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SCS1113: COMPUTATIONAL MATHEMATICSCHAPTER 7

EIGEN VALUE & EIGEN VECTOR

1nzah@utm.my: 2013/2014 - 22IntroductionLet A be a square matrix of dimension nxn and let v be a vector of dimension n.The product,Y=Avcan be viewed as a linear transformation from n-dimensional into itself.3We want to find scalars ( ) for which there exists a nonzero vector v such thatAv = vthat is, the linear transformation T(v) = Av maps v onto the multiple v.When this occurs, we call v an eigen vector that corresponds to the eigen value (), and together they form the eigen pair (), v for A.Introduction (cont.)4The identity matrix [I] can be used to express equationAv = vasAv = Ivwhich is then rewritten in the standard form for a linear system as,(A - I) v = 0

Introduction (cont.)5This linear system, (A - I) v = 0has nontrivial solutions if and only if the matrix A - I is singular, that is|A - I| = 0When the determinant is expanded, it becomes a polynomial of degree n, which is called the characteristic polynomial, p()=|A - I| Introduction (cont.)6Eigen Value & Eigen Vector If A is an nxn matrix, then its n eigen values 1, 2,, n are the roots of the characteristic polynomial p()= det (A - I)If is an eigen value of A and the nonzero vector v has the property that Av = v then v is called an eigen vector of A corresponding to the eigen value .

7Example 7.1Let

The characteristic polynomial of the matrix A

Therefore, the eigen values of A are = 6 and = 1.

8Let,

Then,

The characteristic polynomial of the matrix B

The eigen values are = -1; = 5 and = 2.

Example 7.29Eigen ValuesTo find eigen values when the dimension n is small:Find the coefficient of the characteristic polynomialFinds its rootFind the nonzero solutions of the homogeneous linear system (A - I) v = 0Finding the determinant of an nxn matrix is computationally expensive and finding good approximations to the roots of p() is also difficult.

10Theorem 7.1Assume that A is nxn matrix and let

are eigen values of A

11Gerschgorins Circle TheoremAssume that A is nxn matrix, and let Bi denote the disk in the complex plane with center aii and radius ri

If then all the eigen values of A lie in the set S.

12Example 7.3Let

13Example 7.3 (cont.)Row 1

centerradius

14Example 7.3 (cont.)Circle 063centerradius, 315Row 2

centerradiusExample 7.3 (cont.)16Circle 03centerradius, 251Example 7.3 (cont.)17Row 3

centerradiusExample 7.3 (cont.)18Circle0-4centerradius, 3-1-7Example 7.3 (cont.)19

-7-10156Example 7.3 (cont.)20Power MethodThe power method is an iterative technique used to determine the dominant eigen value of a matrix that is, the eigen value with the largest magnitude.By modifying the method slightly, it can also be used to determine other eigen values.One useful feature of the power method is that it produces not only an eigen value, but also the associated eigen vector.

21If 1 is an eigen value of A that is larger in absolute value than any other eigen value, it is called the dominant eigen value.

An eigen vector v1 corresponding to 1 is called a dominant eigen vector.

Power Method (cont.)22To apply the power method, we assume that the nxn matrix A has n eigen values 1, 2,, n with an associated collection of linearly independent eigenvectors {v(1), v(2),.. v(n)}

We assume that A has precisely one eigen value, 1 that is largest in magnitude, so that

Power Method (cont.)23Start with vector, v(0)

Power Method AlgorithmGenerate the sequence {v(k)} recursively, using

Where mk+1 is the coordinate of Av(k) of largest magnitude (in the case of a tie, choose the coordinate that comes first).

Power Method Algorithm (cont.)2425The sequences {v(k)} and {mk} will converge to v and , respectively.

Termination criteria: Select a tolerance (error), >0 and generate v1,.,vk+1 until

Power Method Algorithm (cont.)26Example 7.4 Let

Use the Power method to approximate the most dominant eigen value of the matrix. Let v(0)=(0,0,1)T and iterate until =0.001.

