14.1. THE POWER METHOD FOR EIGENVALUES 365

Divide by largest entry  −54. 115

26. 615

49. 184

 1

−54. 115=

 1.0

−0.491 82

−0.908 88

You can see the vector didn’t change much and so the next scaling factor will not be muchdifferent than this one. Hence you need to solve for λ

1

λ+ 1.2= −54. 115

Then λ = −1. 218 5 is an approximate eigenvalue and 1.0

−0.491 82

−0.908 88

is an approximate eigenvector. How well does it work? 2 1 3

2 1 1

3 2 1

 1.0

−0.491 82

−0.908 88

 =

 −1. 218 5

0.599 3

1. 107 5

(−1. 218 5)

 1.0

−0.491 82

−0.908 88

 =

 −1. 218 5

0.599 28

1. 107 5

You can see that for practical purposes, this has found the eigenvalue closest to −1. 218 5

and the corresponding eigenvector.The other eigenvectors and eigenvalues can be found similarly. In the case of −.4, you

could let α = −.4 and then

(A− αI)−1

=

 8. 064 516 1× 10−2 −9. 274 193 5 6. 451 612 9

−. 403 225 81 11. 370 968 −7. 258 064 5

. 403 225 81 3. 629 032 3 −2. 741 935 5

 .

Following the procedure of the power method, you find that after about 5 iterations, thescaling factor is 9. 757 313 9, they are not changing much, and

u5 =

 −. 781 224 81. 0

. 264 936 88

 .

Thus the approximate eigenvalue is

1

λ+ .4= 9. 757 313 9

which shows λ = −. 297 512 78 is an approximation to the eigenvalue near .4. How well doesit work?  2 1 3

2 1 1

3 2 1

 −. 781 224 8

1. 0

. 264 936 88

 =

 . 232 361 04

−. 297 512 72−.0 787 375 2

 .