Question 6.2.3: For the matrix A in Example 1, find the smallest eigenvalue ...
For the matrix A in Example 1, find the smallest eigenvalue (in absolute value) and an associated eigenvector.
Here we apply the Power Method to
A^{-1}=\frac{1}{78}\left[\begin{array}{rrrrr}-646 & 698 & -420 & -1430 & 620 \\-461 & 487 & -306 & -988 & 448 \\99 & -112 & 70 & 221 & -86 \\-27 & 40 & -64 & -143 & 14 \\-290 & 329 & -178 & -663 & 264\end{array}\right]
generating a sequence of vectors of the form
x _{k+1}=\frac{1}{s_k}A^{-1}x _k (4)
For a change of pace (and another reason to be discussed later), let’s take the initial vector to be x_0=(1,2,3,4,5). Table 5 gives the results of every other iteration.
From the output, we see that A^{-1} has eigenvalue λ = 1, so that \lambda_3=\lambda^{-1}=1 is an eigenvalue of A. We check the vector u = (1.0000, 0.6923, −0.1538, 0.0769, 0.4615) from Table 5 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}{r}1.0000 \\0.6923 \\-0.1538 \\0.0769 \\0.4615\end{array}\right] \approx\left[\begin{array}{r}1.0000 \\0.6920 \\-0.1537 \\0.0768 \\0.4617\end{array}\right] \approx u
Thus u is an eigenvector associated with eigenvalue λ_3 = 1.
| k | x_k | s_k |
| 2 | (1.0000, 0.6842, −0.1565, 0.1178, 0.4734) | −1.253 |
| 4 | (1.0000, 0.6887, −0.1556, 0.0921, 0.4669) | 2.185 |
| 6 | (1.0000, 0.6912, −0.1544, 0.0811, 0.4632) | 1.168 |
| 8 | (1.0000, 0.6920, −0.1540, 0.0780, 0.4620) | 1.037 |
| 10 | (1.0000, 0.6922, −0.1539, 0.0772, 0.4616) | 1.009 |
| 12 | (1.0000, 0.6923, −0.1539, 0.0770, 0.4616) | 1.002 |
| 14 | (1.0000, 0.6923, −0.1538, 0.0769, 0.4615) | 1.001 |
| 16 | (1.0000, 0.6923, −0.1538, 0.0769, 0.4615) | 1.000 |
| Table 5 The Inverse PowerMethod Applied to A^{−1} | ||