Answer: The corresponding eigenvalue is \(3\).

A corresponding eigenvector for eigenvalue \(4\) is \(\begin{pmatrix}1\\-4\\-4\end{pmatrix}\).

For \(\mathbf{A}^5\), take the eigenvectors as the columns of \(\mathbf{P}\), with \(\mathbf{D}\) containing the fifth powers of the corresponding eigenvalues, and \(\mathbf{Q}=\mathbf{P}^{-1}\):

\(\mathbf{P}=\begin{pmatrix}1&1&1\\-4&-2&-4\\-1&-1&-4\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}243&0&0\\0&1024&0\\0&0&1\end{pmatrix}\), \(\mathbf{Q}=\mathbf{P}^{-1}=\begin{pmatrix}-\frac23&-\frac12&\frac13\\2&\frac12&0\\-\frac13&0&-\frac13\end{pmatrix}\).

Hence \(\mathbf{A}^5=\mathbf{P}\mathbf{D}\mathbf{Q}\).

First, multiply \(\mathbf{A}\) by the given eigenvector:

\(\mathbf{A}\begin{pmatrix}1\\-2\\-1\end{pmatrix}=\begin{pmatrix}4&1&-1\\-4&-1&4\\0&-1&5\end{pmatrix}\begin{pmatrix}1\\-2\\-1\end{pmatrix}=\begin{pmatrix}4-2+1\\-4+2-4\\2-5\end{pmatrix}=\begin{pmatrix}3\\-6\\-3\end{pmatrix}=3\begin{pmatrix}1\\-2\\-1\end{pmatrix}.\)

So the corresponding eigenvalue is \(3\).

For the eigenvalue \(4\), we need a non-zero vector \(\mathbf{x}\) such that \((\mathbf{A}-4\mathbf{I})\mathbf{x}=\mathbf{0}\). This gives

\(\mathbf{A}-4\mathbf{I}=\begin{pmatrix}0&1&-1\\-4&-5&4\\0&-1&1\end{pmatrix}.\)

Let \(\mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). Then the first row gives \(y-z=0\), so \(y=z\). The second row gives \(-4x-5y+4z=0\), and using \(z=y\) this becomes \(-4x-y=0\), so \(y=-4x\). Hence \(z=-4x\) too. Taking \(x=1\), a corresponding eigenvector is

\(\begin{pmatrix}1\\-4\\-4\end{pmatrix}.\)

Now the given eigenvectors are

\(\begin{pmatrix}1\\-2\\-1\end{pmatrix}\) for eigenvalue \(3\), \(\begin{pmatrix}1\\-4\\-4\end{pmatrix}\) for eigenvalue \(4\), and \(\begin{pmatrix}1\\-4\\-1\end{pmatrix}\) for eigenvalue \(1\).

So we may choose \(\mathbf{P}\) to have these eigenvectors as columns:

\(\mathbf{P}=\begin{pmatrix}1&1&1\\-2&-4&-4\\-1&-4&-1\end{pmatrix}.\)

The corresponding diagonal matrix for \(\mathbf{A}^5\) has the fifth powers of the eigenvalues on the diagonal:

\(\mathbf{D}=\begin{pmatrix}3^5&0&0\\0&4^5&0\\0&0&1^5\end{pmatrix}=\begin{pmatrix}243&0&0\\0&1024&0\\0&0&1\end{pmatrix}.\)

Finally, since the columns of \(\mathbf{P}\) are eigenvectors, \(\mathbf{Q}=\mathbf{P}^{-1}\), so \(\mathbf{A}=\mathbf{PDP}^{-1}\) for the original diagonalisation, and therefore \(\mathbf{A}^5=\mathbf{P}\mathbf{D}\mathbf{Q}\) with \(\mathbf{Q}=\mathbf{P}^{-1}\).

Thus one valid set is

\(\mathbf{P}=\begin{pmatrix}1&1&1\\-2&-4&-4\\-1&-4&-1\end{pmatrix},\quad \mathbf{D}=\begin{pmatrix}243&0&0\\0&1024&0\\0&0&1\end{pmatrix},\quad \mathbf{Q}=\mathbf{P}^{-1}.\)

Any equivalent choice of ordering of the eigenvectors is also acceptable, provided \(\mathbf{D}\) is rearranged to match.