Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
9-\lambda & 5 & -5\\
0 & -1-\lambda & 0\\
10 & 5 & -6 - \lambda\end{bmatrix}
\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix}$
$\begin{bmatrix}
9-\lambda & 5 & -5\\
0 & -1-\lambda & 0\\
10 & 5 & -6 - \lambda\end{bmatrix}=0$
$\lambda_1=4, \lambda_2=\lambda_3=-1$
2. Find eigenvectors:
For $\lambda=4$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
9-\lambda & 5 & -5\\
0 & -1-\lambda & 0\\
10 & 5 & -6 - \lambda\end{bmatrix}
=\begin{bmatrix}
5 & 5 & -5\\
4 & -5 & 0\\
10 & 5 & 2
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix} \\$
Let $r$ be a free variable.
$\vec{V}=r(1,0,1) \\
E_1=\{(1,0,1)\} \\
\rightarrow dim(E_2)=1$
The eigenvectors span $\{1,0,1)\}$ in $R$
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
9-\lambda & 5 & -5\\
0 & -1-\lambda & 0\\
10 & 5 & -6 - \lambda\end{bmatrix}
=\begin{bmatrix}
10 & 5 & -5\\
0 & 0 & 0\\
10 & 5 & -5
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix} \\$
Let $r,s$ be a free variable.
$\vec{V}=r(\frac{1}{2},0,1)+s(-\frac{1}{2},1,0) \\
E_1=\{(\frac{1}{2},0,1); (-\frac{1}{2},1,0)\} \\
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(\frac{1}{2},0,1); (-\frac{1}{2},1,0)\}$ in $R$
$S=\begin{bmatrix} 1 & \frac{1}{2} & - \frac{1}{2}\\ 0 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \\
\rightarrow S^{-1}AS=D=\begin{bmatrix} 4 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $
