r r
where the sum is taken over all r-tuples j1, j2 ,..., jr J such that
j1 j 2 ... j r 1 j1 , 2 j 2 ,..., r j r .
188
6) If D is an n-linear skew-symmetric function on n n matrices over ,
then D A det A D I .
7) If A is n n triangular matrix then det A A11 A22 ... Ann .
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
n n matrix aij 1 i, j n with det A 0 . X is n 1 column
x1
vector in n-unknowns x1,x2 ,...,xn X . B is given column
x
n
b1
vector B . The system AX B can be represented in the
b
n
following form as a set of n-equations in n-knowns x1,x2 ,...,xn .
a11 x1 a12 x2 ... a1n xn b1
a21x1 a22 x2 ... a2n xn b2
an1x1 an 2 x2 ... ann xn bn
Cofactors and minors Adjoint of an n n matrix A. basic results such
as A. adj A det A. I n
190
An n n real matrix A is invertible, if and only if det A 0 ; for n n
invertible matrix A.
1
A1 adj A
det A
Cromer‟s rule as a method for solving a system AX B whenever
det A 0 .
Uses of determinant as area and volume.
11.1 INTRODUCTION :
Determinants also allow us to determine when vectors are linearly
independent.
1 2
For example, if you consider two vectors and in the plane of
2 4
1
2 , then the determinant of the matrix A C1 C2 where C1 &
2
2
C2 gives you the information, whether C1 & C2 are linearly
4
independent vectors or linearly dependent vectors in terms of the det A ,
1 2
which is equal to 1 4 2 2 = 0.
2 4
1 2
This implies that C1 & C2 are linearly dependent,
2 4
which can be clearly because 2C1 C2 . In other words 2C1 C2 0 .
We try to generalize this result in the following theorem.
Theorem : Let C1,C2 ,...,Cn be column vectors in n and C1,C2 ,...,Cn
are linearly dependent, then D C1,C2 ,...,Cn 0 .
Proof : Suppose that given n-column vectors C1,C2 ,...,Cn in n are
linearly dependent.
scalars 1,2 ,...,n in such that
1c1 2 c2 ... n cn 0 0 n
and some scalars are non-zero.
Assume that j 0 for some j 1 j n
191
n
Cj k Ck
k j j
k 1
n
k
D C1,C2 ,...,C j ,...,Cn D C1,C2 ,..., Ck ,...,Cn
j
k j
k 1
n
k D C1,C2 ,..,Ck ,..,Cn
k j j
k 1
j 1
1 D C1,...,C1,...,Cn 2 D C1,C2 ,...,C2 ,...,Cn ....
j j j
j 1
D C1,C2 ,...,C j 1,C j 1,...,Cn
j
D C1,...,C j 1,C j 1,...,Cn
... n D C1,...,Cn ,...,Cn
j
=0
Since each determinant has a column equal to the jth column.
Colollary : If C1,C2 ,...,Cn are column vectors of n such that
D C1,C2 ,...,Cn 0 then there exist scalars 1,2 ,...,n such that
B 1c1 2 c2 ... n cn for given column vector B in n .
Proof : Since D C1,C2 ,...,Cn 0 C1,C2 ,...,Cn are linearly
independent n-vectors in n (by above theorem).
C1,C2 ,...,Cn 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
C1,C2 ,...,Cn . Thus for B n there are scalars 1,2 ,...,n such that
B 1c1 2c2 ... ncn
Note : The above corollary states that the system of n-equations in n-
unknowns x1,x2 ,...,xn namely,
192
a11x1 a12 x2 ... a1n xn b1
AX B OR has a solution
an1x1 an 2 x2 ... ann xn bn
provided that det A 0 .
We shall also see that the converse of the above theorem is also
true, in other words the determinant.
D C1,C2 ,...,Cn 0 C1,C2 ,...,Cn are linearly dependent.
Before providing this result we prove certain results about on n n matrix
A.
Definition : An n n 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 1 1
2 1 2 .
A 2 1 2 has rank equal to 2.
3 1 3 3 3
1 0 0
B 0 1 0 has rank equal to 3. ( E1,E2 ,E3 are linearly independent.)
0 0 1
1 0 1 0
2 1 0 2
C has rank equal to 3,
3 1 1 2
0 1 2 1
3,1,1,2 1,0,1,0 2,1,0,2
We have seen that an n n matrix A is invertible iff rank A n ,
because if A A1 A2 ... An is an n n matrix then A is invertible iff
A1, A2 ,..., An 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 0 1 0 1 0 1 0
2 1 0 2 2 1 0 2
1) and are row equivalent.
3 1 1 2 0 0 0 0
0 1 2 1 0 1 2 1
2 0 1 1 2 0
2) 1 1 0 and 0 1 1 are column equivalent.
0 1 3 3 0 1
Next we shall state a result for an n n matrix A, which is column
equivalent to a triangular matrix.
Theorem : Let A be an n n matrix, then A is column equivalent to a
b11 00 0
b21 b22 0 0
triangular matrix B of the form B .
bn1 bn 2 bnn
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 n 1 there is nothing to prove.
Assume that n 1 , let A aij ,i 1,2,...,n, j 1,2,...,n
0 0 0
a21 a22 a2n
Case (1) A
an1 an 2 ann
194
a22 a2n
*
Consider A .
a
2n ann n 1 n 1
By Induction hypothesis A* is equivalent to a triangular matrix, say
C22 0 0
C C33 0
C 32 .
Cn 2 Cn3 Cnn
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.
0 0 0
a21 C22 0 0
a31 C32 C33 0 0 A is equivalent to a linear triangular
a Cn3 Cnn
n1 Cn 2
matrix.
Case (2) : Assume that a11 0 .
Performing column operations again, we can reduce each of the
elements a1 j to 0 for j 2,3,...,n .
a11 0 0 0
A is column equivalent to M
a21
p22 p23 p2n
an1 pn 2 pn3 pnn
p22 p2n
Now, consider the n 1 n 1 matrix p .
p pnn
n2
By induction hypothesis, the matrix p is row equivalent to
q22 0 0
q32 q33 0 .
qn 2 qn3 qnn
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
a11 0 0 0
a q 0 0
to 21 22 which is a lower triangular matrix.
an1 qn 2 qn3 qnn
This completes the proof.
We have seen that if the column vectors C1,C2 ,...,Cn are linearly
dependent then D C1,C2 ,...,Cn 0 . Here we prove it‟s converse.
Theorem : If D C1,C2 ,...,Cn 0 then the column vectors C1,C2 ,...,Cn
are linearly dependent.
Proof : Let A C1,C2 ,...,Cn , C j is jth column of the matrix A. By the
previous theorem, A is column equivalent to say
b11 0 0
b21 b22 0 0
B and det A det B
bn1 bn 2 bnn
det B D C1,C2 ,...,Cn 0 .
Since, B is a triangular matrix.
det B b11b22...bnn .
Assume that bkk 0 for some k, 1 k n .
b11 0 0
b21 b22 0 0
B
bkk
bk 1,k 1
bk 1,k
bn1 bn 2 bnn
Reduce bk 1,k 1 to 0 by multiplying kth column by
bk 1,k 1
and
bk 1,k
subtracting from k 1 th column continuing in this manner of reducing
196
bkk to 0 upto k n , using such column operations will lead to a matrix
column equivalent to B,
d 11 0 0 0
Say
D = d K-1,K-1
0
0
0
d n1 dn2 d n-1
Since D is column equivalent to B and B is column equivalent to A.
rank A = rank D .
However, the set D1,D2 ,..,Dn of column of D is linearly
dependent set, as the last column is 0.
rank D n 1
rank A n 1 dimC1,C2 ,...,Cn n 1
C1,C2 ,...,Cn are linearly dependent.
Corollary : A square n n matrix A is invertible iff det A 0 .
Now, we shall see the existence and uniqueness of the system
AX B , where A is an n n matrix with det A 0 .
Given a system of n-equations in n-unknowns
a11 x1 a12 x2 ... a1n xn b1
a21x1 a22 x2 ... a2n xn b2
an1x1 an 2 x2 ... ann xn bn
can be represented in the following form AX B where A is the n n
matrix.