27

(largest magnitude)

(next iteration)

Example 7.4 - Solution28

(next iteration) Example 7.4 - Solution (cont.)29

Example 7.4 - Solution (cont.)30

Example 7.4 - Solution (cont.)31

Example 7.4 - Solution (cont.)32Exercise 7.1Let

Use the Gerschgorins Circle Theorem to determine a region containing all the eigen values of A.Use the Power method to determine the dominant eigen value of the matrix and its associated eigenvector. Iterate until =0.005.

To be continued34The Shifted Power MethodConsider the standard eigenvalue problem Av=vBy subtracting a scalar s from both sides of the standard eigenvalue problem, the eigenvalues of the matrix are shifted:Av sIv =v - svwhich yields (A sI) v = ( s)v

and can be written as Ashiftedv = shiftedv35Ashifted is the shifted matrix,Ashifted = (A sI)shifted = - s is the eigenvalue of the shifted matrix.Shifting matrix A by a scalar s shifts the eigenvalue by s.Shifting a matrix by a scalar does not affect the eigenvectors.The Shifted Power Method (cont.)36Shifting of eigen values of a matrix can be used to find the opposite extreme eigen value, which is eitherthe smallest magnitude eigen value, orthe largest magnitude eigen value of opposite sign.

The Shifted Power Method (cont.)37Opposite Extreme Eigen valueConsider a matrix whose eigen values are all the same sign; =1, 2, 4, and 8.= 8 is the eigen value of largest magnitude.=1 is the opposite extreme eigen value.The eigen value of largest magnitude 1=8 can be found by the power method.Shifting the eigen values by s = 8, yields the shifted eigen values = -7, -6, -4, 0.38Opposite Extreme Eigen valueThe largest magnitude eigen value of the shifted matrix can be found by the power method; shifted, largest = -7.

The smallest eigen value of the original matrix may be found by smallest = shifted, largest + 8 = -7+ 8 =139The Shifted Power MethodConsider a matrix whose eigen values are both positive and negative; =-1, -2, 4, and 8.=8 is the eigen value of largest magnitude.=-2 is the opposite extreme eigen value.The eigen value of largest magnitude 1=8 can be found by the power method.Shifting the eigen values by s = 8, yields the shifted eigen values = -9, -10, -4, 0.

40The Shifted Power Method (cont.)The largest magnitude eigen value of the shifted matrix can be found by the power method; shifted, largest = -10.The largest magnitude eigen value of opposite sign of the original matrix may be found by largest,negative = shifted, largest + 8 = -10+ 8 = -2

41The shifted power method can be summarized as follows:Solve for the eigen value of largest magnitude 1 using the power method.Shift the matrix A by s =1 to obtain the shifted matrix Ashifted.Solve for the eigen value shifted by the power method.Compute the opposite extreme eigen value of matrix A by = shifted + sThe Shifted Power Method (cont.)42Example 7.5Let,

The dominant eigen value of A is 1= 3.0.Use the shifted power method to find the remaining eigen values of A.

43Example 7.5 - SolutionLet B = Ashifted

44Let, v (0) =[0, 1, 0]T and =0.001

Example 7.5 - Solution (cont.)45

(next iteration) Example 7.5 - Solution (cont.)46Example

47The largest magnitude eigen value of the shifted matrix and its eigenvector,

The opposite extreme eigen value of matrix A and its eigenvector,

Example 7.5 - Solution (cont.)48

Example 7.5 - Solution (cont.)49Since the opposite extreme eigenvalue =1, has the same sign as the largest magnitude eigen value of the original matrix 1=3, all eigen values of A are positive and =1 is the smallest eigen value of matrix A.

Example 7.5 - Solution (cont.)50Matrix A has 3 eigen values, 1,2, and 3.Let, |1| |2| |3|

The largest eigen value of matrix A, 1=3The smallest eigen value of matrix A, 3=1

Example 7.5 - Solution (cont.)51Find the intermediate eigen value of matrix A, 2

Example 7.5 - Solution (cont.)52Exercise 7.2Let,

The dominant eigenvalue of D is 1=3.0.Use the shifted power method to find the remaining eigenvalues of D.Let v (0) =[1, 0.5, 0]T and iterate until =0.01.