Answer: (a) Since \(\mathbf{A}\mathbf{e}=\lambda \mathbf{e}\), it follows that \(\mathbf{A}^3\mathbf{e}=\lambda^3\mathbf{e}\). So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^3\) with eigenvalue \(\lambda^3\).

(b) The eigenvalues of \(\mathbf{A}\) are \(-1\), \(1\) and \(5\).

(c) One suitable choice is

\(\mathbf{P}=\begin{pmatrix}1&5&2\\0&2&0\\0&1&3\end{pmatrix},\quad \mathbf{D}=\begin{pmatrix}-3&0&0\\0&-1&0\\0&0&3\end{pmatrix}\),

so that \(\mathbf{A}-2\mathbf{I}=\mathbf{PDP}^{-1}\).

(d) \((\mathbf{A}-2\mathbf{I})^3=-\mathbf{A}^2+13\mathbf{A}-13\mathbf{I}\).

Hence \(a=-1\), \(b=13\), \(c=-13\).

(a) We are given that \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\).

Multiply both sides by \(\mathbf{A}^2\):

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

Also,

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

Therefore

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

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

(b) To find the eigenvalues, solve \(\det(\mathbf{A}-\lambda\mathbf{I})=0\).

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

Expanding the determinant down the first column,

\(\det(\mathbf{A}-\lambda\mathbf{I})=(-1-\lambda)\begin{vmatrix}1-\lambda&0\\-2&5-\lambda\end{vmatrix}\).

So

\(\det(\mathbf{A}-\lambda\mathbf{I})=(-1-\lambda)(1-\lambda)(5-\lambda)\).

Hence the characteristic equation is

\((-1-\lambda)(1-\lambda)(5-\lambda)=0\),

so the eigenvalues are \(\lambda=-1,1,5\).

(c) For \(\mathbf{A}-2\mathbf{I}\), the eigenvalues are those of \(\mathbf{A}\) shifted by \(-2\), so they are \(-3,-1,3\).

We find eigenvectors of \(\mathbf{A}\) corresponding to \(\lambda=-1,1,5\).

For \(\lambda=-1\):

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

Let the eigenvector be \(\begin{pmatrix}x\\y\\z\end{pmatrix}\). Then

\(3y+4z=0,\quad 2y=0,\quad -2y+6z=0\).

From \(2y=0\), we get \(y=0\), and then \(z=0\). Thus \(x\) is free, so one eigenvector is

\(\mathbf{v}_1=\begin{pmatrix}1\\0\\0\end{pmatrix}\).

For \(\lambda=1\):

\(\mathbf{A}-\mathbf{I}=\begin{pmatrix}-2&3&4\\0&0&0\\0&-2&4\end{pmatrix}\).

This gives

\(-2x+3y+4z=0,\quad -2y+4z=0\).

From the second equation, \(y=2z\). Substituting into the first:

\(-2x+3(2z)+4z=0\Rightarrow -2x+10z=0\Rightarrow x=5z\).

Taking \(z=1\), one eigenvector is

\(\mathbf{v}_2=\begin{pmatrix}5\\2\\1\end{pmatrix}\).

For \(\lambda=5\):

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

This gives

\(-6x+3y+4z=0,\quad -4y=0\).

So \(y=0\), and then

\(-6x+4z=0\Rightarrow 3x=2z\).

Taking \(x=2\), \(z=3\), one eigenvector is

\(\mathbf{v}_3=\begin{pmatrix}2\\0\\3\end{pmatrix}\).

Using these eigenvectors as the columns of \(\mathbf{P}\), in the order corresponding to eigenvalues \(-3,-1,3\),

\(\mathbf{P}=\begin{pmatrix}1&5&2\\0&2&0\\0&1&3\end{pmatrix}\).

The corresponding diagonal matrix is

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

Hence

\(\mathbf{A}-2\mathbf{I}=\mathbf{PDP}^{-1}\).

(d) From part (b), the characteristic equation of \(\mathbf{A}\) is

\((\lambda+1)(\lambda-1)(\lambda-5)=0\).

Expanding,

\((\lambda^2-1)(\lambda-5)=\lambda^3-5\lambda^2-\lambda+5=0\).

Therefore

\(\mathbf{A}^3-5\mathbf{A}^2-\mathbf{A}+5\mathbf{I}=\mathbf{0}\),

so

\(\mathbf{A}^3=5\mathbf{A}^2+\mathbf{A}-5\mathbf{I}\).

Now expand:

\((\mathbf{A}-2\mathbf{I})^3=\mathbf{A}^3-6\mathbf{A}^2+12\mathbf{A}-8\mathbf{I}\).

Substitute for \(\mathbf{A}^3\):

\((\mathbf{A}-2\mathbf{I})^3=(5\mathbf{A}^2+\mathbf{A}-5\mathbf{I})-6\mathbf{A}^2+12\mathbf{A}-8\mathbf{I}\).

Collecting terms gives

\((\mathbf{A}-2\mathbf{I})^3=-\mathbf{A}^2+13\mathbf{A}-13\mathbf{I}\).

So the required constants are \(a=-1\), \(b=13\), \(c=-13\).