Answer: (i) Since \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), multiplying by \(\mathbf{A}\) again gives

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda^2\mathbf{e}\).

Multiplying once more by \(\mathbf{A}\),

\(\mathbf{A}^3\mathbf{e}=\mathbf{A}(\lambda^2\mathbf{e})=\lambda^2\mathbf{A}\mathbf{e}=\lambda^3\mathbf{e}\).

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^3\) with corresponding eigenvalue \(\lambda^3\).

(ii) First find \(\mathbf{A}^3\). Since \(\mathbf{A}\) is triangular, its eigenvalues are its diagonal entries \(2\) and \(3\), so \(\mathbf{A}^3+\mathbf{I}\) has eigenvalues \(2^3+1=9\) and \(3^3+1=28\).

Now compute \(\mathbf{A}^3\) directly:

\(\mathbf{A}^2=\begin{pmatrix}2&0\\-1&3\end{pmatrix}\begin{pmatrix}2&0\\-1&3\end{pmatrix}=\begin{pmatrix}4&0\\-5&9\end{pmatrix}\),

so

\(\mathbf{A}^3=\begin{pmatrix}4&0\\-5&9\end{pmatrix}\begin{pmatrix}2&0\\-1&3\end{pmatrix}=\begin{pmatrix}8&0\\-19&27\end{pmatrix}.\)

Hence

\(\mathbf{A}^3+\mathbf{I}=\begin{pmatrix}9&0\\-19&28\end{pmatrix}.\)

To diagonalise this matrix, find its eigenvectors.

For eigenvalue \(9\):

\(\begin{pmatrix}0&0\\-19&19\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\), so \(-19x+19y=0\), hence \(x=y\). Take \(\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}.\)

For eigenvalue \(28\):

\(\begin{pmatrix}-19&0\\-19&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\), so \(x=0\). Take \(\mathbf{v}_2=\begin{pmatrix}0\\1\end{pmatrix}.\)

Using these eigenvectors as columns of \(\mathbf{P}\),

\(\mathbf{P}=\begin{pmatrix}1&0\\1&1\end{pmatrix}, \qquad \mathbf{D}=\begin{pmatrix}9&0\\0&28\end{pmatrix}.\)

Then \(\mathbf{A}^3+\mathbf{I}=\mathbf{PDP}^{-1}\).

(i) Let \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\). Then

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\mathbf{A}\mathbf{e})=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda^2\mathbf{e}\).

Applying \(\mathbf{A}\) once more gives

\(\mathbf{A}^3\mathbf{e}=\mathbf{A}(\mathbf{A}^2\mathbf{e})=\mathbf{A}(\lambda^2\mathbf{e})=\lambda^2\mathbf{A}\mathbf{e}=\lambda^3\mathbf{e}\).

Therefore \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^3\), with corresponding eigenvalue \(\lambda^3\).

(ii) We consider \(\mathbf{B}=\mathbf{A}^3+\mathbf{I}\), where

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

First calculate \(\mathbf{A}^3\):

\(\mathbf{A}^2=\begin{pmatrix}2&0\\-1&3\end{pmatrix}\begin{pmatrix}2&0\\-1&3\end{pmatrix}=\begin{pmatrix}4&0\\-5&9\end{pmatrix}\),

so

\(\mathbf{A}^3=\mathbf{A}^2\mathbf{A}=\begin{pmatrix}4&0\\-5&9\end{pmatrix}\begin{pmatrix}2&0\\-1&3\end{pmatrix}=\begin{pmatrix}8&0\\-19&27\end{pmatrix}.\)

Hence

\(\mathbf{B}=\mathbf{A}^3+\mathbf{I}=\begin{pmatrix}9&0\\-19&28\end{pmatrix}.\)

This is triangular, so its eigenvalues are the diagonal entries \(9\) and \(28\).

For \(\lambda=9\), solve \((\mathbf{B}-9\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\begin{pmatrix}0&0\\-19&19\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{0}\),

so \(-19x+19y=0\), hence \(x=y\). A suitable eigenvector is \(\begin{pmatrix}1\\1\end{pmatrix}\).

For \(\lambda=28\), solve \((\mathbf{B}-28\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\begin{pmatrix}-19&0\\-19&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{0}\),

so \(x=0\). A suitable eigenvector is \(\begin{pmatrix}0\\1\end{pmatrix}\).

Take these eigenvectors as the columns of \(\mathbf{P}\):

\(\mathbf{P}=\begin{pmatrix}1&0\\1&1\end{pmatrix}.\)

Then the corresponding diagonal matrix is

\(\mathbf{D}=\begin{pmatrix}9&0\\0&28\end{pmatrix}.\)

Therefore \(\mathbf{A}^3+\mathbf{I}=\mathbf{PDP}^{-1}\), with

\(\mathbf{P}=\begin{pmatrix}1&0\\1&1\end{pmatrix},\quad \mathbf{D}=\begin{pmatrix}9&0\\0&28\end{pmatrix}.\)