Unit-2

Eigen values and Eigen vectors

Linear transformation-

Suppose that ‘A’ is a m× n matrix. Then a function f(x) = Ax is a linear transform of matrix A. Where f:.

For examples,

Suppose we have a 2×3 matrix as below,

A = , now if we multiply this matrix with vector X = (x,y,z)

Then,

AX = = (x – z , )

Example: suppose you have a matrix A = , then find the linear transformation of A.

Sol. Here we have,

A =

Multiply matrix A by vector (x,y).

X = (x,y)

Ax =

We get, f(x,y) = (2x + y , y , x – 3y)

Which is the linear transformation of matrix A.

Example: find the linear transformation of the matrix A.

A =

Sol. We have,

A =

Multiply the matrix by vector x = (x , y , z) , we get

Ax =

= ( )

f(x , y , z) = ( )

Which is the linear transformation of A.

Orthogonal transformation-

Let us suppose be a vector space of size -2 column vector. This vector space has an inner product defined by < v , w> = linear transformation T: is called orthogonal transformation for every for all v,w belongs to ,

< T(v) , T(w)> = <v , w>

First we will go through some important definitions before studying Eigen values and Eigen vectors.

1. Vector-

An ordered n – touple of numbers is called an n – vector. Thus the ‘n’ numbers x1, x2, ………… xn taken in order denote the vector x. i.e. x = (x1, x2, ……., xn).

Where the numbers x1, x2, ……….., xn are called component or co – ordinates of a vector x. A vector may be written as row vector or a column vector.

If A be an mxn matrix then each row will be an n – vector & each column will be an m – vector.

2. Linear dependence-

A set of n – vectors. x1, x2, …….., xr is said to be linearly dependent if there exist scalars. k1, k2, ……., kr not all zero such that

k1 + x2k2 + …………….. + xr kr = 0 … (1)

3. Linear independence-

A set of r vectors x1, x2, …………., xr is said to be linearly independent if there exist scalars k1, k2, …………, kr all zero such that

x1 k1 + x2 k2 + …….. + xr kr = 0

Important notes-

- Equation (1) is known as a vector equation.
- If the vector equation has non – zero solution i.e. k1, k2, …., kr not all zero. Then the vector x1, x2, ………. xr are said to be linearly dependent.
- If the vector equation has only trivial solution i.e.

k1 = k2 = …….= kr = 0. Then the vector x1, x2, ……, xr are said to linearly independent.

4. Linear combination-

A vector x can be written in the form.

x = x1 k1 + x2 k2 + ……….+xr kr

Where k1, k2, ………….., kr are scalars, then X is called linear combination of x1, x2, ……, xr.

Results:

- A set of two or more vectors are said to be linearly dependent if at least one vector can be written as a linear combination of the other vectors.
- A set of two or more vector are said to be linearly independent then no vector can be expressed as linear combination of the other vectors.

Example 1

Are the vectors , , linearly dependent. If so, express x1 as a linear combination of the others.

Solution:

Consider a vector equation,

i.e.

Which can be written in matrix form as,

Here & no. Of unknown 3. Hence the system has infinite solutions. Now rewrite the questions as,

Put

and

Thus

i.e.

i.e.

Since F11 k2, k3 not all zero. Hence are linearly dependent.

Example 2

Examine whether the following vectors are linearly independent or not.

and .

Solution:

Consider the vector equation,

i.e. … (1)

Which can be written in matrix form as,

R12

R2 – 3R1, R3 – R1

R3 + R2

Here Rank of coefficient matrix is equal to the no. Of unknowns. i.e. r = n = 3.

Hence the system has unique trivial solution.

i.e.

i.e. vector equation (1) has only trivial solution. Hence the given vectors x1, x2, x3 are linearly independent.

Example 3

At what value of P the following vectors are linearly independent.

Solution:

Consider the vector equation.

i.e.

This is a homogeneous system of three equations in 3 unknowns and has a unique trivial solution.

If and only if Determinant of coefficient matrix is non zero.

consider .

.

i.e.

Thus for the system has only trivial solution and Hence the vectors are linearly independent.

Note:-

If the rank of the coefficient matrix is r, it contains r linearly independent variables & the remaining vectors (if any) can be expressed as linear combination of these vectors.

