06.09.2021 Views

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

Linear Algebra, Theory And Applications, 2012a

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

15.1. THE POWER METHOD FOR EIGENVALUES 379<br />

Example 15.1.6 Find the eigenvalues and eigenvectors of the matrix A = ⎝<br />

⎛<br />

2 1 3<br />

2 1 1<br />

3 2 1<br />

This is only a 3×3 matrix and so it is not hard to estimate the eigenvalues. Just get<br />

the characteristic equation, graph it using a calculator and zoom in to find the eigenvalues.<br />

If you do this, you find there is an eigenvalue near −1.2, one near −.4, and one near 5.5.<br />

(The characteristic equation is 2 + 8λ +4λ 2 − λ 3 =0.) Of course I have no idea what the<br />

eigenvectors are.<br />

Lets first try to find the eigenvector and a better approximation for the eigenvalue near<br />

−1.2. In this case, let α = −1.2. Then<br />

⎛<br />

(A − αI) −1 = ⎝<br />

−25. 357 143 −33. 928 571 50.0<br />

12. 5 17. 5 −25.0<br />

23. 214 286 30. 357 143 −45.0<br />

As before, it helps to get things started if you raise to a power and then go from the<br />

approximate eigenvector obtained.<br />

⎛<br />

⎞7 ⎛<br />

−25. 357 143 −33. 928 571 50.0<br />

⎝ 12. 5 17. 5 −25.0 ⎠ ⎝ 1 ⎞ ⎛<br />

⎞<br />

−2. 295 6 × 1011<br />

1 ⎠ = ⎝ 1. 129 1 × 10 11 ⎠<br />

23. 214 286 30. 357 143 −45.0 1 2. 086 5 × 10 11<br />

Then the next iterate will be<br />

⎛<br />

⎝<br />

⎞<br />

−2. 295 6 × 1011<br />

1. 129 1 × 10 11 ⎠<br />

⎛<br />

1<br />

−2. 295 6 × 10 11 = ⎝<br />

Next iterate:<br />

⎛<br />

⎞ ⎛<br />

−25. 357 143 −33. 928 571 50.0<br />

⎝ 12. 5 17. 5 −25.0 ⎠ ⎝<br />

23. 214 286 30. 357 143 −45.0<br />

Divide by largest entry<br />

⎛<br />

⎝<br />

−54. 115<br />

26. 615<br />

49. 184<br />

⎞<br />

⎠<br />

⎛<br />

1<br />

−54. 115 = ⎝<br />

1.0<br />

−0.491 85<br />

−0.908 91<br />

1.0<br />

−0.491 82<br />

−0.908 88<br />

1.0<br />

−0.491 85<br />

−0.908 91<br />

⎞ ⎛<br />

⎠ = ⎝<br />

⎞<br />

⎠<br />

⎞<br />

⎠ .<br />

⎞<br />

⎠<br />

−54. 115<br />

26. 615<br />

49. 184<br />

You can see the vector didn’t change much and so the next scaling factor will not be much<br />

different than this one. Hence you need to solve for λ<br />

1<br />

= −54. 115<br />

λ +1.2<br />

Then λ = −1. 218 5 is an approximate eigenvalue and<br />

⎛<br />

⎝<br />

1.0<br />

−0.491 82<br />

−0.908 88<br />

⎞<br />

⎠<br />

is an approximate eigenvector. How well does it work?<br />

⎛<br />

⎝ 2 1 3<br />

⎞ ⎛<br />

⎞<br />

1.0<br />

2 1 1 ⎠ ⎝ −0.491 82 ⎠ =<br />

3 2 1 −0.908 88<br />

⎛<br />

(−1. 218 5) ⎝<br />

1.0<br />

−0.491 82<br />

−0.908 88<br />

⎞<br />

⎠ =<br />

⎛<br />

⎝<br />

⎛<br />

⎝<br />

−1. 218 5<br />

0.599 3<br />

1. 107 5<br />

−1. 218 5<br />

0.599 28<br />

1. 107 5<br />

⎞<br />

⎠<br />

⎞<br />

⎠<br />

⎞<br />

⎠<br />

⎞<br />

⎠ .

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!