356 CHAPTER 15. NUMERICAL METHODS, EIGENVALUES

 8.0645161×10−2 −9.2741935 6.4516129−.40322581 11.370968 −7.2580645.40322581 3.6290323 −2.7419355

17 1

11

=

 −1.8535×1016

2.3724×1016

6.2874×1015

The first choice for an approximate eigenvector is −1.8535×1016

2.3724×1016

6.2874×1015

 12.3724×1016 =

 −0.781281.0

0.26502

Lets see how well this works as an eigenvector. 2 1 3

2 1 13 2 1

 −0.78128

1.00.26502

=

 0.2325−0.29754−0.07882



(−0.29754)

 −0.781281.0

0.26502

=

 0.23246−0.29754

−7.8854×10−2

Thus this works as an eigenvector with the eigenvalue (−0.29754).

Next we will find the eigenvalue and eigenvector for the eigenvalue near 5.5. In thiscase,

(A−αI)−1 =

 29.2 16.8 23.219.2 10.8 15.228.0 16.0 22.0

 .

As before, I have no idea what the eigenvector is but to avoid giving the impression thatyou always need to start with the vector (1,1,1)T , let u1 = (1,2,3)T . I will use the sameshortcut to get this eigenvector as in the above case. 29.2 16.8 23.2

19.2 10.8 15.228.0 16.0 22.0

16 1

23

=

 1.0987×1029

7.1868×1028

1.0482×1029

Then dividing by the largest entry, a good guess for the eigenvector is 1.0

0.654120.95404

