Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} 2-\lambda & -1 & 0 & 1 \\ 0 & 3-\lambda & -1 & 0\\ 0 & 1 & -1-\lambda & 0\\ 0 & -1 & 0 & 3-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \\ v_4\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \\ 0\end{bmatrix}$
$\begin{bmatrix} 2-\lambda & -1 & 0 & 1 \\ 0 & 3-\lambda & -1 & 0\\ 0 & 1 & -1-\lambda & 0\\ 0 & -1 & 0 & 3-\lambda \end{bmatrix}=0$
$( 2-\lambda)^3(3-\lambda)=0$
$\lambda_1=\lambda_2=\lambda_3=2;\lambda_4=3$
2. Find eigenvectors:
For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 2-\lambda & -1 & 0 & 1 \\ 0 & 3-\lambda & -1 & 0\\ 0 & 1 & -1-\lambda & 0\\ 0 & -1 & 0 & 3-\lambda \end{bmatrix}=\begin{bmatrix} 0 & -1 & 0 & 1 \\ 0 & 1 & -1 & 0\\ 0 & 1 & -1 & 0\\ 0 & -1 & 0 & 1\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \end{bmatrix} $
Let $r,s,t$ be free variables.
$\vec{V}=r(-1,0,1,-1)+s(0,-1,-2,0)+t(1,0,0,0)\\
E_1=\{(-1,0,1,-1);(0,-1,-2,0);(1,0,0,0)\} \\
\rightarrow dim(E_2)=3$
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 2-\lambda & -1 & 0 & 1 \\ 0 & 3-\lambda & -1 & 0\\ 0 & 1 & -1-\lambda & 0\\ 0 & -1 & 0 & 3-\lambda \end{bmatrix}=\begin{bmatrix} -1 & -1 & 0 & 1 \\ 0 & 0 & -1 & 0\\ 0 & 1 & -4 & 0\\ 0 & -1 & 0 & 0 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0 \\ 0 \end{bmatrix} $
Let $r$ be a free variable.
$\vec{V}=r(-1,0,0,0)\\
E_1=\{(-1,0,0,0)\} \\
\rightarrow dim(E_2)=1$
Hence, $S=\begin{bmatrix} -1& -1 & 0 & 1\\ 0 & 0 & -1 & 0\\
0 & 1 & -2 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}$
