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3. Eigenvalues and Eigenvectors.

3.1. General

Many dynamic physical systems can bemodeled by systems of linear differentialequations. The dynamic nature of the systemarises from energy storage elements (rotatingmasses, springs, capacitors, pressurizedcontainers) within the system.

Eigenvalue or characteristic value, problems

are special class of boundary-value problems thatare common in engineering problems such asvibration, elasticity and other oscillating systems

consisting of the above elements.

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Consider a set of linear algebraic equationsof the form

[A] {X} = {C} (1)

The above system of equations are callednon-homogenous (right hand side vector {C}is present). They will have a unique solution.

In contrast,

[A] {X} = 0 (2)

This is a homogeneous equation. Thesolutions are generally unique. Thesimultaneous equations establish relationshipamong the xs that can be satisfied by variouscombinations of values.

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Homogenous equations of the following typeare often encountered in certain kinds ofphysical problems.

a11 x1 + a12 x2 + a13 x3 + .+ a1n xn = x1(3a)

a21 x1 + a22 x2 + a23 x3 + .+ a2n xn = x2(3b)

an1 x1 + an2 x2 + an3 x3 + + ann xn = xn(3n)

Where is undetermined parameter.

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The x values of for which non-zero roots

of the homogenous equations (3) exist, (i.e.),

for which determinant = 0 are called theeigenvalues or the characteristic values ofparameters . When the n eigenvalues are

distinct, to each eigenvalue therecorresponds a set of n roots, called aneigenvector.

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Steps in the determination of eigenvalues andeigenvectors:

1.For a given matrix, say A, form an equation.

A x = x (i.e.) (A I) = 0.

2.Set the determinant of the coefficient matrix(A I ) = 0 and obtain polynomial equation in .

This equation is called characteristic polynomial.3.Determine the roots of this equation. They give

the eigenvalues. They equal the size of the

matrix.4.Substitute these eigenvalues one by one in the

equation ( A I) x = 0 and obtain the associatedeigenvectors.

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Example: 3.1

Determine the eigenvalues and the

corresponding eigenvectors.

Solution:

1 1

2 2

3 3

10 2 1 x x

2 10 1 x = x

2 1 10 x x

10 - 2 1= 2 10 - 1

2 1 10 -

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( )

10 - 1 2 1 2 10 - = 1 0 - - 2 + 1

1 10 - 2 10 - 2 12=(10-){(10 - ) - 1} - 2{2(10 - ) - 2} + 1{2 - 2(10 - )} = 0

3

=(10-) - 7(10 - ) + 6 = 0

-3

(10 -) = 1 ;

2

1 = 13;

2 = 9;

3 = 8;

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Substituting 1 = 13,

3 x1 + 2 x2 + x3 = 0

2 x1 3x2 + x3 = 0

2 x1 + x23 x3 = 0

Solutions of the first two equations for x1, x2with x3 = 1, gives the eigenvector.

(1)1x = 1

(1)2x = 1 (1)3x = 1

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When 2 = 9,

x1 + 2 x2 + x3 = 0

2 x1 + x2 + x3 = 0

2 x1 + x2 + x3 = 0

With , the second eigenvector becomes(2)

31x =

(2)

11x = - ; 3

(2)

2

1x = - ; 3

(2)

3x = 1;

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When 3 = 8,

2x1 + 2 x2 + x3 = 0

2 x1 + 2 x2 + x3 = 02 x1 + x2 + 2 x3 = 0

With , the third eigenvector becomes(3)

3x = 1

;

(3)

1

3

x = -2

(3)

2x = 1; (3)

3x = 1;

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Physical Background:

A two mass system is shown in Fig. Assumeeach spring has an initial length l and spring

constant k. The masses are displace by x1, x2from initial position as shown in the figure.

