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TERM PAPER

MATHEMATICS

MTH-101

Topic:Prove the following statements and illustrate them with examples

of your own choice. Here, are the (not necessarily distinct) eigen

values of a given matrix A=

(i)Trace of A equals the sum of its eigen values

(ii) has the eigen values , , and the same eigen vectors

as A.

DOA:2.10.09

DOR:20.10.09

DOS:01.12.09

Submitted to: Submitted by:

Miss. Priyanka Rohit Kundliya

Deptt. Of Mathematics Reg.No: 10901525

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ROLL NO: M4901A15

CLASS: M4901 (ME)

ACKNOWLEDGEMENT

I would really like to thank the authors of the following books:

ENGINEERING MATHEMATICS-B.S.GREWAL

ENGINEERING MATHEMATICS-H.K.DASS

MATHEMATICS FOR STUDENTS-R.D.SHARMA

NCERT MATHS(VOL 1 AND 2)

UNIVERSITY MATHEMATICS

I would like to thank my mathematics teacher Miss.Priyanka for

constantly guiding me.

I would also like to thank the librarians for helping me to search

the books containing the topic of my term paper.

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1. INTRODUCTION

In mathematics, eigenvalue, eigenvector, and eigenspace are related concepts in thefield oflinear algebra. They are derived from the German word "eigen" which

means "proper" or "characteristic." An eigenvalue of a square matrix is a scalar that

is usually represented by the Greek letterl(pronounced lambda). As one might

suspect, an eigenvector is a vector. Moreover, it is required that an eigenvector be a

non-zero vector, in other words, an eigenvector cannot be the zero vector. An

eigenvector will be denoted by the small letterx. All eigenvalues and eigenvectors

satisfy the equation AX=lXfor a given square matrix, A. Eigenvalues are a special

set of scalars associated with a linear system of equations (i.e., a matrix equation)

that are sometimes also known as characteristic roots, characteristic values

(Hoffman and Kunze 1971), proper values, or latent roots (Marcus and Minc 1988,

p. 144). An eigenvector of a square matrix A is a nonzero vector K such that AK = lK for

some number called an eigenvalue of A. The vector K is called an eigenvector

corresponding to eigenvalue .To find the eigenvalues of A one has to find the roots

of det(A lI) = 0.To find the eigenvectors corresponding to eigenvalue, the linear

system (A lI)K = 0 is to be solved.

If Av=v with v nonzero, then is called an eigenvalue of A and v is called an

eigenvector of A corresponding to eigenvalue .

Agenda: Understanding the action of A by seeing how it acts on eigenvectors.

(A-I)v=0 system to be satisfied by and v

For a given , only solution is v=0 except if det(A-I)=0; in that case a nonzero

v that satisfies (A-I)v=0 is guaranteed to exist, and that v is an eigenvector.

det(A-I) is a polynomial of degree n with leading term (-)n . This polynomial

is called the characteristic polynomial of the matrix A.

det(A-I)=0 is called the characteristic equation of the matrix A

There will be n roots of the characteristic polynomial, these may or may not be

distinct.

For each eigenvalue the corresponding eigenvectors satisfy (A-I)v=0 ; the

(nonzero) solutions v are the eigenvectors, and these are the nonzero vectors in

the null space of (A-I). For each eigenvalue, we calculate a basis for the null

space of (A-I) and these represent the corresponding eigenvectors, with it

being understood that any linear combination of these vectors will produce

another eigenvector corresponding to that .

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http://en.wikipedia.org/wiki/Mathematicshttp://en.wikipedia.org/wiki/Linear_algebrahttp://mathworld.wolfram.com/LinearSystemofEquations.htmlhttp://mathworld.wolfram.com/MatrixEquation.htmlhttp://en.wikipedia.org/wiki/Mathematicshttp://en.wikipedia.org/wiki/Linear_algebrahttp://mathworld.wolfram.com/LinearSystemofEquations.htmlhttp://mathworld.wolfram.com/MatrixEquation.html

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If is a nonrepeated root of the characteristic polynomial, then the null space of

(A-I) has dimension one, so there is only one corresponding eigenvector (that

is, a basis of the null space has only one vector), which can be multiplied by any

nonzero scalar.

Observation: If i is an eigenvalue of A with vi a corresponding eigenvector,

then for any , (doesnt have to be an eigenvalue) we have (A-I)v i=(i-) vi.

1.TRACE OF A MATRIX A EQUALS THE SUM OF

ITS

EIGEN VALUES

If any square matrix of order n, can form the matrix A- lI, where I is the nthorder of unit matrix. The determinant of this matrix equated to zero, i.e.,

| A-lI | = 0, is called the characteristic equation of A. On expanding the

determinant, the characteristic equation takes the form:

(-1)n ln + k1ln 1 + k2l

n 2 + +kn = 0,

where ks are expressible in terms of the elements a ij.The roots of the equation

are called the eigen values or latent roots or characteristic roots of the matrix A.

