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