Answer:

(a) If \the first eigenvalue is \the and \the corresponding eigenvector is \te, then \te satisfies \text{[?]}[?]

Let \te be a corresponding eigenvector of \textbf{A} with eigenvalue a. Then

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

Apply \mathbf{A} again to both sides:

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

So \te is also an eigenvector of \mathbf{A}^2, with corresponding eigenvalue \lambda^2.

Now suppose \the is an eigenvalue of \mathbf{A}. Then for some eigenvector \te,

\(\mathbf{A}\mathbf{e}=3\mathbf{e}.\)

Using the result above, \mathbf{A}^2\mathbf{e}=9\mathbf{e} and \mathbf{A}^4\mathbf{e}=81\mathbf{e}. Therefore

\((\mathbf{A}^4+3\mathbf{A}^2+2\mathbf{I})\mathbf{e}=81\mathbf{e}+3(9\mathbf{e})+2\mathbf{e}=110\mathbf{e}.\)

Hence \(110\) is an eigenvalue of \(\mathbf{A}^4+3\mathbf{A}^2+2\mathbf{I}\).