X x1x2 ...xn is a column vector in n .
t
B b1b2 ...bn is a column vector in n .
t
B n is fixed column vector.
197
If A is invertible A1 exists.
By premultiplying the system AX B by A1 we get
A1 AX A1 B .
A1 A X A1B
IX A1B
X A1B
This implies that the system has a unique solution namely
1
X A B provided that A is invertible matrix.
We know that if det A 0 A is invertible hence det A 0
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 n n matrix over A i j denotes the
n 1 n 1 matrix obtained from A by deleting ith row and jth column
of A.
A aij 1 i, j n , for 1 i, j n
i j
The cofactor of aij is the scalar given by 1 det A i j .
Example :
1 1 0
1) Consider a 3 3 matrix A 0 2 2 .
1 3 1
2
The cofactor of a11 111 det A 1 1 2 2 6 4
3 1
2 2
Here A 1 1 .
3 1
198
The cofactor of a21 121 det A 2 1 1 0
10 1
0 1
1 0
Here A 2 1 .
3 1
1 1 2 0
0 1 1 2
2) Consider a 4 4 matrix A
3 2 1 0
1 0 0 3
The cofactor of a42
4 2
1 det A 4 2
1 2 0
1 2 0 2
0 1 2 1 2 0
1 0 3 0
3 1 0
2 2 0 6 2 12 10
Definition : Let A be an n n matrix then the n 1 n 1 matrix.
A i j obtained by deleting i row and j column of the matrix A is
th th
called as the i, j minor of the matrix A.
Example :
3 1 0
1) Consider a 3 3 matrix A 1 2 5
1 0
2
the 1,1 minor of the matrix A
2 5
A 1 1
2 0
1 1 0 2
2 1 3 0
2) Consider a 4 4 matrix A
0 2 2 1
3 4 1 3
199
1 1 0
The 3,4 minor of the matrix A = 2 1
3
3 4 1
Definition : Let A be an n n matrix. A aij 1 j, j n . Let Cij
denote the cofactor of aij for all i, j a i, j n .
Cij 1
i j
det A i j for 1 i, j n write C Cij 1 i, j n the
C is an n n matrix called as the matrix of cofactors of A.
(Note : Cij = the cofactor of aij )
Example :
2 1 3
1) For a 3 3 matrix A 0 1 1
1 2 0
11 1 1
C11 1 det A 1 1 2
2 0
1 2 0 1
C12 1 det A 1 2 1
1 0
1 3 0 1
C13 1 det A 1 3
1 2
C21 6,C22 3,C23 5,C31 4,C32 2,C33 2
2 1 1
The matrix of cofactors of A C 6 3 5
4 2 2
Definition : Let A be an n n matrix over then the transpose of the
matrix of cofactors of A is called as adjoint of A denoted by adj A.
Example :
2 6 4
1) In above example adj A CT 1 3 2 our aim behind
1 5 2
finding out adjoint of the matrix A is to simply steps in order to obtain the
inverse of the matrix A, whenever det A 0 .
200
Check your progress :
1) Find adj A for the following matrices
1 1 2 0
1 1 0 1 2 3
0 3 5 2
i) 2 1 3 ii) 2 1 2 iii)
1 2 2 3 1
2 0 4
5 3
1 0 5 4
1 1 2 0
2 3 3 1
2) For the 4 4 matrix A given by A , find
4 5 0 3
2 1 3 2
i) i, j minor of A for all i, j , 1 i, j 4
ii) the cofactor of aij for all i, j . 1 i, j 4
11.2 PROPERTIES OF DETERMINANTS AND
CRAMER’S RULE :
Now, we have already proved that a determinant function defined
n
i j
on n n matrices is given by E j A 1 aij det A i j whenever
i 1
det is a determinant function.
But the uniqueness of a determinant function tells us that, if we fix
any column index j.
n
i j
det A 1 aij det A i j (the expansion by minors of the jth
i 1
column)
n
det A aij cij
i 1
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 B 0 & B i j A i j .
n
i j
det B 0 1 bij det B i j
i 1
201
n
i j
1 aik det A i j
i 1
n
aik Cij
i 1
C ij 1
i j
det A i j &bij aik 1 i n
n
If j k aik Cij 0
i 1
n
This means that aik Cij ik det A jk 0if j k , 1if j k
i 1
n
Cij aik det A
. jk
i 1
n
adj A ji aik det A
. jk
i 1
adj A. A det A.I
Theorem : For on n n matrix A over .
A adj A det AI n
Proof : We already know that adj A. A det AI n .
For an n n matrix A.
Applying this result for At .
adjAt . At det At . I det A I
adj At . At det A I
Taking transpose on both the sides,
t
adj At . At det A. I
adj At det A. I
t t
At
202
A. adj At det A. I
But adj At adj A
t
t
A adj A det A. I
t
A. adj A det A. I
Theorem : Let A be an n n matrix over , then A1 exists if and only if
det A 0 whenever A is invertible, the unique inverse for A is
1 1
A1 det A adj A adj A
det A
Proof : For an n n matrix A over , we know that
A. adj A det A. I .
If A is invertible A1 exists premultiplying by A1 on both the sides,
we get,
A1 Aadj A A1 det A I
A1. A adj A det A A1
1
A1 adj A
det A
Conversely, we know that for an n n matrix over ,
A. adj A det A I n and adj A. A det A I n
1
If det A 0 A adj A I n
det A
1
Also adj A A I n
det A
1
A1 exists and A1 adj A
det 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
A1, A2 ,..., An such that D A1, A2 ,..., An 0 .
b1 x1
b2 x
Let B be a column vector such that AX B where X 2 is a
bn xn
column vector consisting of n-unknowns, then for each j.
det M j
xj
det A
Where M j is the n n matrix obtained from A by replacing the jth
column of A by B.
M j A1, A2 ,...,B,..., An
jth column
Proof : For 1 j n , consider det M j D A1, A2 ,...,B,..., An
We know that x1 A1 x2 A2 ... xn An B .
det M j D A1,...,x1A1 x2 A2 ... xn An ,..., An
x1D A1,..., A1,..., An x2 D A1, A2 ,..., A2 ,..., An ...
x j D A1,..., A j ,..., An ... xn D A1, A2 ,..., An ,..., An
x j D A1, A2 ,..., An
Since every term in this sum except jth them is 0, because two
columns are same.
det M j
x j
det A
Note : The above theorem gives us the way of obtaining the unique
solution of the system AX B where A is an n n matrix.
A aij 1 i, j n, X x1,x2 ,...,xn ,B b1,b2 ,...,bn
t t
OR
204
a11 x1 a12 x2 ... a1n xn b1
a21x1 a22 x2 ... a2n xn b2
an1x1 an 2 x2 ... ann xn bn
which is a system of n-equations in n-unknowns Cramer‟s rule
gives the method of solving the system, when det A 0 .
Example :
1) Consider the following system of linear equations
3x 2 y 4 z 1
2 x y z 0
x 2 y 3z 1
3 2 4
Solution : Let A 2 1 1 det A 5
1 2 3
1 2 4
1
x 0 1 1
1
5 5
1 2 3
3 1 4
1
y 2 0 1 0
5
1 1 3
3 2 1
1 2
z 2 1 0
5 5
1 2 1
Check your progress :
1) For an n n matrix A, show that A adj adj A , whenever det A 1 .
