364 CHAPTER 14. NUMERICAL METHODS, EIGENVALUES

Clearly the uk are not changing much. This suggests an approximate eigenvector for thiseigenvalue which is close to −.855 is the above u3 and an eigenvalue is obtained by solving

1

λ+ .855= −513. 42,

which yields λ = −0.856 95 Lets check this. 1 2 3

2 1 4

3 4 2

 1. 0

−. 587 78−. 227 14

 =

 −0.856 98

0.503 66

0.194 6

 .

−0.856 95

 1. 0

−. 587 77−. 227 14

 =

 −0.856 95

0.503 69

0.194 65

Thus the vector of 14.4 is very close to the desired eigenvector, just as −. 856 9 is very closeto the desired eigenvalue. For practical purposes, I have found both the eigenvector and theeigenvalue.

Example 14.1.6 Find the eigenvalues and eigenvectors of the matrix A =

 2 1 3

2 1 1

3 2 1

 .

This is only a 3×3 matrix and so it is not hard to estimate the eigenvalues. Just getthe characteristic equation, graph it using a calculator and zoom in to find the eigenvalues.If you do this, you find there is an eigenvalue near −1.2, one near −.4, and one near 5.5.(The characteristic equation is 2 + 8λ + 4λ2 − λ3 = 0.) Of course I have no idea what theeigenvectors are.

Lets first try to find the eigenvector and a better approximation for the eigenvalue near−1.2. In this case, let α = −1.2. Then

(A− αI)−1

=

 −25. 357 143 −33. 928 571 50.0

12. 5 17. 5 −25.0

23. 214 286 30. 357 143 −45.0

 .

As before, it helps to get things started if you raise to a power and then go from theapproximate eigenvector obtained. −25. 357 143 −33. 928 571 50.0

12. 5 17. 5 −25.0

23. 214 286 30. 357 143 −45.0

7 1

1

1

 =

 −2. 295 6× 1011

1. 129 1× 1011

2. 086 5× 1011

Then the next iterate will be −2. 295 6× 1011

1. 129 1× 1011

2. 086 5× 1011

 1

−2. 295 6× 1011=

 1.0

−0.491 85

−0.908 91

Next iterate: −25. 357 143 −33. 928 571 50.0

12. 5 17. 5 −25.0

23. 214 286 30. 357 143 −45.0

 1.0

−0.491 85

−0.908 91

 =

 −54. 115

26. 615

49. 184

