Answer
See below
Work Step by Step
1. Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix}
1-\lambda & 0 & -1\\
0 & 1-\lambda & 1\\
-1 & 1 & -\lambda
\end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\0\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 1 & -1\\
1 & 1-\lambda & 1\\
-1 & 1 & 1-\lambda
\end{bmatrix}=0$
$(\lambda+1)(1-\lambda)(\lambda-2)=0$
$\lambda_1=-1,\lambda_2=1,\lambda=2$
2. Find eigenvectors: For $\lambda=-1$ let $B=A-\lambda_1I$ $B=\begin{bmatrix}
2 & 0 & -1\\
0 & 2 & 1\\
-1 & 1 & 1 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$
Let $r$ be a free variable.
$\vec{V}=r(1,-1,2)\\
E_1=\{(\frac{1}{\sqrt 6},-\frac{1}{\sqrt 6},\frac{2}{\sqrt 6})\} \rightarrow dim(E_1)=1$
The eigenvectors span $\{(\frac{1}{\sqrt 6},-\frac{1}{\sqrt 6},\frac{2}{\sqrt 6})\} $ in $R$
For $\lambda=1$ let $B=A-\lambda_1I$
$B=\begin{bmatrix}
0 & 0 & -1\\
0 & 0 & 1\\
-1 & 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.
$\vec{V}=r(1,1,0) \\ E_2=\{(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0)\}$ in $R$
For $\lambda=2$ let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-1 & 0 & -1\\
0 & -1 & 1\\
-1 & 1 & -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.
$\vec{V}=r(-1,1,1) \\ E_2=\{(-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3})\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3})\} $ in $R$
Hence, $S=\begin{bmatrix}
\frac{1}{\sqrt 6} & \frac{1}{\sqrt 2} & -\frac{1}{\sqrt 3}\\
-\frac{1}{\sqrt 6} & \frac{1}{\sqrt 2} & \frac{1}{\sqrt 3} \\
\frac{2}{\sqrt 6} & 0 & \frac{1}{\sqrt 3}
\end{bmatrix}$
then $S^TAS =diag(-1,1,2)$