2) Use adjoint to find the inverse of
a b
c , given that ad bc 0
d
i)
1 2 0
ii) 1 1 1
1 2 1
205
5 0 0 0
7 2 0 0
3) Compute
9 4 1 0
9 2 3 1
4) Solve the following systems using Cramer‟s rule.
i) 3x y z 0 ii) 2 x y z 1
x yz 0 x 3 y 2z 0
iii) x y 2z 1 iv) 2 x 3 y 4 z a
x y z 2 5x 6 y 7 z b
2 x y z 5 8x 9 y 9 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
u u1,u2 , 1,2 both in 2 such parallelogram is denoted by
P u, .
u+v
o u
P u, u 0 , 1
Let A u, denote the area of P u,
A u, 0
Here we introduce the concept of oriented area denoted by
u 1
A u, A u, if 1 0
u2 2
206
u 1
A u, if 1 0
u2 2
u 1
A u, has same sign as 1
u2 2
We shall show that A u, is actually the determinant
u 1
1 D u, . Consider the following axioms about area, which
u2 2
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 G1 and G2 are two regions which are non-intersecting or their
intersection has area 0, then A G1 G2 A G1 A G2
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 0
A e1,e1 1 where e1 , e2 , then by the uniqueness theorem
0 1
of the determinant function.
A u, D u,
Therefore we need to check the following properties of
A : 2 2 .
1) A is linear in each variable u,
2) A u,u 0, A u, A ,u
3) A e1,e2 1
207
1) Firstly show that following.
i) A nu, n A u, ,n
1 1
ii) A u, A u,. n
n n
iii) For C ,C 0 A cu, CA u,
iv) For any C , A cu, CA u, & A0 u,c CA u,
v) A u , A u,
vi) A u , A u,
vii) A cu d , CA u,
viii) A u1 u2 , A u1, A u2 ,
2) A u1u 0 is clear from the definition of oriented area, let
D u, 0 .
A u, A u, A ,u
A ,u A ,u A u,
A u, A ,u
The case D u, 0 is same.
(Here u, are linearly independent vectors.)
3) A e1,e2 is the area of a unit square which is equal to
1 A e1,e2 1
(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
Z 3V G Z V G .
3) If G1 and G2 are two regions, which are disjoint or such that their
intersection has volume zero, then V G1 G2 V G1 V G2 .
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 V e1,e2 ,e3 1 where
1 0 0
e1 0 ,e2 1 ,e3 0
0 0 1
Thus we need to show that
1) V is linear in each variable u,,
2) V u,, 0 if u or , u,, ,u,
u,, u,,
u,, ,,u
3) V e1,e2 ,e3 1
1) Firstly show the following.
i) u,, for n
V nu,, nV
1 m
ii) For n 3 , V u,, V u,,
n 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) V u1 u2 ,, V u1,, V u2 ,,
V u,1 2 , V u,1, V u,2 ,
V u,,1 2 V u,,1 V u,,2
Consider linearly independent vectors u,, .
There are numbers, a,b,c,...,,, such that
u1 au b c
u2 u .
V u1 u2 ,, V a u b c ,,
a V u,,
aV u,, V u,,
V au b c,, V u ,,
V u1,, V u2 ,,
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 T : 2 2 such
that u Te1 and Te2 .
Let u 1e1 2e2 & 1e1 2e2 .
210
1
Then m T w.r.t. the standard basis on both sides is 1 , denote
2 2
m T by det T .
A u, = absolute value of det u,
= absolute value of det m T
det T
If C is a unit square, then
C t1e1 t2e2 0 t1,t2 1
T C T t1e1 t2e2 0 t1,t2 1
t1Te1 t2Te2 0 t1,t2 1
t1u t2 P u,
Theorem : Let S : 2 2 be a linear map, then Area of
S P u, det S (Area of P u, ).
Proof : Since P u, is spanned by two vectors u, . S P u, is
spanned by S u ,S .
S
u Su
There is a linear map T : 2 2 such that Te1 u and
Te2 , 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 .
A S P u, det S T
det S det 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
Te1, Te2 and Te3 .
Write u u1e1 u2e2 u3e3
1e1 2e2 3e3
1e1 2e2 3e3
u1 1 1
m T u2 2 2
u 3 3
3
V u,, det u,,
det m T det T
If C is a unit cube, then
C 1e1 2e2 3e3 0 j 1
T C T 1e1 2e2 3e3
1e1 2e2 3e3 0 i 1 P u,,
Theorem : Let S : 3 3 be a linear map, then volume of
S P u,, det 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
V S P u,, det S T det S . det 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
a11 x1 a12 x2 ... a1n xn b1
a21x1 a22 x2 ... a2n xn b2
OR AX B
an1x1 an 2 x2 ... ann xn bn
The system has a unique solution namely X A1B provided that the
co-efficient matrix. A has non-vanishing determinant.
4) Basic results like Aadj A det A. I n , for an n n matrix A over
.
5) use of adj A to find inverse of an invertible n n matrix A over as
1
A1 adj A .
det A
6) Cromer‟s rule as a method for solving the system AX B , whenever
det A 0 , in this case the unique solution is given in terms of the
determinant function as follows.
det M j
xj for 1 j n
det A
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.
T : 2 3 defined by
214
T x, y x, y, x y we shall see the images of standard basis
1 0
element e1 . e2 of 2 under this linear transformation.
0 1
1
Te1 T 1, 0,1
0
0
Te2 T 0,1, 1
1
Since Te1,Te2 3 and 1,0,0 , 0,1,0 , 0,0,1 is a basis for 3 .
Te1 a11 1,0,0 a21 0,1,0 a31 0,0,1
Te2 a12 1,0,0 a22 0,1,0 a32 0,0,1
For some scalars a11,a21,a31,a12 , a22, a32 which are uniquely
determined.
We found that
a11 1 a12 0
a21 0 and a22 1
a31 1 a32 1
If we form a matrix of order 3 2 as follows:
1 j 2
A aij 1 i 3
1 0
A 0 1
1 1
32
x
If we consider AX, where X 1
x2 21
1 0
x
then we get 0 1 1 x1, x2 ,x1 x2
1 1 x2
T x1, x2 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
B 1, 2 ,..., n and let V be an m-dimensional vector space with
order basis B 1 , 2 ,..., m .
Let T : V V be a linear transformation. Since T j 1 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 a1 j ,a2 j amj
we produce mn numbers in all
m
Write T j aij ui
i 1
1 j n
This gives rise to an m n matrix. A aij 1 i m
aij
a2 j
t
th
A has j column aij ,a2 j ,..., amj
amj
a11
a
21 is the first column of A.
am1
a12
a22 is the second columns of A.
am 2
216
a1n
a2 n is the nth column of A.
amn
Call this matrix A to be the matrix of T with respect to the ordered basis
B
B, B of V ,V respectively, denote this matrix by m T B .
B
Note: m T B is uniquely determined for given ordered bases B, B of V
and V respectively.
B
We shall use this procedure of finding m T B to establish a one-
to-one correspondence between L V ,V and M mn .
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 mn .
Proof:
Let T L V ,V ,T : V V is a linear transformation.
m
T j aij i for each j, 1 j n .
i 1
B
This gives rise to the matrix m T B .
Define a function : L V ,V M mn by
B
T m T B
We show that for each A M mn these is a unique linear map from V to
V , whose matrix is precisely A.
Let A aij ,1
i m;1 j n .
Let y1 a1111 a2112 ... am11m
y2 a12 11 a22 12 ... am11m
217
yn am11 a2n12 ... amn1m
Here y1, y2 ,..., yn are vectors on V . We know that a unique linear
transformation T : V V such that y j T j for j =1, 2 …, n
B
m T B A .
: L V ,V M mn is onto map.
Let X B denotes co-ordinate vector in n of a vector V .
X B T denotes co-ordinate vector in m of a vector T V .
We show that
B
X B T m T B X B
x1
n
x
Let V , x j j X B 2
j i
xn
Let T A aij
Let Ai the ith row of the matrix A.
ai1, ai 2 ,..., ain .
x1
x
Ai X B ai1, ai 2 ,..., ain 2
xn
n
aij x j
j 1
a11 a12 a1n x1
a21 a22 a2n x2
X B
am1 am 2 amn x3
218
n
aij x j
j 1
n
a2 j x j
j 1
n
amj x j
j 1
n n
Also, T T x j j x j T j
j 1 j 1
n m m n
x j aij i aij x j i
j 1
j 1 i 1 j 1
n
aij x j
j 1
X B T X B
n
mj j mT B X B
a x
j 1 B
B
X B T m T B X B
We use this result to prove injectivity of .
