# A General Method using Geometry to Find Eigen Vectors and Eigen Values of Matrix of Size 3×3 Text Only Version

#### A General Method using Geometry to Find Eigen Vectors and Eigen Values of Matrix of Size 3×3

Rania. B. M. Amer1, M. Afwat2

1,2Department of Engineering Mathematics and Physics, Faculty of Engineering, Zagazig University,

P.O. 44519, Zagazig, Egypt

Abstract:- In this paper, the general model symmetric matrix 3×3 is expressed. Furthermore, the general model of non symmetric matrix 3×3 is discussed. The examples are presented to verify the results.

MSC: 05C50 , 15Axx, 15A18

INTRODUCTION

Eigen values are greatest importance in dynamic problems(Luenberger 1979), (Johansen 1988),(Haftka and Adelman 1989) and many engineering application (Thomson 1996). The eigenvectors (Joy 2000) denoted by and eigenvalues of a any matrix A that is satisfied Ax = x. If A is an n x n matrix, then is an n x 1 vector, and is a constant.

The matrix A has Eigen vectors and Eigen values are written as:

1 1 0

1 1

= ( ) , = ( )

1 0

The matrix V is called the modal matrix of A. Since D, as a diagonal matrix, has Eigen values 1, . . , which are the same as those of A then the matrices D and A are said to be similar. The transformation of A into D using V 1A V = D is said to be a similarity transformation.

TO FIND THE MODEL MATRIX AND THE EIGEN VALUES OF A SYMMETRIC MATRIX:

[ ]

We express the given matrix to demonstrate a surface of second degree

[ ] [ ] []

Then

2 + 2 + 2 + 2 + 2 + 2 + = 0 (1)

With center(0,0,0), i.e = = = 0

We take a function: ( )2 + ( )2 + ( )2 and we find its extreme points on the given surface, using Lagrange multiplier . Then

(, , ) = ( 0)2 + ( 0)2 + ( 0)2 + (2 + 2 + 2 + 2 + 2 + 2 + )

We find:

= 0, = 0, = 0 , = 0, then:

=

=

=

+ +

We can find:

+ +

=

+ +

2 + 2 + 2 + 2 + 2 + 2 + = 0

(2 2) + ( ) + = 0, (2)

(2 2) + ( ) + = 0, (3)

(2 2) + ( ) + = 0, (4)

From equation (4), we get:

=

(22)+()

(5)

Multiply equation (2) by , Multiply equation (3) by , Multiply equation (4) by and adding, then

[ 2 + 2] + [2 + 2] +

[2 + 2 + ] = 0 (6)

From equation (6), we get:

[22+]

= (7)

[2+2]+[2+2]

From equations (5) and (7):

(2 2) + ( )

[2 2 + ]

=

[ 2 + 2] + [2 + 2]

The cubic equation in z is:

[(2 + 2)] + [( 2 + 2) + ( )(2 + 2) (2 2

+ )]2 + [(2 + 2) + ( )( 2 + 2) + (2 2 + )]2 + [( 2 + 2)]3 = 0 (8)

0 = (2 + 2)

1 = [( 2 + 2) + ( )(2 + 2) (2 2 + )]

2 = [(2 + 2) + ( )( 2 + 2) + (2 2 + )]2

3 = [( 2 + 2)]3

=

Then the cubic equation (8) can be written in its reduced form:

1

30

3 + ( + ) + (

+ + 2 2 ) = 0 (9)

0 2 1

For Cardan's method:

3 2 3 1

+

+ 2

+ 3

+

+ 2 2

= 3

2 1

20

0 = 3

2 3 1,

20

=

2 + 21 + 320

=

30

2 + 1

30

= [2 + (3 + 3)],

=

By the following relation

< 0, 0 = 180 arccos() = arccos(),

> 0, 0 = arccos(),

0

1 = 3 ,

= 23

1 = 1 + ,

2 = (1 + 120) + ,

3 = (1 + 240) + .

