Determining $\det(\mathbf{A})$ using the characteristic polynomial 
Let the 3x3 matrix be $ \mathbf{A} = \begin {bmatrix} 3&1&0\\1&3&0\\0&0&1 \end {bmatrix}$.
a) Determine its eigenvalues and eigenvectors.
b) Do the eigenvectors form an orthonormal set? Justify your answer.
c) Show the diagonalization of $\mathbf{A}$.
d) Determine $\mathbf{A}^{10}$ without using $\mathbf{A}$.
e) By using the characteristic polynomial of part a), determine $\det(\mathbf{A})$.

a) From the characteristic equation I found eigenvalues as 1, 2 and 4. And from the equation $(\lambda I - A)x = 0$, I found the set of eigenvectors as
$$\left\{(x_1, x_2, x_3) \mid x_1 = x_2 = 0, x_3 \neq 0\right\} \cup \left\{(x_1, x_2, x_3) \mid x_1 = -x_2 \neq 0, x_3 = 0 \right\} \cup \left\{ (x_1, x_2, x_3) \mid x_1 = x_2 \neq 0, x_3 = 0\right\}.$$
b) I said "For a set to be orthonormal it must be an orthogonal set and all vectors in the set must have the unit norm. But not all pairs of eigenvectors of $A$ gives zero when we take their inner product. For example $(0,0,1)$ is an eigenvector of $A$, so is $(0,0,2)$, and their inner product gives 2, not 0. Besides, it's obvious that not all eigenvectors have the unit norm. So, the eigenvectors of $A$ does not form an orthonormal set."
I'll skip the solutions of parts c) and d) because the part with which I actually have trouble is part e). I don't know what to do in e). Can somebody help me?
 A: Running from the beginning, we have that:
$$\chi_{\mathbf{A}}=\begin{vmatrix}3-\lambda & 1 & 0 \\ 1 & 3 - \lambda & 0 \\ 0 & 0 & 1-\lambda\end{vmatrix}=(1-\lambda)(\lambda^{2}-6\lambda+8)$$
Thus we have that: $$\lambda=\{1,2,4\}$$
As you found. Now we can find the eigenvectors for each eigenvalue by solving $(\mathbf{A}-\lambda \mathbf{I})\vec{x}=\vec{0}$, doing this we get the three principle eigenvectors:
$$\vec{v}_{1}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix},\quad \vec{v}_{2}=\frac{1}{\sqrt{2}}\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix}, \quad \vec{v}_{3}=\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$$
We note that all of these vectors are normalized, and taking the inner product we have:
$$\vec{v}_{i} \cdot \vec{v}_{j}=\delta_{ij}, \qquad \forall i,j \in \{1,2,3\}$$
Where $\delta_{ij}$ is the Kronecker delta. And thus they form an orthonormal set. 
For the next part we note that the spectral theorem implies that:
$$\mathbf{A}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}$$
Where $\mathbf{Q}$ is the matrix with eigenvectors as columns and $\mathbf{\Lambda}$ is the diagonal matrix defined by: $\mathbf{\Lambda}=\operatorname{diag}(\lambda_{1},\dots,\lambda_{n})$. Thus we have:
$$\mathbf{A}=\frac{1}{2}\begin{pmatrix}1 & -1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4\end{pmatrix}\begin{pmatrix}1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix}$$
For the next bit of the question we note that in general we have:
$$\mathbf{A}^{n}=\left(\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}\right)^{n}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}\mathbf{Q}\cdots\mathbf{Q}^{-1}\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}=\mathbf{Q}\mathbf{\Lambda}^{n}\mathbf{Q}^{-1},\quad \forall n \in \mathbb{N}$$
Thus we have for your matrix:
$$\begin{align*}\mathbf{A}^{10}&=\frac{1}{2}\begin{pmatrix}1 & -1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix}\begin{pmatrix}1^{10} & 0 & 0 \\ 0 & 2^{10} & 0 \\ 0 & 0 & 4^{10}\end{pmatrix}\begin{pmatrix}1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix} \\
&= \begin{pmatrix}524800 & 523776 & 0 \\ 523776 & 524800 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{align*}$$
We note that for the last part we have that for an $n\times n$ matrix, we have that the coefficient of $\lambda^{0}$ is given by $\det(\mathbf{A})$, thus we have:
$$\det(\mathbf{A})=8$$
