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Linear Algebra, Theory And Applications, 2012a

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380 NUMERICAL METHODS FOR FINDING EIGENVALUES<br />

You can see that for practical purposes, this has found the eigenvalue closest to −1. 218 5<br />

and the corresponding eigenvector.<br />

The other eigenvectors and eigenvalues can be found similarly. In the case of −.4, you<br />

could let α = −.4 andthen<br />

⎛<br />

(A − αI) −1 = ⎝ 8. 064 516 1 × ⎞<br />

10−2 −9. 274 193 5 6. 451 612 9<br />

−. 403 225 81 11. 370 968 −7. 258 064 5 ⎠ .<br />

. 403 225 81 3. 629 032 3 −2. 741 935 5<br />

Following the procedure of the power method, you find that after about 5 iterations, the<br />

scaling factor is 9. 757 313 9, they are not changing much, and<br />

⎛<br />

⎞<br />

−. 781 224 8<br />

u 5 = ⎝ 1. 0 ⎠ .<br />

. 264 936 88<br />

Thus the approximate eigenvalue is<br />

1<br />

=9. 757 313 9<br />

λ + .4<br />

which shows λ = −. 297 512 78 is an approximation to the eigenvalue near .4. How well does<br />

it work? ⎛<br />

⎝ 2 1 3 ⎞ ⎛<br />

⎞ ⎛<br />

⎞<br />

−. 781 224 8 . 232 361 04<br />

2 1 1 ⎠ ⎝ 1. 0 ⎠ = ⎝ −. 297 512 72 ⎠ .<br />

3 2 1 . 264 936 88 −.0 787 375 2<br />

⎛<br />

⎞ ⎛<br />

⎞<br />

−. 781 224 8<br />

. 232 424 36<br />

−. 297 512 78 ⎝ 1. 0 ⎠ = ⎝ −. 297 512 78 ⎠ .<br />

. 264 936 88 −7. 882 210 8 × 10 −2<br />

It works pretty well. For practical purposes, the eigenvalue and eigenvector have now been<br />

found. If you want better accuracy, you could just continue iterating.<br />

Next I will find the eigenvalue and eigenvector for the eigenvalue near 5.5. In this case,<br />

⎛<br />

(A − αI) −1 = ⎝<br />

29. 2 16. 8 23. 2<br />

19. 2 10. 8 15. 2<br />

28.0 16.0 22.0<br />

As before, I have no idea what the eigenvector is but I am tired of always using (1, 1, 1) T<br />

and I don’t want to give the impression that you always need to start with this vector.<br />

Therefore, I shall let u 1 =(1, 2, 3) T . Also, I will begin by raising the matrix to a power.<br />

⎛<br />

⎝<br />

29. 2 16. 8 23. 2<br />

19. 2 10. 8 15. 2<br />

28.0 16.0 22.0<br />

⎞9 ⎛<br />

⎠ ⎝<br />

1<br />

2<br />

3<br />

⎞ ⎛<br />

⎠ = ⎝<br />

⎞<br />

⎠ .<br />

⎞<br />

3. 009 × 10 16<br />

1. 968 2 × 10 16 ⎠ .<br />

2. 870 6 × 10 16<br />

Divide by largest entry to get the next iterate.<br />

⎛<br />

⎞<br />

⎛<br />

3. 009 × 1016<br />

⎝ 1. 968 2 × 10 16 ⎠<br />

1<br />

2. 870 6 × 10 16 3. 009 × 10 16 = ⎝ 1.0<br />

0.654 1<br />

0.954<br />

⎞<br />

⎠<br />

Now<br />

⎛<br />

⎝<br />

29. 2 16. 8 23. 2<br />

19. 2 10. 8 15. 2<br />

28.0 16.0 22.0<br />

⎞ ⎛<br />

⎠ ⎝ 1.0 ⎞ ⎛<br />

0.654 1 ⎠ = ⎝<br />

0.954<br />

62. 322<br />

40. 765<br />

59. 454<br />

⎞<br />

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