Answer: For an eigenvector \(\mathbf{e}\) of \(\mathbf{A}\) with eigenvalue \(\lambda\), we have \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\). Then

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

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^2\) with corresponding eigenvalue \(\lambda^2\).

The eigenvalues of \(\mathbf{B}\) are \(-2\), \(1\), and \(4\).

Hence the eigenvalues of \(\mathbf{B}^2\) are \(4\), \(1\), and \(16\), and the eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\) are obtained by applying \(x^4+2x^2+3\) to each eigenvalue of \(\mathbf{B}\):

For \(x=-2\): \((-2)^4+2(-2)^2+3=16+8+3=27\).

For \(x=1\): \(1^4+2(1)^2+3=1+2+3=6\).

For \(x=4\): \(4^4+2(4)^2+3=256+32+3=291\).

Therefore the eigenvalues are \(27\), \(6\), and \(291\).

Let \(\mathbf{e}\neq \mathbf{0}\) be an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\). Then

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

Multiplying both sides by \(\mathbf{A}\) gives

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

Hence \(\mathbf{e}\) is also an eigenvector of \(\mathbf{A}^2\), with corresponding eigenvalue \(\lambda^2\).

Now consider

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

The characteristic equation is

\(\det(\mathbf{B}-\lambda\mathbf{I})=0.\)

Expanding, this gives

\((1-\lambda)(\lambda-4)(\lambda+2)=0.\)

So the eigenvalues of \(\mathbf{B}\) are

\(\lambda=-2,\ 1,\ 4.\)

To find the eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\), use the fact that if \(\lambda\) is an eigenvalue of \(\mathbf{B}\), then \(\mathbf{B}^2\) has eigenvalue \(\lambda^2\), and more generally a polynomial in \(\mathbf{B}\) has the corresponding polynomial in \(\lambda\) as an eigenvalue.

So the required eigenvalues are

\(\lambda^4+2\lambda^2+3\)

for \(\lambda=-2,1,4\).

• For \(\lambda=-2\): \((-2)^4+2(-2)^2+3=16+8+3=27\).
• For \(\lambda=1\): \(1^4+2(1)^2+3=1+2+3=6\).
• For \(\lambda=4\): \(4^4+2(4)^2+3=256+32+3=291\).

Therefore the eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\) are \(27\), \(6\), and \(291\).