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