B B
Let T1 B m T2 B
B
X B T1 m T1 B X B
B
m T2 B X B X B T2
T1 T2 for all V
T1 T2
is an injective map
There is a on to one correspondence between L V ,V and M mn .
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
m S T B m S B m T B
Proof:
Consider,
B
m S T B X B
X B S T
X B S T
B
m S B X B T
B B
m S B m T B X B
B B B
Using comparison m S T B m S B m T B
e.g.
Consider a linear transformation
T Id : V V T (V is finite dim vector space)
Then, for 1 j n
n
j id j aij i
i 1
Since 1, 2 ,..., n is a basis of V.
Therefore, a jj 1 and aij 0 if i j for 1 i n
1 j n
m Id ij 1 i nI n (the n n identity matrix)
1 j n
Corollary:
220
Let V be an n-dimensional vector space and T : V V be a linear
transformation. Let A m T B , where B is a fixed ordered basis
B
1, 2 ,..., 3 of V. Then T is non-singular iff A is non-singular, and in
B A1 .
B
that case m T 1
Proof:
Suppose that T is non-singular. Then T T 1 Id T 1 T .
B B
B B
m T T 1 m Id B m T 1 T
B
m T m T 1 I m T 1 m T
A m T 1 I m T 1 A
Hence m T A is invertible and A1 m T 1 conversely,
Suppose that A is invertible. Then A1 exists and also there exists a
transformation.
S : V V such that m S B A1
B
I AA1 m T m(S ) m T S
Similarly m S T m(S ) m T A1 A I
m T S I m S T
However I m Id
By one-to-one correspondence between L V ,V and M nn
T S Id S T
T 1 S
Thus T is invertible and by uniqueness of inverse T 1 S .
Corollary: Let V be a vector space with ordered bases B and B .
B
Then m Id B m Id B I
B
B
m Id B m Id B
B
In the previous theorem take V V V and S T Id .
Replace BbyB . Then
B
m Id B m Id B m Id B and
B B
221
B B
m Id B m Id B m Id B
B
Since m Id B I for all bases.
B
B B
m Id B m Id B I m Id B m Id B
B B
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
B
m T B N 1m T B N
B
Proof:
Write T : V V as follows:
T Id T Id
B B
m T B m Id T Id B
B
m Id T B m Id B
B
B
m Id B m T B m Id B
B B
B
Put N m Id B
B B
Since m Id B m Id B T m Id B m Id B
B B
N is invertible matrix and N 1 m Id B
B
B
m T B N 1 m T B N
B
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
m s T B m S B m T B and for any scalar ,
B B
m T B C m T B
Solution:
Let B 1, 2 ,..., 3 and B 1 , 2 ,..., m be bases for V ,V
respectively.
222
Let m S aij 1 i mm T bij 1 i m
1 j n
1 j n
and m S T Cij 1 i m
1 j n
m
Then S T j Cij i for each j = 1, 2, n
i 1
also S T j S j T j
m m
aij i bij i
i 1 i 1
m
aij bij i
i 1
aij aij bij for all i, j
1 i m;1 j n
Hence m S T m S m T
biji bij i
m m
Also, T j T j
i 1 i 1
m T bij bij m T
2) Let T : P3 P5 the linear transformation given by
T p x 1 2 x x2 p x . Find the matrix of T relative to the
standard bases of P3 and P3 .
Solution : We know that P3 has standard basis B 1,x,x 2 ,x3 and
P5 has standard basis B1 1,x,x2 ,x3 ,x 4 ,x5 .
T 1 1 2 x x 2
T x x 2 x 2 x3
T x 2 x 2 2 x3 x 4
T x3 x3 2 x 4 x5
223
1 0 0 0
2 1 0 0
B1 1 2 1 0
m T B
0 1 2 1
0 0 1 2
0 0 0 1
T 1 T 2 T 3 T 4
3) Calculate the matrix of the linear transformation T : 4 P1 given
by T x1,x2 ,x3 ,x4 x1 x3 x x2 x4 relative to the basis.
B 1 1,1,1,1
,2 1,1,1, 0,0 ,4 1,0,0,0
0 ,3 1,1,
of 4 and basis B1 11 1 x,12 1 x of P1 .
Solution : T 1 T 1,1,1,1
2 2 x 211 012
3 1
T 2 T 1,1,1,
0 2 x 11 12
2 2
T 3 T 1,1,
0,0 1 x 11 012
1 1
T 4 T 1,0,0,0 1 11 12
2 2
3 1
2 1
B1 2
M T B
2
0 1 1
0
2 2
4) Let T : 3 3 be a linear transformation defined by
T x, y,z x,x y,0 . Find matrix of T ,m T
i) with respect to a natural basis B on both sides.
ii) with respect to basis B 1,1, 0 , 0,1,1 on both
,1,1,1
sides.
iii) with respect to basis B e1,e2 ,e3 for the domain and basis
B1 1,1,
0 , 0,1,1 for the co domain.
,1,1,1
Solution :
224
i) Let e1,e2 ,e3 be a natural basis for 3 on both the sides,
e1 1,0,0 , e2 0,1,
0 ,e3 0,0,1
then Te1 T 1,0,0 1,1,
0 1.e1 1.e2 0.e3
Te2 T 0,1,
0 0,1,
0 0.e1 1.e2 0.e3
Te3 T 0,0,1
0,,0 0.e1 0.e2 0.e3
1 0 0
m T 1 1 0 w.r.t natural basis for 3 on both the sides.
0 0 0
ii) Let u1 1,1,
0 ;u2 0,1,1
;u3 1,1,1
Tu1 T 1,1,
0 1,2,0 1u1 2u2 3u3
Tu2 T 0,1,1
0,1,
0 1u1 2u2 3u3
Tu3 T 1,1,
1,2,0 1u1 2u2 3u3
i ,i ,i are scalars.
1,2,0 1,1,0 0,2 ,2 3,3,3
1 3 ,1 2 3 ,2 3
1 2,2 1,3 1
Tu1 2u1 1u2 1 u3
Similarly,
Tu2 0,1,
0 1u1 1u2 1 u3 ,
Tu3 1,2,0 2u1 1u2 1 u3
2 1 2
m T 1 1 1 w.r.t. basis B u1,u2 ,u3 on both the
1 1
sides.
iii) Te1 1,1,
0 1u1 2 u2 3u3
Te2 0,1,
1u1 2 u2 3u3
Te3 0,,0 1u1 2 u2 3u3
225
Where i ,i ,i are all scalars.
Te1 1,1,
0 1 1,1,
0 2 0,1,1
3 1,
0 1 3 ,1 2 3 ,2 3
1,1,
1 1,2 0,3 0
Te1 1,1,
0 1.u1 0.u2 0.u3
Similarly,
Te2 0,1,
1.u1 1.u2 1 u3
Te3 0,,0 0.u1 0.u2 0.u3
1 1 0
m T 0 1 0 w.r.t. basis e1,e2 ,e3 of 3 and basis
0 1 0
B u1,u2 ,u3 of 3 (co domain).
5) Let V M 22 (set of 2 2 matrices over ). Let T : V V
1 1
be defined by T A A . Find the matrix of T with respect to
1 1
1 0 0 1 0 0 0 0
basis E1 ,E2 ,E3 ,E4 of
0 0 0 0 1 0 0 1
M 22 .
Solution :
1 1 1 0 1 0
T E1 1.E1 0.E2 1.E3 0.E4
1 1 0 0 1 0
1 1 0 1 0 1
T E2 0.E1 1.E2 0.E3 1.E4
1 1 0 0 0 1
1 1 0 0 1 0
T E3 1.E1 0.E2 1.E3 0.E4
1 1 1 0 1 0
226
1 1 0 0 0 1
T E4 0.E1 1.E2 0.E3 1.E4
1 1 0 1 0 1
1 0 1 0
The matrix of the transformation is 0 1 0 1
. Now it‟s
1 0 1 0
0 1 0 1
your turn to solve certain exercises.