(2 2) + ( )

To find 1, 2 , 3 1, , 3

1 2 3

The model matrix is [1 2 3]

1 2 3

To find the Eigen values:

= ; = 1,2,3

1 2 3 0

[ ] [1 2 3] = 

1 2 3 0

Then

1 =

2 =

1 + 1 + 1

,

1

2 + 2 + 2

,

2

Example (1):

3 =

3 + 3 + 3

3

11 6 2

Find the model matrix and Eigen values of the matrix[6 10 4]

2 4 6

Solution:

Then

= 11, = 10, = 6, = 6, = 2, = 4 0,

A0 = 320, A1 = 480y, 2 = 4802, A3 = 320y3 , k = 0.5y,

+

+

320y3+0.54803 2(0.5)24803

= 3 = 0,

2320

=

2 + 21 + 320

30

4802 + 2 0.5 4802 + 3 (0.5)2 3202

=

3 320

= 0.75002

= 0.6495 3, = 0,

= cos = ,

= 0 = , = 23 = 1.7321

Then

0 2 1 3 6

1 = 1, 2 = 2 2, 3 = 0.5 3 ,

(2 2) + ( )1 1

1 =

1 1

1 1

= 0.5 1,

(2 2) + ( )22

2 =

2 2

2 2

= 22,

(2 2) + ( )33

Then the model matrix is:

3 =

3 3

3 3

= 3,

1 2 3

0.5 2 1

[1 2 3] = [ 1 1 1 ]

1 2 3

1 2 0.5

To check the results:

– By using Maple program

>

– By using Matlab program

1 =

2 =

3 =

1 + 1 + 1

1

2 + 2 + 2

2

3 + 3 + 3

3

= 3,

= 6,

= 18

>> A=[11 -6 2;-6 10 -4;2 -4 6]

>> [V,D] = eig(A)

V = 0.3333 -0.6667 0.6667

0.6667 -0.3333 -0.6667

0.6667 0.6667 0.3333

D =

3.0000 0 0

0 6.0000 0

0 0 18.0000

TO FIND THE MODEL MATRIX AND THE EIGEN VALUES OF A GIVEN NON SYMMETRIC MATRIX:

= [ ]

We express the given matrix to demonstrate a surface of second degree

= [ ] [ ] []

Then

= 2 + 2 + 2 + ( + ) + ( + ) + ( + ) + = 0 (10) With center(0,0,0), i.e = = = 0

We take a function: ( )2 + ( )2 + ( )2 and we find its extreme points on the given surface, using Lagrange multiplier . Then

(, , ) = 2 + 2 + 2 + (2 + 2 + 2 + ( + ) + ( + ) + ( + ) + )

We find:

= 0, = 0, = 0 , = 0, then:

2

2

= = = 2 + ( + ) + ( + ) ( + ) + 2 + ( + )

We can find:

2

= ( + ) + ( + ) + 2

2 + 2 + 2 + ( + ) + ( + ) + ( + ) + = 0

From equation (4), we get:

( + )(2 2) + 2( ) + ( + ) ( + ) = 0, (2) ( + )(2 2) + 2( ) + ( + ) ( + ) = 0, (3) ( + )(2 2) + 2( ) + ( + ) ( + ) = 0, (4)

(+)(22)+2()

=

(+)(+)

(5)

Multiply equation (2) by ( + )( + ), multiply equation (3) by ( + )( + ), multiply equation (4) by ( + )( +

) and adding, then

[2( )( + )( + ) ( + )( + )2 + ( + )( + )2] +

[( + )( + )2 2( )( + )( + ) ( + )( + )2] + [( + )( + )2 + ( + )( + )2 + 2( )( + )( + )] = 0

From equation (6), we get:

[( + )( + )2 ( + )( + )2 2( )( + )( + )]

(6)

=

[2( )( + )( + ) ( + )( + )2 + ( + )( + )2] + [( + )( + )2 2( )( + )( + ) ( + )( + )2]

From equations (5) and (6):

(+)(22)+2() = [(+)(+)2(+)(+)22()(+)(+)]

(+)(+)

Then

[2()(+)(+)(+)(+)2+(+)(+)2]+ [(+)(+)22()(+)(+)(+)(+)2]

