Answer: (a) If \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), then multiplying by \(\mathbf{A}\) gives \(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda^2\mathbf{e}\). So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^2\) with eigenvalue \(\lambda^2\).

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

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

(a) Suppose \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), where \(\mathbf{e}\neq \mathbf{0}\). Then

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

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

For \(\mathbf{B}=\begin{pmatrix}1&3&0\\2&0&2\\1&1&2\end{pmatrix}\), the eigenvalues are found from the characteristic equation

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

Evaluating the determinant gives

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

So the eigenvalues of \(\mathbf{B}\) are \(\lambda=-2,\,1,\,4\).

Now let \(\mu\) be an eigenvalue of \(\mathbf{B}\) with eigenvector \(\mathbf{e}\). Then

\(\mathbf{B}\mathbf{e}=\mu\mathbf{e}.\)

So

\(\mathbf{B}^2\mathbf{e}=\mu^2\mathbf{e}\) and \(\mathbf{B}^4\mathbf{e}=\mu^4\mathbf{e}.\)

Therefore

\((\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I})\mathbf{e}=(\mu^4+2\mu^2+3)\mathbf{e}.\)

So the eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\) are obtained by substituting the eigenvalues of \(\mathbf{B}\):

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

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