Check your progress :
1) T : 2 2 is defined by T1 1 2 and T2 1 2 , where
1 e1 2e2 and 2 e1 e2
i) write matrix of T, relative to basis 1,2 on both the sides.
ii) write matrix of T, relative to basis e1,e2 on both the sides.
1 2
2) Let M and T be the linear operator defined on the set of
3 4
1 0 0 1 0 0
2 2 matrices having basis E1 ,E2 ,E3 ,
0 0 0 0 1 0
0 0
E4 by T A M . A , find the matrix of T.
0 1
3) Let B1 u1 1,1,1 0 ,u3 1,0,0 be a basis of 3
,u2 1,1,
and B2 1 1,3 ,2 2,5 be a basis of 2 . Let T : 3 2
be defined by T x, y,z 3x 2 y 4 z,x 5 y 3z . Find m T B2 .
B
1
4) Consider the set e3t ,te3t ,t 2e3t a basis of a vector space V of
functions f :: . Let D be the differential operator on V.
D f t f 1 t . Find the matrix of D in the given basis.
(Hint : Find D e3t ,D te3t ,D t 2e3t , write all these as a linear
combination of e3t ,te3t ,t 2e3t .)
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 B1,B2 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 B,B1 , what can
B1
be the relation between m T B and m T 1 . To find the answer we state
B
B
the following theorem.
Theorem : Let V be a vector space over with basis
B 1,2 ,...,n . Let T L V ,V be such that
m T B A aij 1 i, j n . If B is another n n matrix such that
B
B N 1 AN for some invertible matrix N nij , then another basis
B1
B1 1,2 ,...,n of V such that m T 1 B .
B
Proof : Since A m T B
B
n
T j aij i
i 1
Define a transformation S corresponding to the matrix
N nij 1 i, j n with respect to basis B, in other words
n
S j nij i .
i 1
Since N 1 exists S 1 exists.
Let j S j , since 1,...,n is a basis of V and S is non-singular
transformation with the fact that B1 1, 2 ,..,n is also a basis of V.
B1
We claim that m T 1 B
B
Let
B1
m T 1 D ij 1 i, j n
B
228
n n
T j ij i ij S i
i 1 i 1
n n
ij nki k
i 1 k 1
n n
nki ij k
i 1 k 1
n n
nki ij k
(i)
k 1 i 1
n
Also T j T nij i
i 1
n n n
nij T j nij aki k
i 1 i 1 i 1
n n
aki nij k
i 1 i 1
n n
aki nij k (ii)
i 1 i 1
From (i) and (ii), we conclude that
ND AN , in other D N 1 AN B
B m T B1
1
In the above theorem, we saw that, how one can connect the two
matrices A, B relative to bases B and B1 of V respectively, in terms of on
invertible matrix N.
Definition : Two matrices A,B M n are said to be similar if there
exists a non-singular matrix C in M n such that B C 1 AC .
1
Here we can observe that B C 1 AC A CBC 1 C 1 BC 1 .
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 A,B M n 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
n
j nij j gives you the columns of N as the co-ordinates of -basis
i 1
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 P3 with basis
B 1 1,2 x,3 x 2 ,4 x3
and B1 1 1,2 1 x,3 x x 2 ,4 x 2 x3
B1
Find A m D B ,B m D 1 , the transition matrix N and verify that
B
B
B B1 AN where D : P3 P3 is a differential operator.
Solution :
0 1 0 0
B
m DB 0
0
0
0
2
0
0
3
A
0 0 0 0
D 10 0.1 0.x 0.x 2 0.x3
D x 1 1.1 0.x 0.x 2 0.x3
D x 2 2 x 0.1 2.x 0.x 2 0.x3
D x3 3x 2 0.1 0.x 3.x 2 0.x3
230
0 1 1 1
B1 0 0 2 1
and m D 1 B
B 0 0 0 3
0 0 0 0
D 100.1 0.1 x 0. x x 2 0. x 2 x3
D 1 x 11.1 0.1 x 0. x x 2 0. x 2 x3
D x x 2 2 x 1.1 2.1 x 0. x x 2 0. x 2 x3
D x 2 x3 2 x 3x 2 1.1 1 .1 x 3. x x 2 0. x 2 x3
1 1 0 0
0 1 1 0
Also the transition matrix N
0 0 1 1
0 0 0 1
1.1 0.x 0.x 2 0.x3
1
2 1 x 1.1 1.x 0.x 2 0.x3
3 x x 2 0.1 1.x 1.x 2 0.x3
4 x 2 x3 0.1 0.x 1.x 2 1.x3
It can be easily seen that N 1 AN B .
2) Let D be the differentiation operator on P3 . Find the matrix B
representing D with respect to basis B 1,x,x 2 and find the matrix A
representing D w.r.t. B1 1,2 x,4 x 2 2 .
Solution :
D 1 0 0 1 0 x 0 x2
D x 1 1 1 0 x 0 x2
D x 2 2 x 0 1 2 x 0 x2
231
0 1 0
B 0 0 2
0 0
0
Now,
D 1 0.1 0.2 x 0. 4 x 2 2
D 2 x 2.1 0.2 x 0. 4 x 2 2
D 4 x 2 2 0.1 4.2 x 0. 4 x 2 2
0 2 0
B1
A m D 1 0 0 4
B 0 0
0
To obtain the transition matrix N, we write
11.1 0.x 0.x 2
2 x0.1 2.x 0.x 2
4 x 2 2 2 .1 0.x 4 x 2
1
1 0
2
1 2 2
N 0 2
1
0 N 0
1
0
0 2
0 4
0 1
0
4
N 1BN A
Check your progress :
1) Let S : 3 3 be the linear transaction
S x, y,z y x,0,z x .
a) Write the standard and matrix representation for operator T.
b) Let B u1 1,0,1 0 be another
,u2 0,1,1 ,u3 1,1,
basis of 3 . Find the transition matrix N from the basis B to the
standard basis e1,e2 ,e3 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
G 1 1,1,
1 ,2 1,1, of 3 .
1 ,3 0,1,1
2) Suppose that the linear mapping T : 3 3 is represented by the
1 2 3
matrix A 1 1 2 relative to the standard basis of 3 .
2 2 2
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) R : 2 2 .
R x, y xcos ysin,xsin ycos
4) Find the matrix for the reflection in the plane along a line e making an
angle with positive x-direction given by,
T x, y xcos ysin,xsin ycos
12.5 SUMMARY :
In the chapter we learned the following concepts.
1) There exists a one-to-one correspondence between the set of m n real
matrices and the set L V ,V 1 where dimV n and dimV 1 m
assuming given ordered bases B 1,..,n and B1 11,...,1m of
V and V 1 respectively.
2) For a linear operator T : V V on a finite dimensional vector space
B1
V, if B&B1 are given bases of V then m T B and m T 1 are
B
B
similar matrices, in other words, there exists an invertible matrix N s.t.
B1
m T 1 N 1m T B . N .
B
B
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,
m STT BB
1 B1 B1
m S B m T B
233
B1 B1
Similarly, m T B . m T B where T is a linear
transformation defined by T .T for all V .
4) Given two similar matrices A,B M n , where dimV n , such
B1
that A m T B and B m T 1 for bases B and B1 of V.
B
B
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 B 0
13.4 Summary
13.0 OBJECTIVES :
This chapter would help you understand the following concepts:
Rank of the linear transformation LA : n m defined by
LA X X and it‟s connection with the rank of on m n
matrix A.
The dimension of the solution space of the system of linear
equations AX 0 where A is an m n matrix and X is n 1
column sector or a system of m linear equations in n unknowns
consisting of a set of equations:
a11x1 a12 x2 .... a1n xn b1
a21x1 a22 x2 a2n xn b2
am1 x1 am2 x2 .... amn xn bm
Where aij and bi are real numbers and x1 ,x2 ,...,xn are unknowns
The dimension of the solution space of the system AX 0 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:
a11x1 a12 x2 .... a1n xn b1
a21x1 a22 x2 a2n xn b2
am1 x1 am 2 x2 .... amn xn bm
Where aij and bi are real numbers and x1 ,...,xn are n-unknowns
a11a12 ...a1n
a21 a22 ...a2n
The m n matrix A aij 1i m
1 j n
am1am 2 ...amn mn
is called the co-efficient matrix of the system or the matrix associated
x1 b1
x2 b
with the system Let X and B 2 then the system
xn bm
can be written as a single matrix equation AX = B.
q1
A solution to the system is a vector Q n
in such that
q
n
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
0
0
0 such solution is called the trivial solution.
