Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
4-\lambda & 6\\
6 & 9-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
4-\lambda & 6\\
6 & 9-\lambda
\end{bmatrix}=0$
$(4- \lambda)(9-\lambda)-36=0$
$\lambda^2-13\lambda=0$
$\lambda_1=0, \lambda_2=13$
2. Find eigenvectors:
For $\lambda=0$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
4-\lambda & 6\\
6 & 9-\lambda
\end{bmatrix}=\begin{bmatrix}
4 & 6 \\
6 & 9
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow 2v_1+3v_2=0$
Let $r$ be a free variable.
$\vec{V}=r(-3,2) \\
E_1=\{(-3,2)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{-3,2)\}$ in $R$
For $\lambda=0$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
4-\lambda & 6\\
6 & 9-\lambda
\end{bmatrix}=\begin{bmatrix}
-9 & 6 \\
6 & -4
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow 3v_1-2v_2=0$
Let $r$ be a free variable.
$\vec{V}=r(2,3) \\
E_2=\{(2,3)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{2,3)\}$ in $R$
Hence, $S=\frac{1}{\sqrt 13}\begin{bmatrix}
-3 & 2\\
2 & 3
\end{bmatrix} =\begin{bmatrix}
-\frac{3}{\sqrt 13} & \frac{2}{\sqrt 13}\\
\frac{2}{\sqrt 13} & \frac{3}{\sqrt 13}
\end{bmatrix} $
