Answer:
\(\lambda^{-1}\) is an eigenvalue of \(\mathbf{A}^{-1}\) with eigenvector \(\mathbf{e}\), and \(\lambda+\lambda^{-1}\) is an eigenvalue of \(\mathbf{A}+\mathbf{A}^{-1}\).
For \(\lambda=1\), one corresponding eigenvector is \(\begin{pmatrix}-1\\-4\\1\end{pmatrix}\).
The eigenvalues corresponding to \(\begin{pmatrix}0\\1\\0\end{pmatrix}\) and \(\begin{pmatrix}1\\2\\1\end{pmatrix}\) are \(2\) and \(3\), respectively.
One valid choice is
\(\mathbf{P}=\begin{pmatrix}-1&0&1\\-4&1&2\\1&0&1\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}8&0&0\\0&\frac{125}{8}&0\\0&0&\frac{1000}{27}\end{pmatrix}.\)
Suppose \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), where \(\mathbf{A}\) is non-singular and \(\mathbf{e}\ne\mathbf{0}\).
Multiplying by \(\mathbf{A}^{-1}\) gives
\(\mathbf{e}=\lambda\mathbf{A}^{-1}\mathbf{e}.\)
Since \(\lambda\ne0\), divide by \(\lambda\):
\(\mathbf{A}^{-1}\mathbf{e}=\lambda^{-1}\mathbf{e}.\)
Therefore \(\lambda^{-1}\) is an eigenvalue of \(\mathbf{A}^{-1}\), with the same eigenvector \(\mathbf{e}\).
Also,
\(\mathbf{A}\mathbf{e}+\mathbf{A}^{-1}\mathbf{e}=\lambda\mathbf{e}+\lambda^{-1}\mathbf{e},\)
so
\(\left(\mathbf{A}+\mathbf{A}^{-1}\right)\mathbf{e}=\left(\lambda+\lambda^{-1}\right)\mathbf{e}.\)
Hence \(\lambda+\lambda^{-1}\) is an eigenvalue of \(\mathbf{A}+\mathbf{A}^{-1}\).
For the eigenvalue \(1\), solve \(\left(\mathbf{A}-\mathbf{I}\right)\mathbf{x}=\mathbf{0}\):
\(\begin{pmatrix}1&0&1\\-1&1&3\\1&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.\)
The first row gives \(x+z=0\), so \(z=-x\). The second row gives \(-x+y+3z=0\), so \(y=4x\). Taking \(x=-1\) gives the eigenvector
\(\begin{pmatrix}-1\\-4\\1\end{pmatrix}.\)
Now check the two given eigenvectors:
\(\mathbf{A}\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\2\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix},\)
so the corresponding eigenvalue is \(2\).
Also,
\(\mathbf{A}\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}3\\6\\3\end{pmatrix}=3\begin{pmatrix}1\\2\\1\end{pmatrix},\)
so the corresponding eigenvalue is \(3\).
The eigenvalues of \(\mathbf{A}+\mathbf{A}^{-1}\) are therefore \(2\), \(\frac{5}{2}\) and \(\frac{10}{3}\). Hence the eigenvalues of \(\left(\mathbf{A}+\mathbf{A}^{-1}\right)^3\) are
\(2^3=8,\qquad \left(\frac{5}{2}\right)^3=\frac{125}{8},\qquad \left(\frac{10}{3}\right)^3=\frac{1000}{27}.\)
Using the corresponding eigenvectors as columns, one valid choice is
\(\mathbf{P}=\begin{pmatrix}-1&0&1\\-4&1&2\\1&0&1\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}8&0&0\\0&\frac{125}{8}&0\\0&0&\frac{1000}{27}\end{pmatrix}.\)
Therefore \(\left(\mathbf{A}+\mathbf{A}^{-1}\right)^3=\mathbf{PDP}^{-1}\).