0 n1
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) =
aAP bAQ a0 b0 0 .
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 T : n m be defined by,
n
aij q j
q1 A1Q j 1
T( Q ) T
q A Q n
n m
mj j
a q
j 1
A1
Where A Ai ' s are rows of the matrix A.
A
m
ap1 bq1 A1 aP bQ
Then T aP bQ T
ap bq A aP bQ
n n m
237
aA1P bA1Q
aA P bA Q
m m
A1P A1Q
a b aT P bT Q
A P A Q
m m
T is a linear transformation.
Also, Q ker T if T( Q ) 0 , in other words.
n
aij q j a11q1 a12 q2 a1n qn
j 1 0
a21q1 a22 q2 a2n qn
iff
n 0
mj j
a q am1q1 am 2 q2 amn qn
j 1
Q S (the solution set of the system AX = 0)
ker T Sdimker T dim S
We know that ImT under the standard basis e1,..., em of m is the
matrix A, that is m T A .
By Rank – nullity theorem,
n dimker T dimIm T
dim S dimImT dim S rank A
dim S n r
Corollary: If m n , the system AX 0 has a non-trivial solution.
Proof: Suppose m n we know that rank A m and rank A n rank
A m n.
dim S n rank A 0. Thus S 0
a vector X 0 n s.t. X 0 S , that is AX 0 has a
nontrivial solution X 0 .
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
x y 0
x yz 0
1 1 0
Here m =2 & n = 3, A
1 1 1 23
1, 1,0 is a non-trivial solution of this system.
Definition:
If A aij 1 i m is the matrix associated with the system of
1 j n
a11 a a1n b1
equations. AX B the m n 1 matrix is
a
m1 am 2 amn bm
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 Q n is a solution of the system AX = B.
a11 a12 a1n q1 b1
=
a q b
m1 am 2 amn n m
b1 a11 q1 q12 q2 a1n qn
b a q a q a q
m m1 1 m2 2 mn n
a11 a12 a1n
a21 a22
q1 q qn
2
am1 am 2 amn
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 A1, A2 , An be columns of the matrix A. There are r of them, which
are linearly independent. Assume that Ai1 ,..., Air be linearly independent.
Let if possible, suppose that B is not a linear combination of Ai1 ,..., Air .
The columns Ai1 ,..., Air B are linearly independent
rank A, B r 1
This is a contradiction
B must be a linear combination of A1, A2 ,..., An , hence there are
numbers 1, 2 ,...n such that
B 1 A1 2 A2 ... n An
1
Let P B AP
n
The system AX B has a solution namely P.
13.3 SOLUTION OF THE NON-HOMOGENEOUS
SYSTEM AX = B, WHERE B 0 :
Given an non homogeneous system of the form AX = B where
B 0, there is a method that describes all solutions of this system.
Theorem:
Suppose the system AX = B has a particular solution X 0 . Then a
vector X in n is also a solution iff X X 0 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 X X 0 Y ,
Where X 0 is a particular solution of the non-homogenous system
AX = B.
Then AX A X 0 Y AX 0 AY B 0 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 X 0 is also a
solution,
A X X 0 AX AX 0 B B 0
240
X X 0 is a solution of the corresponding homogeneous system, call it
Y. Then X X 0 Y ,
Here Y satisfies AX = 0
Note: When we are asked to find all solutions of the non-homogeneous
system AX = B. B 0, 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 in 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 B 0. Suppose X 0 is the unique solution of this
system. Let Y be any solution of the corresponding homogeneous system
AX = 0 AY = 0.
A X 0 Y AX 0 AY B indicating that X 0 Y is also a solution of
AX = B.
The uniqueness implies that X 0 Y X 0 Y 0.
Thus the homogeneous system AX = 0 has only trivial solution.
dim S 0 n rank A 0 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 n,n 1
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 X 0 Y , where Y is
any solution of AX = 0.
241
But rank A n dim S 0
S 0
X 0 Y can take only one value, namely
X0 0 X0 .
Thus the solution of the system AX = B is unique.
We would make use of certain symbols, which are interpreted as follows:
Ri R j : Exchange ith row and j th row
Ri (k ) : Multiply ith row by some non zero k .
Ri k R j : Multiply ith row by a scalar k and add it to j th 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
x 2 y 3z 0
3x 4 y 4 z 0
7 x 10 y 12 z 0
Solution:
The given system can be represented in the following form.
1 2 3 x 0
3 4 4 y 0
7 10 12 z 0
We try to calculate the rank of the matrix
1 2 3 1 2 3 1 2 3
R2 3R1 R3 2 R2
3 4 4 0 2 5 0 2 5
R3 7 R1
7 10 12 0 4 9 0 0 1
1 2 3
the rank of 0 2 5 is 3
0 0 1
1 2 3
Since A is row equivalent to 0 2 5
0 0 1
242
rank (A) = 3. Hence the only solution is the trivial solution, namely 0.
2) Solve
2 x 2 y 5z 3w 0
4x y z w 0
3x 2 y 3z 4w 0
x 3 y 7 z 6w 0
Solution:
The system can be written as follows:
2 2 5 3 x 0
4 1 1 1 y 0
3 2 3 4 z 0
1 3 7 6 w 0
1 3 7 6
R1 R4 4 1 1 1 R2 4 R1,R3 3R1
3 2 3 4 R4 2 R1
2 2 5 3
1 3 7 6 x 0
0 11 27 23 y 0
0 7 18 14 z 0
0 4 9 9 w 0
R2 (3) R4 and R3 (2) R4 gives
1 3 7 6 x 0
0 1 0 4 y 0
0 1 0 4 z 0
0 4 9 9 w 0
R3 (1) R2 and R4 4R2 gives
1 3 7 6 x 0
0 1 0 4 y
0
0 0 0 0 z 0
0 0 9 7 w 0
The rank of the associated matrix is 3.
243
dim S 4 3 1
All the solutions of the system are scalar multiples of one single non
zero vector.
The last matrix equation gives
x 3 y 7 z 6w 0
y 4w 0
9 z 7w 0
7
y 4w&z w,x 3 y 7 z 6w
9
49 5
12w w 6w w
9 9
5 7
x, y, z, w w, 4w, w, w
9 9
w
5,36, 7,9
9
Thus the solution space is S 5,36,7,9 /
3) Show that the only real value for which the following equations
have nonzero solution is 6.
x 2 y 3z x;
3x y 2 z y;
2 x 3 y z z .
Solution:
The system can be rewritten as.
1 2 3 x 0
3 1 2 y 0
2 3 1 z 0
4 3 5 x 0
R1 R2 gives 3 1 2 y 0
2 1 z 0
3
R1 R3gives
244
6 6 6 x 0
3 1 2 y 0
2 1 z 0
3
6 6 6
If 6 , the coefficient matrix 3 1 2
2 1
3
0 0 0
becomes 3 5 2 , whose rank is 2.
2 3 5
dim S 3 2 1
the system has non-trivial solutions
If the associated matrix becomes, after operation
1 1 1
1
R1 , 3 1 2
6
2 3 1
R2 3R1,R3 2R1gives
1 1 1
0 2 1 If 2 , this matrix is
0 1 1
1 1 1
0 0 1 , which has rank 3.
0 1 1
In this case these is only one solution namely, the trivial (zero) solution.
If 2, after performing operation R2 2 R3
1 1 1 1 1 1
0 2
0
1 becomes 0 0 2 3 3
1 1 0 1 1
245
2 3 3 0 for any value of real
1 1 1
1
R2 gives 0 0 1
2
3 3 0 1 1
1 1 1
R1 R2 and R3 R2 gives 0 0 1
0 1
the rank of this matrix is 3, irrespective of whether 0or 0 .
the system can have trivial solution only.
