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Linear Algebra, Theory And Applications, 2012a

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15.1. THE POWER METHOD FOR EIGENVALUES 377<br />

Now solve<br />

⎛<br />

⎝<br />

12 −14 11<br />

−4 11 −4<br />

3 6 4<br />

⎞ ⎛<br />

⎠ ⎝<br />

x<br />

y<br />

z<br />

⎞ ⎛<br />

⎠ = ⎝<br />

and divide by the largest entry, 1. 051 5 to get<br />

⎛<br />

1.0<br />

⎞<br />

u 3 = ⎝ .0 266<br />

−. 970 61<br />

⎠<br />

Solve<br />

⎛<br />

⎝<br />

12 −14 11<br />

−4 11 −4<br />

3 6 4<br />

⎞ ⎛<br />

⎠ ⎝ x ⎞ ⎛<br />

y ⎠ = ⎝<br />

z<br />

and divide by the largest entry, 1. 01 to get<br />

⎛<br />

1.0<br />

⎞<br />

u 4 = ⎝ 3. 845 4 × 10 −3<br />

−. 996 04<br />

⎠ .<br />

1.0<br />

. 184<br />

−. 76<br />

⎞<br />

⎠<br />

1.0<br />

.0 266<br />

−. 970 61<br />

These scaling factors are pretty close after these few iterations. Therefore, the predicted<br />

eigenvalue is obtained by solving the following for λ.<br />

1<br />

λ +7 =1.01<br />

which gives λ = −6. 01. You see this is pretty close. In this case the eigenvalue closest to<br />

−7 was−6.<br />

How would you know what to start with for an initial guess? You might apply Gerschgorin’s<br />

theorem.<br />

Example 15.1.5 Consider the symmetric matrix A =<br />

⎛<br />

⎝ 1 2 2 1 3<br />

4<br />

⎞<br />

⎠ . Find the middle<br />

3 4 2<br />

eigenvalue and an eigenvector which goes with it.<br />

Since A is symmetric, it follows it has three real eigenvalues which are solutions to<br />

⎛ ⎛ ⎞ ⎛ ⎞⎞<br />

1 0 0 1 2 3<br />

p (λ) = det⎝λ<br />

⎝ 0 1 0 ⎠ − ⎝ 2 1 4 ⎠⎠<br />

0 0 1 3 4 2<br />

= λ 3 − 4λ 2 − 24λ − 17 = 0<br />

If you use your graphing calculator to graph this polynomial, you find there is an eigenvalue<br />

somewhere between −.9 and−.8 and that this is the middle eigenvalue. Of course you could<br />

zoom in and find it very accurately without much trouble but what about the eigenvector<br />

which goes with it? If you try to solve<br />

⎛ ⎛<br />

⎝(−.8) ⎝ 1 0 0<br />

⎞ ⎛<br />

0 1 0 ⎠ − ⎝<br />

0 0 1<br />

1 2 3<br />

2 1 4<br />

3 4 2<br />

⎞⎞<br />

⎛<br />

⎠⎠<br />

⎝<br />

x<br />

y<br />

z<br />

⎞<br />

⎠<br />

⎞ ⎛<br />

⎠ = ⎝<br />

there will be only the zero solution because the matrix on the left will be invertible and the<br />

same will be true if you replace −.8 with a better approximation like −.86 or −.855. This is<br />

0<br />

0<br />

0<br />

⎞<br />

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