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Linear Algebra, Theory And Applications, 2012a

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15.1. THE POWER METHOD FOR EIGENVALUES 381<br />

Then the next iterate is<br />

⎛<br />

⎝<br />

62. 322<br />

40. 765<br />

59. 454<br />

⎞ ⎛<br />

⎠ 1<br />

62.322 = ⎝ 1.0<br />

⎞<br />

0.654 1 ⎠<br />

0.953 98<br />

This is very close to the eigenvector given above and so the next scaling factor will also be<br />

close to 62.322. Thus the approximate eigenvalue is obtained by solving<br />

1<br />

λ − 5.5 =62.322<br />

An approximate eigenvalue is λ =5. 516 and an approximate eigenvector is the above<br />

vector. How well does it work?<br />

⎛<br />

⎝ 2 1 3<br />

⎞ ⎛<br />

2 1 1 ⎠ ⎝ 1.0<br />

⎞ ⎛ ⎞<br />

5. 516<br />

0.654 1 ⎠ = ⎝ 3. 608 1 ⎠<br />

3 2 1 0.953 98 5. 262 2<br />

It appears this is very close.<br />

15.1.3 Complex Eigenvalues<br />

⎛<br />

5. 516 ⎝ 1.0 ⎞ ⎛<br />

0.654 1 ⎠ = ⎝<br />

0.953 98<br />

5. 516<br />

3. 608<br />

5. 262 2<br />

What about complex eigenvalues? If your matrix is real, you won’t see these by graphing<br />

the characteristic equation on your calculator. Will the shifted inverse power method find<br />

these eigenvalues and their associated eigenvectors? The answer is yes. However, for a real<br />

matrix, you must pick α to be complex. This is because the eigenvalues occur in conjugate<br />

pairs so if you don’t pick it complex, it will be the same distance between any conjugate<br />

pair of complex numbers and so nothing in the above argument for convergence implies you<br />

will get convergence to a complex number. Also, the process of iteration will yield only real<br />

vectors and scalars.<br />

Example 15.1.7 Find the complex eigenvalues and corresponding eigenvectors for the matrix<br />

⎛<br />

⎝ 5 1 −8 0 6<br />

0<br />

⎞<br />

⎠ .<br />

0 1 0<br />

Here the characteristic equation is λ 3 − 5λ 2 +8λ − 6=0. One solution is λ =3. The<br />

other two are 1 + i and 1 − i. I will apply the process to α = i to find the eigenvalue closest<br />

to i.<br />

⎛<br />

−.02− . 14i 1. 24 + . 68i −. 84 + . 12i<br />

⎞<br />

(A − αI) −1 = ⎝ −. 14 + .02i .68 − . 24i .12 + . 84i ⎠<br />

.02+. 14i −. 24 − . 68i .84 + . 88i<br />

Then let u 1 =(1, 1, 1) T for lack of any insight into anything better.<br />

⎛<br />

⎞ ⎛<br />

−.02− . 14i 1. 24 + . 68i −. 84 + . 12i<br />

⎝ −. 14 + .02i .68 − . 24i .12 + . 84i ⎠ ⎝ 1 ⎞ ⎛<br />

. 38 + . 66i<br />

1 ⎠ = ⎝ . 66 + . 62i<br />

.02+. 14i −. 24 − . 68i .84 + . 88i 1 . 62 + . 34i<br />

⎞<br />

⎠<br />

⎞<br />

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