The sum of the eigen values of a matrix is the sum of the elements of the

principal diagonal.

Consider the square matrix,

A=

so that |A lI | =

= - l3 + l2 ( a11+ a22 + a33) - .. -----(i)

Ifl1,l2,l3be the eigen values of A,then:

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|A lI | = (-1)3 (l-l1) (l-l2) (l-l3)

= - l3 + l2 ( l1+ l2 + l3) - .. -----(ii)

Equating the right hand side of the equations (i) and (ii) and comparing the

coefficients of l2:

l1+ l2 + l3 = a11+ a22 + a33

Examples of the fact that trace of a matrixA is equal to the sum of its

eigen values:

Lets consider a 3*3 matrix: A =

Now, | A-lI | = 0

Therefore, = 0

l3+7l2+36=0

(l+2) (l-3) (l-6) = 0

l= -2,3,6

Therefore, l1+ l2 + l3 = -2+3+6 = 7

Also, a11+ a22 + a33 = 1+5+1 = 7

Thus, l1+ l2 + l3 = a11+ a22 + a33

Lets consider a 3*3 upper triangular matrix: B =

Now, | B-lI | = 0

Therefore, = 0

(1-l) (3-l) (4-l) = 0

l= 1,3,4

Therefore, l1+ l2 + l3 = 1+3+4 = 8

Also, a11+ a22 + a33 = 1+3+4 = 8

Thus, l1+ l2 + l3 = a11+ a22 + a33In this case it can be seen that the eigen values of the triangular matrix are just the

diagonal elements of the matrix.

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Lets consider a 3*3 diagonal matrix: C =

Now, | C-lI | = 0

Therefore, = 0

(2-l) (3-l) (5-l) = 0

l= 2,3,5

Therefore, l1+ l2 + l3 = 2+3+5 = 10

Also, a11+ a22 + a33 = 2+3+5= 10

Thus, l1+ l2 + l3 = a11+ a22 + a33In this case also, it can be seen that the eigen values of the diagonal matrix are just the

diagonal elements of the matrix.

2. HAS THE EIGEN VALUES ,..

,AND THE SAME EIGEN VECTORS AS A.

Eigen vector : If X= and A=

Then the linear transformation Y=AX .(i)

carries the column vector X into the column vector Y by means of the square matrix A.Let

X be

such a vector which transforms into lX by means of the transformation(i).

Then, lX=AX

AX-lIX = 0

[A -lI ] X = 0 (ii)

This matrix equation represents n homogeneous linear equations

(a11-l)x1+ a12 x2+..+ a1n xn = 0

a11x1+ (a22-l) x2+..+ a2n xn = 0

an1x1+ an2 x2+..+ (ann-l)xn = 0

.(iii)

which will have a non trivial solution only if the coefficient matrix is singular,

i.e. if

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| A-lI | = 0

This is called the characteristic equation of the transformation and is same as the

characteristic equation of the matrix A. It has n roots and corresponding to each root,the

equation (ii) [or (iii)] will have a non-zero solution.

X=[x1,x2,. Xn] , which is known as the eigen vector or latent root.

Example of the fact a matrix [A-KI] has the eigen values ,..

,and the same eigen vectors as A ,where are the eigen

values of matrix A and K is any real number :

Lets consider a 2*2 matrix: A =

Now, | A-lI | = 0

Therefore, = 0

L2+7l+6=0

(l-6) (l-1) = 0

l= 6,1

Let x,y,z be the components of X for the eigen value l=6;

| A - 6I | X = 0

= 0

= 0

which gives only one independent equation: x+4y = 0

Therefore, x/4 =y/1 giving the eigen vector (4,1)

Similarly,let x,y,z be the components of X for the eigen value l=1;

| A - I | X = 0

= 0

= 0

which gives only one independent equation: x+y = 0

Therefore, x/1 =y/(-1) giving the eigen vector (1,-1)

Now let, B = A- KI,where K is any real number.Let K be 1.

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Thus, B = -

B =

Now, | B-lI | = 0

= 0

(4-l) (1-l) - 4 = 0

4 - 4l-l+l2-4=0

l= 0,5Therefore, l1= 1-1=0

l2= 6-1=5

Let x,y,z be the components of X for the eigen value l=5;

| B - 5I | X = 0

= 0

= 0

which gives only one independent equation: x+4y = 0

Therefore, x/4 =y/1 giving the eigen vector (4,1)

Similarly,let x,y,z be the components of X for the eigen value l=0;

| B - 0I | X = 0

= 0

= 0

which gives only one independent equation: x+y = 0

Therefore, x/1 =y/(-1) giving the eigen vector (1,-1)

Thus, has the eigen values ,.. ,and the same eigen vectors as

A.

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