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Using the Newtons second law,

(1)

(2)The above equation can be expressed as,

(3)

(4)

2

11 1 2 12

d x

m = -kx + k(x - x )dt2

222 2 12

d xm = -k(x - x ) - kx

dt

11 1 2m x - k(-2x + x ) = 0

2 2 1 2 m x - k(x - 2x ) = 0

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The solution of Eqn.(3) can take the form

(5)

Where,

Ai in amplitude of vibrating mass i.

is frequency of vibration = 2 / t

(6)

Substituting in Eqn.(3) and (4)

(7)

(8)

i ix = A sin(t)

2

i ix = -A sin(t)

21 2

1 1

2k k- A - A = 0 m m

2

1 12 2

k 2k- A + - A = 0

m m

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Example: 3.2

Evaluate the eigenvalues and eigenvectors where

m1 = m2 = 40 kg and k = 200 N/mSolution:

Substituting the values in Eqn. (7) and (8)

(10 2) A1 5 A2= 0 5 A1 + (10-

2) A2= 0

The determinant of this system is

( )2

2 2 - 20 + 75 = 0

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Solving 2 = 15 and 5

1 = 3.873 / s

2 = 2.236 / s

These values can be used to determine the periodsof vibration.

For the first mode t1 = 1.62 s

Second mode t2 = 2.81 sUnique set of values cannot be obtained forunknowns A1 and A2 .

Their ratios can be obtained by substituting 1 and

2.For the first mode A1 = A2

Second mode A1 = A2The mode shapes are shown in the next .

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Mode shapes

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3.2 Leverrier-Faddeev method of generating acharacteristic polynomial:

Leverrier-Faddeev method is an efficientmethod to generate the characteristicpolynomial based on the traces as explainedbelow.

(1) Make matrix A1 = A

1 2 1

A = 2 3 2

1 2 4

1

1 2 1

A = 2 3 2

1 2 4

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(2) Calculate the trace T1 of this matrix (Traceis defined as the sum of the diagonal

elements).T1 = 1 + 3 + 4 = 8

(3) Obtain matrix A by subtracting T1 = 8 from

diagonal elements. Calculate A2 = AA

(4) Calculate the trace of A2

and let

T2 = (1/2) (trace of A2) = (1/2) (2711) =10

2

1 2 1 7 2 1 2 6 1

2 3 2 2 5 2 6 7 0

1 2 4 1 2 4 1 0 1

A

= =

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(5) Obtain matrix A by subtracting T2 =10from the diagonal elements of A2 andcalculate A2 and calculate A3 = AA.

(6) Finally obtain T3 = (1/3) (trace of A3)

= (1/3) (333) = 3

(7) The characteristic polynomial is given by

(8) Find the roots of this equation to obtain

3

1 2 1 8 6 1 3 0 0

2 3 2 6 3 0 0 3 0

1 2 4 1 0 1 0 0 3

A

= =

3 21 2 3T T T 3 28 10 3 + +

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3.3. Power Method:

The power method is an iterative approach

that can be employed to determine the largesteigenvalue. With a minor modification, it can alsobe employed to determine the smallest and theintermediate values. It has the additional benefit

that the corresponding eigenvector is obtained asa by-product of the method.

(a) Determination of the Largest Eigenvalue.

In order to implement the power method, thesystem being analyzed must be expressed in theform

[ ]{ } { }A X = X

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Example 3.3: The system is given by

Then, assuming the xs on the left-hand side

of the equation are equal to 1

1 2 1

1 2 3 2

2 3 3

+3.556x -1.778x = x

-1.778x +3.556x -1.778x =x-1.778x +3.556x =x

+3.556(1) -1.778(1) = 1.778

-1.778(1) +3.556(1) -1.778(1) = 0.000

-1.778(1) +3.556(1) = 1.778

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Next, the right-hand side is normalized by1.778 to make the largest element equal to 1.

Thus, the first estimate of the eigenvalue is1.778. This iteration can be expressedconcisely in matrix form as

1.778 10 = 1.778 0

1.778 1

3.556 -1.778 0 1 1.778 1

-1.778 3.556 -1.778 1 = 0 = 1.778 0

0 -1.778 3.556 1 1.778 1

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The next iteration consists of multiplying [A] by

to give

Therefore, the eigenvalue estimate for thesecond iteration is 3.556 which can beemployed to determine the error estimate

The process can then be repeated.

[ ]T

1 0 1

3.556 -1.778 0 1 3.556 1

-1.778 3.556 -1.778 0 = -3.556 = 3.556 -1

0 -1.778 3.556 1 3.556 1

0 00 0

3.556 - 1.778= 100 = 50

3.556a

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Third iteration:

Where (which is high because of thesign change).

