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