The system can have a non-trivial solution if 6 .
4) Show that the equations
2 x 6 y 11 0;
6 x 20 y 6 z 3 0;
6 y 18z 1 0 are not consistent.
Solution:
The augmented matrix A B IS
2 6 0 11 2 6 0 11
R2 3R1
6 20 6 3 0 2 6 30
0 6 18 1 0 6 18 1
2 6 0 11
R3 3R2
0 2 6 30
0 0 0 91
2 6 0 2 6 0 11
rank 6 20 6 2 and rank 0 2 6 30 3
0 6 18 0 0 0 91
Hence the system has no solution. In other words the system is
inconsistent.
5) Show that the equations
x 2 y z 3;
3x y 2 z 1;
246
2 x 2 y 3z 2;
x y z 1 are consistent and solve them.
Solution:
1 2 1 3
3 1 2 1
The augmented matrix A B is
2 2 3 2
1 1 1 1
1 2 1 3
R2 R4 ,R3 2 R4 2 0 1 2
0 0 1 4
1 1 1 1
1 2 1 3
R2 R3 ,R4 R3 2 0 0 2
0 0 1 4
1 1 0 5
1 2 0 7
1
R1 R3 ,R2 1 0 0 1
2
0 0 1 4
1 1 0 5
0 2 0 8
R1 R2 ,R4 R2 1 0 0 1
0 0 1 4
0 1 0 4
0 0 0 0
R1 2 R4 1 0 0 1
0 0 1 4
0 1 0 4
Since rank A = rank A B 3 (number of) variables.
247
The system has precisely one solution which is same as the solution of
the system
0 0 0 0
x
1 0 0 1
y
0 0 1 4
z
0 1 0 4
x 1, y 4,z 4
6) For what values of , the simultaneous equations.
x y z 6;
x 2 y 3z 10;
x 2 y z have
(i) no solution (ii) a unique solution
(iii) an infinite number of solutions?
Solution:
1 1 1 6
A B 1 2 3 10
1 2
1 1 1 1
R3 R2
1 2 3 10
0 0 3 10
1 1 1 6
R2 R1
0
1 2 4
0 0 3 10
(i) If 3 and 10 then rank A = 2 but rank A B 3, then the
given system has no solution.
(ii) If 3 Rank A rank A B 3, 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
x y 4 z 6;
x 2 y 2 z 6;
x y z 6 for different values of .
Solution:
1 1 4 6
A B 1 2 2 6
1 1 6
1 1 4 6
R2 R1
0
1 6 0
1 1 6
1 0 10 6
R1 R2
0 1 6 0
1 1 6
1 0 10 6
R3 R2
0 1 6 0
0 7 6
7 0 10 6
R3 R1 1
10
1 6 0
0
7 9
0 0
10 5
7
the system is consistent iff
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.
x 4 y 7 z 14;
3x 8 y 2 z 13;
249
7 x 8 y 26 z 5
2) Solve the following systems
(i) x y z 6;
x 2 y 3z 10;
x 2 y 4z 1
(ii) 2 x 3 y z 9;
x 2 y 3z 6;
3x y 2 z 8
(iii) x 2 y 5z 9;
3x y 2 z 5;
2 x 3 y z 3;
4 x 5 y z 3
3) Find the dimension of the solution space of the system.
2 x y z 0;
3x 2 y 0;
x y z
4) For the system of equations
ax y z 4
x by z 3
x 2by z 4
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 AX 0 is given by
n-rank A.
4) The general solution of the non homogeneous system AX B , B 0
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 p A t in the real variable to associated with each
matrix A.
We will also combine the above two concepts and obtain a
polynomial pT t for a linear transformation T: V → V, pT is the
“characteristic polynomial” of the transformation T. We are interested in
the roots of the characteristic polynomial pT because the roots carry
important information about the transformation T. We will study the
characteristic polynomial pT 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 : B EAE 1 (1)
251
We indicate similarity between the matrices A and B by the notation :
AB (2)
41 37
Thus, for example, A and B are similar because there is
25 06
32
the invertible matrix E which gives the equality B EAE 1 .
11
But note that a pair of matrices A, B is not always in similarity
41 37
relation. For example, if A and B then A, B are not
25 06
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. B EAE 1
Now note that EAE 1 is invertible, its inverse being EA1E 1 . 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 AB 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.
1 0
Proof : Recall, the identity matrix I 1
0 1
Is invertible with I 1 . Consequently for any A M n, we have.
A I A IA I I A I IAI 1 , This shows that A A proving (1).
252
We suppose A B. Then by the definition of , there exists invertible E
such that BEAE 1 We put E-1 = F so that F is an invertible matrix, (its
inverse F1 being equal to E). now the equality BEAE 1 gives
FB FEAE 1 FE AF IAF AF , i.e. AF FB
Multiplying on the right of each side of this equation by F 1(E ) we
get : ( AF )F 1FBF 1
i.e. A( FF 1)FBF 1
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
BEAE 1 (*)
Also, since B C, there exists and invertible F such that CFBF 1
(**)
Combining (*) and (**) we get,
CFBF 1F(EAE 1)F 1( FE ) A( FE ) 1
Since ( FE ) 1 E 1F 1 showing that FE = D is invertible matrix
with D 1E 1F 1 . Thus we have CDAD1 .
A C proving (3).
Let GL (n, ) = { E M (n, ) : E is invertible}.
Also for A M (n, ) we put :
[A] = { B M (n, ) : A B}.
Clearly, [A] is the subset of M (n, ) consisting of all B M (n, ) which
are of the form B = EAE-1 for some E GL (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 A M (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 G GL(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 E GL (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. G 1 is invertible with F-1 = G. E 1 Thus
the equality X = E.G-1. C. G. X 1 becomes X F. C. F 1 and therefore we
have : X [C]
This being true for every X [A] we get [A] [C]
Next, let Y [C]. Therefore there exists E GL (n, ) such that Y = E. C.
E-1 = E. G. A. G-1. E-1 using (*)
But again E. G GL (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
p A 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
pA t pB t . We use the notion of the determinant det (A) of a square
254
matrix A to get the polynomial p A t . Recall if A = aij then
1i , j n
det(A) is the sum: det (A) = ()a1(1) .a2(2) a n(n) (*)
The sum is ranging over all the permutations
: ,n n being the signature of the permutation
.
To get the desired polynomial p A 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
1
det ( E 1 ) = .
det( E )
Let A = aij be any n x n matrix. For any t R we consider the
n n matrix : A tI aij t ij
1i , j n
Clearly det (A – tI) = ai i ti i is a polynomial in t, its
S n
degree being n. We denote this polynomial by p A t . We prove that
p A 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 pB t = det (E (A – T I) E-1)
= det (E). det (A – t I). (det (E)-1
By the multiplicative property (II) quoted above
pB t = det (A – t I) = p A t
This completes the proof that if AB then PA(t) PB(t)
Definition 3 : The polynomial PA(t) is the Characteristic polynomial of
the matrix A.
ab
For example for a 2 x 2 matrix A we have :
cd
255
ab t
AtI
cd t
a-tb
cd-t
and therefore det (A – t I) = (a – t) (d – t) – bc
= t2 – (a + d) t + ad – bc.
= t2 – tr (A). t + det (A)
abc
Also, for a 3 x 3 matrix A d e f
g hi
a tbc
p A (t )det d e t f
ghit
= t3 – (a + e + i) t3 + ……0
14.3 CHARACTERISTIC POLYNOMIAL OF A
LINEAR TRANSFORMATION
We consider a linear transformation T : V → V. Let B{e1e2 , , en}
basis of the vector space.
Recall, the vector basis B associates with T a n x n matrix. Thus, if
n
T( e j )= aij e j for1 jn, then the coefficients aij determine the n x n
i 1
matrix A = [ aij ].