Characteristic equation:-

Let A he a square matrix, be any scaler then is called characteristic equation of a matrix A.

Note:

Let a be a square matrix and ‘’ be any scaler then,

1) is called characteristic matrix

2) is called characteristic polynomial.

The roots of a characteristic equations are known as characteristic root or latent roots, eigen values or proper values of a matrix A.

Eigen vector:-

Suppose be an eigen value of a matrix A. Then a non – zero vector x1 such that.

… (1)

Such a vector ‘x1’ is called as eigen vector corresponding to the eigen value .

Properties of Eigen values:-

- Then sum of the eigen values of a matrix A is equal to sum of the diagonal elements of a matrix A.
- The product of all eigen values of a matrix A is equal to the value of the determinant.
- If are n eigen values of square matrix A then are m eigen values of a matrix A-1.
- The eigen values of a symmetric matrix are all real.
- If all eigen values are non – zen then A-1 exist and conversely.
- The eigen values of A and A’ are same.

Properties of eigen vector:-

- Eigen vector corresponding to distinct eigen values are linearly independent.
- If two are more eigen values are identical then the corresponding eigen vectors may or may not be linearly independent.
- The eigen vectors corresponding to distinct eigen values of a real symmetric matrix are orthogonal.

Example-1:

Determine the eigen values of eigen vector of the matrix.

Solution:

Consider the characteristic equation as,

i.e.

i.e.

i.e.

Which is the required characteristic equation.

are the required eigen values.

Now consider the equation

… (1)

Case I:

If Equation (1) becomes

R1 + R2

Thus

independent variable.

Now rewrite equation as,

Put x3 = t

&

Thus .

Is the eigen vector corresponding to .

Case II:

If equation (1) becomes,

Here

independent variables

Now rewrite the equations as,

Put

&

.

Is the eigen vector corresponding to .

Case III:

If equation (1) becomes,

Here rank of

independent variable.

Now rewrite the equations as,

Put

Thus .

Is the eigen vector for .

Example 2

Find the eigen values of eigen vector for the matrix.

Solution:

Consider the characteristic equation as

i.e.

i.e.

are the required eigen values.

Now consider the equation

… (1)

Case I:

Equation (1) becomes,

Thus and n = 3

3 – 2 = 1 independent variables.

Now rewrite the equations as,

Put

,

i.e. the eigen vector for

Case II:

If equation (1) becomes,

Thus

Independent variables.

Now rewrite the equations as,

Put

Is the eigen vector for

Now

Case II:-

If equation (1) gives,

R1 – R2

Thus

independent variables

Now

Put

Thus

Is the eigen vector for

Two square matrix and A of same order n are said to be similar if and only if

for some non-singular matrix P.

Such transformation of the matrix A into with the help of non singular matrix P is known as similarity transformation.

Similar matrices have the same Eigen values.

If X is an Eigen vector of matrix A then is Eigen vector of the matrix

Reduction to Diagonal Form:

Let A be a square matrix of order n has n linearly independent Eigen vectors which form the matrix P such that

Where P is called the modal matrix and D is known as spectral matrix.

Procedure: let A be a square matrix of order 3.

Let three Eigen vectors of A are corresponding to Eigen values

Let

{by characteristics equation of A}

Or

Or

Note: The method of diagonalization is helpful in calculating power of a matrix.

.Then for an integer n we have

We are using the example of 1.6*

Example1: Diagonalize the matrix

Let A=

The three Eigen vectors obtained are (-1,1,0), (-1,0,1) and (3,3,3) corresponding to Eigen values .

Then and

Also we know that

Example2: Diagonalize the matrix

Let A =

The Eigen vectors are (4,1),(1,-1) corresponding to Eigen values .

Then and also

Also we know that

Statement-

Every square matrix satisfies its characteristic equation, that means for every square matrix of order n,

|A - | =

Then the matrix equation-

Is satisfied by X = A

That means

Example-1: Find the characteristic equation of the matrix A = and Verify cayley-Hamlton theorem.

Sol. Characteristic equation of the matrix, we can be find as follows-

Which is,

( 2 - , which gives

According to Cayley-Hamilton theorem,

…………(1)

Now we will verify equation (1),

Put the required values in equation (1) , we get

Hence the cayley-Hamilton theorem is verified.

