Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-9-\lambda & 0\\
4 & -9-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
-9-\lambda & 0\\
4 & -9-\lambda
\end{bmatrix}=0$
$(-9- \lambda)^2=0$
$\lambda_1=\lambda_2=-9$
2. Find eigenvectors:
For $\lambda=-9$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-9-\lambda & 0\\
4 & -9-\lambda
\end{bmatrix}=\begin{bmatrix}
0 & 0 \\
4 & 0
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow x_1=$
Let $r$ be a free variable.
$\vec{V}=r(0,x_2) \\
E_2=\{(0,x_2)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{0,x_2)\}$ in $R$
Hence, $A$ is not diagonalizable