If C{ f1, f 2 ,, f n} is another vector basis of V then C associates (in a
similar way) another matrix B bij 1 i, j n with T, where
n
T f j = bij fi (1 jn)
i1
We claim that the matrices A and B are similar. For, let C[cij ]1i, jn
n
be the matrix given by f j ckj ek (1
jn)
k 1
Clearly C is an invertible n x n matrix. We prove : CBC 1 A = and
therefore AB
256
Towards this claim we apply T to the equation
n
f j ckj ek 1 jn
k1
n
To get, T ( f j ) ckj T (ek ).
k 1
n
NowT ( f j ) bkj f k
k 1
n n
bkj ck e
k 1 1
n n
( ck bkj )e
1 k 1
n n n n n
and ckj T (ek ) ckj ak e ( ak ckj )e
k1 k 1 1 1 k 1
Putting together these results, we get,
n n n n
( ck bkj )e ( ak ckj )e
1 k 1 1 k 1
For the range 1 jn . Comparing the coefficients of each e we get.,
n n
ck bkj ak ckj 1, jn
k 1 k 1
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 AB we
have
AtI CBC 1t
CBC-1CtIC 1
C(B-tI)C 1
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 p A t pB t
We use this fact to associate with a linear transformation T : V → V a
polynomial.
Choose a vector basis {e1e2 en} and let A be the matrix of T with
respect to the basis B. Then we consider p A 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
43
A and therefore the characteristic polynomial pT(t) of T is :
14
4 t3
pT (t )det = (4 – t)2 – 3 = t2 – 8t + 13.
14 t
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: AB . The Similarly AB 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 AB then for any A, B in M (n, )
258
III. If AB , C then AC A, B, C being arbitrary elements of
M (n1)
[A] denotes the subset of M (n, ) consisting of all B with AB . If
GL(n, ) denotes the subset of M(n, ) consisting of all invertible
elements, then [A] is given by [ A]{EAE 1 :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
A M(n, ), PA(t) is the characteristic polynomial of the matrix.
Characteristic polynomial has the property: If AB , 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 AB 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
103 310 500
A123 B140 C050
014 001 005
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 . E v V : v 0orv 0,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 av bw 0 , then there is nothing to prove. Therefore, we
assume av bw 0 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
av bw , 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 V 2 , with its usual (2-dimensional) real vector space
structure; Let T : 2 2 be given by T x, y (2 x, x 3 y);( x, y) 2.
Then the linear transformation T has, 2 and 3 as its eigenvalues. For, the
vector v 1, 1 satisfies T 1, 1 2 1, 1 while the vector w 0,1
satisfies T 0,1 3 0,1
Here is another example of a linear transformation not having any
(real) eigenvalues. Again we take V 2 and the linear transformation
S : 2 2 is given by S x, y y, x , for every x, y 2 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), a 0 under the assumption
v 0,0 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, a 2a and a 0 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 e1, e2 en of V and consider the
linear transformation T : V V
given by T ek kek , 1 k n That is if v v1e1 v2e2 vnen is any
element of V, then T v v1e1 2v2e2 kvk ek nvnen . Clearly, T
is a linear transformation having the distinct eigenvalues 1, 2, 3,
….n, the respective eigen-vectors being e1, e2 , e3 en .
In passing, notice that the matrix T of T with respect to the
chosen basis e1, e2 , en which consists of eigen-vectors is the diagonal
matrix:
1
2 0
T k
0
n
and consequently, the matrix (with respect to the same vector basis of V)
of any power T of T is :
1 1
2 2 0
T T
k
0
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 E E 0
Proof: Pick an arbitrary v E E
We prove v 0
Assuming the contrary i.e.v 0 , we get that v is an eigen vector
of T with the two distinct eigen values and .
T v v v
and therefore v v 0 i.e. v 0 with 0 which implies
v 0 a contraction. Therefore E E 0 .
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 E e1e2 , en of V. Now we
consider an eigenvalue of T and an eigen-vetor belonging to .
T v v
Suppose T aij is the matrix of T with respect to the chosen vector
n
basis E . (Thus we have T e j aij ei )
i 1
n
Therefore T v T x j e j x1 xn being the components of
j 1
v with respect to E.
n n
n n n
T(v) x jT e j x j aij ei aij x j ei
j 1 j 1 i 1 i 1 j l
n
Also, v xiei . Therefore the equation T v v gives;
i 1
263
n n n
aij x j ei xi ei
i 1 j 1 i 1
Equating the coefficients of ei 1 i n , we get ;
n
aij x j xi 1 i n
j 1
Thus the equation T v v gives rise to the system of simultaneous,
linear equations (in x1 xn involving the unknown quantity );
n
aij x j xi 1 i n
j 1
n
But we can write xi ij x j and therefore the above equations (*)
j 1
can be rewritten in the form :
n
aij ij x j 0 1 i n
j 1
Now this system has a “non trivial” solution x1 xn [non-trivial in the
sense that x1, x2 , , xn 0,0, 0 remembering v should not be the zero
vector] if and only if
det aij ij 0
Note that det aij ij is a polynomial in of degree n. It is the
characteristic polynomial pT of the linear transformation T.
Thus, we solve the polynomial equation pT 0 or equivalently
put det aij ij 0 to get all the eigen-values 12 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 :
a11x1 a12 x1 a1n xn i x1
a21x1 a22 x2 a2n xn i x2
an1x1 an2 x2 an1xn i xn
264
A non-zero solution of these equation will give an eigen vector
vi x1, x2 xn .
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 T : 2 2 given by
T x, y 2 x 3 y,x 4 y x, y .
2
Solution:
Clearly the matrix T of T with respect to the standard basis
2 3
e1 1,0 , e2 0,1 of 2 is : T 1
4
Therefore the characteristic polynomial pT of T is :
2 3
det 0
1 4
i.e. pT 2 4 3 0
pT 2 6 5 0
Its roots are 1 5, 1 . Thus, the eigen-values of T are 1 5, 2 1.
Therefore, the simultaneous linear equations determining the eigen-vector
v x, y belonging to the eigen-value 1 5 are
2 x 3 y 5x
x 4y 5y
It has v 11 as a non-zero solution.
Therefore v 11 is an eigen-vector of T belonging to the eigen-value 5
of T.
Next, we consider the simultaneous equations
2x 3 y x
x 4y 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 T : 3 3 be the linear transformation given by
T x, yz ( x,2 y z, 2 y 5z ) . Find the eigen-values and eigen-vectors.
Solution; The matrix of T with respect to the standard basis
e1 1,0,0 , e2 0,1,0 ,e3 0,0,1 of 3 is;
1 0 0
T 0 2 1
0 2 5
and therefore the characteristic polynomial pT of T is:
1 0 0
pT det 0 2 1 1 3 4
2 5
Therefore, the eigen-values are 1 1, 2 3,3 4 . Now, an eigen-
vector u x, y, z of T belonging to the eigen-value 1 1 satisfies the
equation : x 0 0 1x
0 2 y z 1y
0 2 y 5z 1z
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 u 1,0,0 .
Next, an eigen vector belonging to 2 3 is v x, y, z which must
satisfy the equation :
x 0 0 3x
0 2 y z 3y
0 2 y 5 z 3z
A solution of this system is an eigen vector v 0,1,1 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 : x 0 0 4x
0 2y z 4y
0 2 y 5z 4 z
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 E E 0 . 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 T 2 . More generally k is an eigen value of T k .
2) What way the eigen spaces E ,E 2 E 3 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) T : 2 2 ,T x, y 5x 2 y,2 x 2 y
(ii) T : 2 2 ,T x, y 4 x 6 y
(iii) T : 2 2 ,T xy 10 x 4 y,18
x 12 y
(iv) T : 2 2 ,T xy 3x 4 y,4 x 3 y
(v) T : 2 2T xy x 2 y,3 y
(vi) T : 3 3 ,T x yz 3x, 8 y,4 z
(vii) T : 3 3 ,T x, y,z 3x 5 y 3z,4 y 6 z, z
(viii) T : 3 3 ,T x, y, z 7 y,0, 2 z
267
y
(ix) T : 3 3 ,T x, y, z 2 x z, ,x 4 z
2
(x) T : 3 3 ,T x, y, z ax y,x ay z, y az
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