Answer
$S=\begin{bmatrix}
2 & 1\\
1 & -2
\end{bmatrix}$
Work Step by Step
Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 4\\
4 & -5-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 4\\
4 & -5-\lambda
\end{bmatrix}=0$
$(1-\lambda)(-5-\lambda)-16=0$
$(\lambda +7)(\lambda -3)=0$
$\lambda_1=-7, \lambda_2=3$
For $\lambda=-7$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1-\lambda & 4\\
4 & -5-\lambda
\end{bmatrix}=\begin{bmatrix}
8 & 4\\
4 & 2
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow v_1-2v_2=0$
Let $r$ be a free variable.
$\vec{V}=r(2,1) \\
E_1=\{(2,1)\}
\rightarrow dim(E_1)=1$
The eigenvectors span $\{2,1)\}$ in $R$
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1-\lambda & 4\\
4 & -5-\lambda
\end{bmatrix}=\begin{bmatrix}
-2 & 4\\
4 & -8
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow -2v_1+v_2=0$
Let $s$ be a free variable.
$\vec{V}=s(1,-2) \\
E_2=\{(1,-2)\}
\rightarrow dim(E_1)=1$
The eigenvectors span $\{1,-2)\}$ in $R$
Then, we obtain $S=\begin{bmatrix}
2 & 1\\
1 & -2
\end{bmatrix}$
