For the matrix A in Example 1, find the smallest eigenvalue (in absolute value) and an associated eigenvector.

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Accepted Answer

Here we apply the Power Method to

[latex]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} ight][/latex]

generating a sequence of vectors of the form

[latex]x _{k+1}=\frac{1}{s_k}A^{-1}x _k[/latex] (4)

For a change of pace (and another reason to be discussed later), let’s take the initial vector to be [latex]x_0=(1,2,3,4,5)[/latex]. Table 5 gives the results of every other iteration.

From the output, we see that [latex]A^{-1}[/latex] has eigenvalue λ = 1, so that [latex]\lambda_3=\lambda^{-1}=1[/latex] 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

[latex]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} ight]\left[\begin{array}{r}1.0000 \0.6923 \-0.1538 \0.0769 \0.4615\end{array} ight] \approx\left[\begin{array}{r}1.0000 \0.6920 \-0.1537 \0.0768 \0.4617\end{array} ight] \approx u[/latex]

Thus u is an eigenvector associated with eigenvalue λ[latex]_3[/latex] = 1.

Question 6.2.3: For the matrix A in Example 1, find the smallest eigenvalue ...

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