Answer: If \(\mathbf e\) is a common eigenvector of \(\mathbf A\) and \(\mathbf B\), with eigenvalues \(\lambda\) and \(\mu\), then \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

For \(\mathbf C\), the eigenvalues are \(-1,1,2\), with corresponding eigenvectors

\(\begin{pmatrix}1\\0\\0\end{pmatrix},\quad \begin{pmatrix}1\\-2\\0\end{pmatrix},\quad \begin{pmatrix}1\\6\\3\end{pmatrix}.\)

Since \(\mathbf D\begin{pmatrix}1\\6\\3\end{pmatrix}=-2\begin{pmatrix}1\\6\\3\end{pmatrix}\), \(\mathbf{CD}\) has eigenvector \(\begin{pmatrix}1\\6\\3\end{pmatrix}\) with eigenvalue \(-4\).

If

\(\mathbf A\mathbf e=\lambda\mathbf e,\qquad \mathbf B\mathbf e=\mu\mathbf e,\)

then

\(\mathbf{AB}\mathbf e=\mathbf A(\mathbf B\mathbf e)=\mathbf A(\mu\mathbf e)=\mu\mathbf A\mathbf e=\mu\lambda\mathbf e=\lambda\mu\mathbf e.\)

So \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

For the matrix \(\mathbf C\), solve \((\mathbf C-k\mathbf I)\mathbf v=\mathbf0\) for each eigenvalue. This gives

\(k=-1:\quad \mathbf v=\begin{pmatrix}1\\0\\0\end{pmatrix},\)

\(k=1:\quad \mathbf v=\begin{pmatrix}1\\-2\\0\end{pmatrix},\)

\(k=2:\quad \mathbf v=\begin{pmatrix}1\\6\\3\end{pmatrix}.\)

Now check the last vector with \(\mathbf D\):

\(\mathbf D\begin{pmatrix}1\\6\\3\end{pmatrix}=-2\begin{pmatrix}1\\6\\3\end{pmatrix}.\)

Thus \(\begin{pmatrix}1\\6\\3\end{pmatrix}\) is a common eigenvector of \(\mathbf C\) and \(\mathbf D\), with eigenvalues \(2\) and \(-2\), respectively.

By the result proved at the start, it is an eigenvector of \(\mathbf{CD}\) with eigenvalue

\(2(-2)=-4.\)