Question 6.2.4: For the matrix A given in Example 1, find the eigenvalue tha...

For the matrix A given in Example 1, find the eigenvalue that is closest to c = 4.

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We start by setting

$B=A-4 I_5=\left[\begin{array}{rrrrr}241 & -254 & -252 & -46 & -224 \\161 & -172 & -174 & -32 & -148 \\-39 & 40 & 41 & 7 & 38 \\27 & -28 & -32 & -10 & -26 \\110 & -113 & -110 & -21 & -105\end{array}\right]$

Now we apply the Inverse Power Method to B, starting out with the vector x $_0$ = (5, 4, 3, 2, 1). Table 6 includes every fifth iteration.

We can see that λ = 0.5 is an eigenvalue for B $^{−1}$ , so that λ = 2 is an eigenvalue for B. By shifting back, we find that λ4 = 2 + 4 = 6 is an eigenvalue for A, with associated eigenvector u = (1, 0.5, 0, 0, 0.5).We check this by computing

$A u=\left[\begin{array}{rrrrr}245 & -254 & -252 & -46 & -224 \\161 & -168 & -174 & -32 & -148 \\-39 & 40 & 45 & 7 & 38 \\27 & -28 & -32 & -6 & -26 \\110 & -113 & -110 & -21 & -101\end{array}\right]\left[\begin{array}{c}1.0 \\0.5 \\0 \\0 \\0.5\end{array}\right]=\left[\begin{array}{l}6 \\3 \\0 \\0 \\3\end{array}\right]=6 u$

k	x $_k$	s $_k$
5	(1.0000, 0.5363, −0.0292, 0.0141, 0.4931)	0.9416
10	(1.0000, 0.4936, 0.0051, −0.0026, 0.5013)	0.4616
15	(1.0000, 0.5008, −0.0006, 0.0003, 0.4998)	0.5053
20	(1.0000, 0.4999, 0.0001, 0.0000, 0.5000)	0.4993
25	(1.0000, 0.5000, 0.0000, 0.0000, 0.5000)	0.5001
30	(1.0000, 0.5000, 0.0000, 0.0000, 0.5000)	0.5000
35	(1.0000, 0.5000, 0.0000, 0.0000, 0.5000)	0.5000
Table 6 The Shifted Inverse PowerMethod (Example 4)