Answer: (a) Let \(\mathbf{e}\) be an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\). Then \(\mathbf{Ae}=\lambda \mathbf{e}\). So

\((\mathbf{A}+k\mathbf{I})\mathbf{e}=\mathbf{Ae}+k\mathbf{Ie}=\lambda\mathbf{e}+k\mathbf{e}=(\lambda+k)\mathbf{e}.\)

Hence \((\mathbf{A}+k\mathbf{I})\) has eigenvalue \(\lambda+k\) with corresponding eigenvector \(\mathbf{e}\).

(b) For \(\mathbf{B}=\begin{pmatrix}2&2&-3\\2&2&3\\-3&3&3\end{pmatrix}\), the given eigenvalues are \(-3\) and \(4\).

Corresponding eigenvectors are

for \(\lambda=-3\): \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\),

for \(\lambda=4\): \(\begin{pmatrix}1\\1\\0\end{pmatrix}\).

Given that \(\begin{pmatrix}1\\-1\\-2\end{pmatrix}\) is an eigenvector of \(\mathbf{B}\), the corresponding eigenvalue is \(6\).

(c) Since \(\mathbf{C}=\begin{pmatrix}-1&2&-3\\2&-1&3\\-3&3&0\end{pmatrix}=\mathbf{B}-3\mathbf{I},\) each eigenvalue of \(\mathbf{B}\) is reduced by \(3\) for \(\mathbf{C}\), while the eigenvectors stay the same.

Therefore the eigenvalues of \(\mathbf{C}\) are \(-6\), \(1\) and \(3\).

The corresponding eigenvectors are

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

(a) If \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\), then \(\mathbf{Ae}=\lambda\mathbf{e}\).

Now

\((\mathbf{A}+k\mathbf{I})\mathbf{e}=\mathbf{Ae}+k\mathbf{Ie}.\)

Since \(\mathbf{Ie}=\mathbf{e}\), this becomes

\((\mathbf{A}+k\mathbf{I})\mathbf{e}=\lambda\mathbf{e}+k\mathbf{e}=(\lambda+k)\mathbf{e}.\)

So \(\mathbf{e}\) is still an eigenvector, and the new eigenvalue is \(\lambda+k\).

(b) To find an eigenvector for an eigenvalue, solve \((\mathbf{B}-\mu\mathbf{I})\mathbf{x}=\mathbf{0}\).

For \(\mu=-3\), we solve \((\mathbf{B}+3\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{B}+3\mathbf{I}=\begin{pmatrix}5&2&-3\\2&5&3\\-3&3&6\end{pmatrix}.\)

A suitable non-zero solution is \(\mathbf{x}=\begin{pmatrix}1\\-1\\1\end{pmatrix}\), so an eigenvector for \(-3\) is \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\).

For \(\mu=4\), we solve \((\mathbf{B}-4\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{B}-4\mathbf{I}=\begin{pmatrix}-2&2&-3\\2&-2&3\\-3&3&-1\end{pmatrix}.\)

A suitable non-zero solution is \(\mathbf{x}=\begin{pmatrix}1\\1\\0\end{pmatrix}\), so an eigenvector for \(4\) is \(\begin{pmatrix}1\\1\\0\end{pmatrix}\).

For the given eigenvector \(\begin{pmatrix}1\\-1\\-2\end{pmatrix}\), let the eigenvalue be \(\mu\). Then \(\mathbf{B}\mathbf{x}=\mu\mathbf{x}\).

Calculate

\(\mathbf{B}\begin{pmatrix}1\\-1\\-2\end{pmatrix}=\begin{pmatrix}2(1)+2(-1)-3(-2)\\2(1)+2(-1)+3(-2)\\-3(1)+3(-1)+3(-2)\end{pmatrix}=\begin{pmatrix}6\\-12\\-18\end{pmatrix}.\)

This is equal to \(6\begin{pmatrix}1\\-1\\-2\end{pmatrix}\), so the eigenvalue is \(6\).

(c) We notice that

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

So if \(\mu\) is an eigenvalue of \(\mathbf{B}\), then \(\mu-3\) is the corresponding eigenvalue of \(\mathbf{C}\), with the same eigenvector.

The eigenvalues of \(\mathbf{B}\) are \(-3\), \(4\), \(6\), so the eigenvalues of \(\mathbf{C}\) are

\(-6,\ 1,\ 3\).

Matching eigenvectors are unchanged:

\(\lambda=-6\) has eigenvector \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\),

\(\lambda=1\) has eigenvector \(\begin{pmatrix}1\\1\\0\end{pmatrix}\),

\(\lambda=3\) has eigenvector \(\begin{pmatrix}1\\-1\\-2\end{pmatrix}\).