414 CHAPTER 15. NUMERICAL METHODS, EIGENVALUES

Divide by the largest entry to obtain a good aproximation. 2.5558×1021

−1.2779×1021

−3.6562×1015

 12.5558×1021 =

 1.0−0.5

−1.4306×10−6

Now begin with this one. 5 −14 11

−4 4 −43 6 −3

 1.0

−0.5−1.4306×10−6

=

 12.000−6.0000

4.2918×10−6

Divide by 12 to get the next iterate. 12.000

−6.00004.2918×10−6

 112

=

 1.0−0.5

3.5765×10−7

Another iteration will reveal that the scaling factor is still 12. Thus this is an approxi-mate eigenvalue. In fact, it is the largest eigenvalue and the corresponding eigenvectoris(

1.0 −0.5 0). The process has worked very well.

15.1.1 The Shifted Inverse Power MethodThis method can find various eigenvalues and eigenvectors. It is a significant generaliza-tion of the above simple procedure and yields very good results. One can find complexeigenvalues using this method. The situation is this: You have a number α which is closeto λ , some eigenvalue of an n× n matrix A. You don’t know λ but you know that α iscloser to λ than to any other eigenvalue. Your problem is to find both λ and an eigenvectorwhich goes with λ . Another way to look at this is to start with α and seek the eigenvalueλ , which is closest to α along with an eigenvector associated with λ . If α is an eigenvalueof A, then you have what you want. Therefore, I will always assume α is not an eigenvalueof A and so (A−αI)−1 exists. The method is based on the following lemma.

Lemma 15.1.3 Let {λ k}nk=1 be the eigenvalues of A. If xk is an eigenvector of A for the

eigenvalue λ k, then xk is an eigenvector for (A−αI)−1 corresponding to the eigenvalue1

λ k−α. Conversely, if

(A−αI)−1y =1

λ −αy (15.3)

and y ̸= 0, then Ay = λy.

Proof: Let λ k and xk be as described in the statement of the lemma. Then

(A−αI)xk = (λ k−α)xk

and so 1λ k−α

xk = (A−αI)−1xk.Suppose 15.3. Then y = 1λ−α

[Ay−αy] . Solving for Ayleads to Ay = λy. ■

Now assume α is closer to λ than to any other eigenvalue. Then the magnitude of 1λ−α

is greater than the magnitude of all the other eigenvalues of (A−αI)−1. Therefore, thepower method applied to (A−αI)−1 will yield 1

λ−α. You end up with sn+1 ≊ 1

λ−αand

solve for λ .