Fourth iteration:

Where (again inflated because of thesign change).

3.556 -1.778 0 1 5.334 -0.75

-1.778 3.556 -1.778 -1 = -7.112 = -7.112 1

0 -1.778 3.556 1 5.334 -0.75

a =150%

3.556 -1.778 0 -0.75 -4.445 -0.714

-1.778 3.556 -1.778 1 = 6.223 = 6.223 1

0 -1.778 3.556 -0.75 -4.445 -0.714

a =214%

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Fifth iteration:

Where (which is high because of the sign change).

Sixth iteration:

Where (again inflated because of the sign change).

Thus the normalizing factor is converging on the value of 6.074.

Hence 6.074 is eigenvalue and [0.707 1 0.707]T is

eigenvector.

a =2.1%

a =0.35%

3.556 -1.778 0 -0.714 -4.317 -0.708

-1.778 3.556 -1.778 1 = 6.095 = 6.095 1

0 -1.778 3.556 -0.714 -4.317 -0.708

3.556 -1.778 0 -0.708 -4.296 -0.707

-1.778 3.556 -1.778 1 = 6.074 = 6.074 1

0 -1.778 3.556 -0.708 -4.296 -0.707

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(b) Determination of the smallest Eigenvalue

There are often cases in engineering where we areinterested in determining the smallest eigenvalue. Thiscan be done by applying the power method to thematrix inverse of [A]. For this case, the power methodwill converge on the largest value of 1 /, in other

words, the smallest value of.Example 3.4: Power method for Lowest eigenvalue.Employ the power method to determine the lowesteigenvalue for

[ ]

3.556 -1.778 0 A = -1.778 3.556 -1.778

0 -1.778 3.556

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Solution: The matrix inverse is

Using the same format as in Example (3.3), the

power method can be applied to this matrix.

First iteration:

[ ]

-1

0.422 0.281 0.141

A = 0.281 0.562 0.281

0.141 0.281 0.422

0.422 0.281 0.141 1 0.844 0.7510.281 0.562 0.281 1 = 1.124 = 1.124 1.000

0.141 0.281 0.422 1 0.844 0.751

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Second iteration:

whereThird iteration:

where

0.422 0.281 0.141 0.751 0.704 0.715

0.281 0.562 0.281 1.000 = 0.984 = 0.984 1.0000.141 0.281 0.422 0.751 0.704 0.715

a =14.6%

0.422 0.281 0.141 0.715 0.684 0.709

0.281 0.562 0.281 1.000 = 0.964 = 0.964 1.000

0.141 0.281 0.422 0.715 0.684 0.709

a = 4%

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Thus, after only three iterations, theresult is converging on the value of 0.955,which is the reciprocal of the smallesteigenvalue 1.0472.

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3.4. Other methods:A wide variety of additional methods are

available for solving eigenvalue problems. Most ofthem are based on a two-step process involvingtransformation of the original matrix to a simpler form

(e.g. tridiagonal) which retains all original eigenvaluesand iterative methods to determine these eigenvalues.Jacobi method:

It transforms a symmetric matrix to a diagonal

matrix by eliminating off-diagonal terms in asymmetric fashion. It is an iterative approach toobtain off-diagonal terms sufficiently small.

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Givens method:

This method also produces tridiagonal matrix in

finite number of steps compared to iterative methodas Jacobi. This method uses rotation matrix fororthogonal transformation. It is more efficient thanJacobi.

Householders method:

It is also similar to Givens approach, but more

efficient as it reduces whole rows and columns ofoff-diagonal elements to zero.

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Once the tridiagonal system is obtained fromGivens method or Householders method, theremaining steps involve finding the eigenvlaues. Adirect way to do this is to expand the determinant.The result is a sequence of polynomial which canbe evaluated iteratively for the eigenvalues.

Strum sequence, Gram Schmidt Deflation,Simultaneous iteration method, Subspace iteration,Lanczos method are other methods used forsolving eigenvalue problems.

In many practical problems, first few naturalfrequencies are of interest to designer. Lanczosmethod serves this purpose saving time by

computing the frequencies of interest only