Answer
$S=\begin{bmatrix}
\frac{1}{\sqrt 5}& \frac{2}{\sqrt 5}\\
-\frac{2}{\sqrt 5} & \frac{1}{\sqrt 5}
\end{bmatrix}$
Work Step by Step
Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda & 2\\
2 & -1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
2-\lambda & 2\\
2 & -1-\lambda
\end{bmatrix}=0$
$(2-\lambda)(-1-\lambda)-4=0$
$\lambda^2-\lambda-6=0$
$\lambda_1=-2, \lambda_2=3$
For $\lambda=-2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 2\\
2 & -1-\lambda
\end{bmatrix}=\begin{bmatrix}
4 & 2\\
2 & 1
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow v_1-2v_2=0$
Let $r$ be a free variable.
$\vec{V}=r(1,-2) \\
E_1=\{(1,-2)\}
\rightarrow dim(E_1)=1$
The eigenvectors span $\{1,-2)\}$ in $R$
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 2\\
2 & -1-\lambda
\end{bmatrix}=\begin{bmatrix}
-1 & 2\\
2 & -4
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow -2v_1+v_2=0$
Let $s$ be a free variable.
$\vec{V}=s(2,1) \\
E_2=\{(2,1)\}
\rightarrow dim(E_1)=1$
The eigenvectors span $\{2,1)\}$ in $R$
Then, we obtain $S=\frac{x}{\sqrt 5}\begin{bmatrix}
1 & 2\\
-2 & 1
\end{bmatrix}=\begin{bmatrix}
\frac{1}{\sqrt 5}& \frac{2}{\sqrt 5}\\
-\frac{2}{\sqrt 5} & \frac{1}{\sqrt 5}
\end{bmatrix}$