0 = ( + )[( + )( + )2 2( )( + )( + ) ( + )( + )2]

1 = [( + )[2( )( + )( + ) ( + )( + )2 + ( + )( + )2]

+ 2( )[( + )( + )2 2( )( + )( + ) ( + )( + )2 ]

( + )[( + )( + )2 ( + )( + )2 2( )( + )( + )]]

2 = [( + )[( + )( + )2 2( )( + )( + ) ( + )( + )2]

+ 2( )[2( )( + )( + ) ( + )( + )2 + ( + )( + )2]

+ ( + )[( + )( + )2 ( + )( + )2 2( )( + )( + )]]2

3 = [( + )[2( )( + )( + ) ( + )( + )2 + ( + )( + )2]]3

=

Then the cubic equation (8) can be written in its reduced form:

1 30

3 + ( + ) + (

+ + 2 2 ) = 0 (9)

0 1 2

3 2 3 1

Using Cardan's method we find the three roots of equation (9) , ,

We obtain the three roots of equation (8):

1 2 3

1 = + , 2 = + , 3 = + ,

By the following relation

1 2 3

To find 1, 2 , 3 1, , 3

1 2 3

( + )(2 2) + 2( )

= ; = 1,2,3

( + ) ( + )

The model matrix is = [1 2 3]

1 2 3

 1 2 3 [ ( + )/2 ( + )/2] [11 2 3] = [0 02 3 0]
 1 2 3 [ ( + )/2 ( + )/2] [11 2 3] = [0 02 3 0]

To find the Eigen values:

+ + ) + ( + )

+ + ) + ( + )

(

(

=

=

2

2

2

2

, = 1,2,3

, = 1,2,3

Then

Example:

Find the model matrix and the Eigen values of non symmetric matrix

11 4 1

= [8 10 6]

3 2 6

= 11, = 10, = 6, = 4, = 1, = 8, = 6, = 3, = 2

Since

0 = 5120

1 = 7680

2 = 7680 2

3 = 51203

1

=

30

= 0.5000

+

+ 2

+ 3

+

+ 2 2

= 3

2 1

20

0 = 3

2 3

20

1

= 0 ,

=

2 + 21 + 320

30

2 + 1

=

30

= 0.7500 y2

Then the three roots are real

= [2 + (2 + 3)] = 0.6495 3

= 2 + 3 = 0.4219 < 0

= = 0

< 0, 0 = 180 arccos() = arccos(),

> 0, 0 = arccos() = 90 ,

0

1 = 3 = 30,

= 23 = 1.7321

1 = 1 + = 1 ,

2 = (1 + 120) + = 2 2 ,

3 = (1 + 240) + = 0.5 3 .

( + )(2 2) + 2( )11

= 1 1

= 0.5

Then the model matrix is:

1 ( + )1 ( + )1 1

2 = 2y2,

3 = 3

1 2 3

0.5 2 1

= [1 2 3] = [ 1 1 1 ],

+ + ) + ( + )

+ + ) + ( + )

1 2 3

So

1 2 0.5

=

(

2

2

=

(

2

2

, = 1,2,3

, = 1,2,3

1 = 3,

2 = 6,

3 = 18

CONCLUSION:

The examples is discussed in the above sections provide us a way to generalize the model matrix for non symmetric matrix. The

a h g

non symmetric matrix [ ] is equivelant to the symmetric matrix [h b f ],

.

.

Where h = f+d , g = e+g , f = k+h

g f k

2 2 2

REFERENCES

1. Haftka, R. T. and H. M. Adelman (1989). "Recent developments in structural sensitivity analysis." Structural optimization 1(3): 137-151.

2. Johansen, S. (1988). "Statistical analysis of cointegration vectors." Journal of economic dynamics and control 12(2-3): 231-254.

3. Joy, K. (2000). "On-Line Geometric Modeling Notes. Eigenvalues and eigenvectors." Computer Science Department, University of California, Davis.

4. Luenberger, D. G. (1979). Introduction to dynamic systems: theory, models, and applications, Wiley New York.

5. Thomson, W. (1996). Theory of vibration with applications, CRC Press.