Example-2: Find the characteristic equation of the the matrix A and verify Cayley-Hamilton theorem as well.

A =

Sol. Characteristic equation will be-

= 0

( 7 -

(7-

(7-

Which gives,

Or

According to cayley-Hamilton theorem,

…………………….(1)

In order to verify cayley-Hamilton theorem , we will find the values of

So that,

Now

Put these values in equation(1), we get

= 0

Hence the cayley-hamilton theorem is verified.

Inverse of a matrix by Cayley-Hamilton theorem-

We can find the inverse of any matrix by multiplying the characteristic equation with .

For example , suppose we have a characteristic equation then multiply this by , then it becomes

Then we can find by solving the above equation.

Example-1: Find the inverse of matrix A by using Cayley-Hamilton theorem.

A =

Sol. The characteristic equation will be,

|A - | = 0

Which gives,

(4-

According to Cayley-Hamilton theorem,

Multiplying by

That means

On solving ,

11

=

=

So that,

Example-2: Find the inverse of matrix A by using Cayley-Hamilton theorem.

A =

Sol. The characteristic equation will be,

|A - | = 0

=

= (2-

= (2 -

=

That is,

Or

We know that by Cayley-Hamilton theorem,

…………………….(1)t,

Multiply equation(1) by , we get

Or

Now we will find

=

=

Hence the inverse of matrix A is,

Power of a matrix by Cayley-Hamilton theorem-

Any positive integral power of matrix A is linearly expressible in the terms of those of lower degree, where m is the positive integer and n is the degree of characteristic equation, such that m>n

Example-1: Find of matrix A by using Cayley-Hamilton theorem.

Sol. First we will find out the characteristic equation of matrix A,

|A - | = 0

We get,

Which gives,

(

We get,

Or I ……………………..(1)

In order to find find we take cube of eq. (1)

We get,

729I we know that-

729 we know that- value of I =

Example-2: Find of matrix A by using Cayley-Hamilton theorem.

A =

Sol. Here we have ,

A =

Characteristics equation will be,

|A - | = 0

On factorization , we get

()()() = 0

Hence eigen values are – 1,2,3

Suppose,

………………….(1)

Now put the values of in equation (1), we get

Put

a+b+c = 1……………..(2)

4a+2b+c = 16……………(3)

9a+3b+c = 81…………….(4)

On solving these three eq. , we get

a = 25 , b = -60 , c = 36

Replace by A in eq.(1), we get

= O +

Put the corresponding values,

= 25

=

=

=

Which is the recuired answer.

Quadratic form can be expressed as a product of matrices.

Q(x) = X’ AX,

Where,

X’ is the transpose of X,

Which is the quadratic form.

Example: find out the quadratic form of following matrix.

A =

Solution: Quadratic form is,

X’ AX

Which is the quadratic form of a matrix.

Example: find the real matrix of the following quadratic form:

Sol. Here we will compare the coefficients with the standard quadratic equation,

We get,

A real symmetric matrix ‘A’ can be reduced to a diagonal form M’AM = D ………(1)

Where M is the normalized orthogonal modal matrix of A and D is its spectoral matrix.

Suppose the orthogonal transformation is-

X =MY

Q = X’AX = (MY)’ A (MY) = (Y’M’) A (MY) = Y’(M’AM) Y

= Y’DY

= Y’ diagonal ()Y

Which is called canonical form.

Step by step method for Reduction of quadratic form to canonical form by orthogonal transformation –

1. Construct the symmetric matrix A associated to the given quadratic form .

2. Now for the characteristic equation-

|A - | = 0

Then find the eigen values of A. Let be the positive eigen values arranged in decreasing order , that means ,

3. An orthogonal canonical reduction of the given quadratic form is-

4. Obtain an order system of n orthonormal vector consisting of eigen vectors corresponding to the eigen values .

5. Construct the orthogonal matrix P whose columns are the eigen vectors

6. The required change of basis is given by X = PY

7. The new basis { } is called the canonical basis and its elements are principal axes of the given quadratic form.

Example: Find the orthogonal canonical form of the quadratic form.

5

Sol. The matrix form of this quadratic equation can be written as,

A =

We can find the eigen values of A as –

|A - | = 0

= 0

Which gives,

The required orthogonal canonical reduction will be,

8 .