Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} -4-\lambda & 3 & 0\\ -6 & 5-\lambda & 0\\ 3 & -3 & -1 - \lambda\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix}$
$\begin{bmatrix} -4-\lambda & 3 & 0\\ -6 & 5-\lambda & 0\\ 3 & -3 & -1-\lambda\end{bmatrix}=0$
$\lambda_1=-1, \lambda_2=2$
2. Find eigenvectors:
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -4-\lambda & 3 & 0\\ -6 & 5-\lambda & 0\\ 3 & -3 & -1\end{bmatrix}=\begin{bmatrix} -3 & 3 & 0 \\ -6& 6 & 0\\
3 & -3 & 0\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix} \\$
Let $r,s$ be free variables.
$\vec{V}=r(1,1,0)+(0,0,1)s \\
E_1=\{(1,1,0)+(0,0,1)\} \\
\rightarrow dim(E_2)=2$
The eigenvectors span $\{(1,1,0)+(0,0,1)\}$ in $R$
For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -4-\lambda & 3 & 0\\ -6 & 5-\lambda & 0\\ 3 & -3 & -1-\lambda \end{bmatrix}=\begin{bmatrix} -6 & 3 & 0 \\ -6 & 3 & 0 \\ 3 & -3 & -3\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0\end{bmatrix} \\$
Let $r$ be a free variable.
$\vec{V}=s(-1,-2,1) \\
E_1=\{(-1,-2,1)\} \\
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(-1,-2,1)\}$ in $R$
Hence, $S=\frac{1}{\sqrt 13}\begin{bmatrix} 1 & 0 & -1\\ 1 & 0 & -2 \\ 0 & 1 & 1 \end{bmatrix} \\
\rightarrow S^{-1}AS=D=\begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{bmatrix} $
