Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda & 3 \\
2 & 1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
2-\lambda & 3 \\
2 & 1-\lambda
\end{bmatrix}=0$
$\left (\lambda -4\right ) ( \lambda +1)=0$
$\lambda_1=4, \lambda_2=-1$
2. Find eigenvectors:
For $\lambda=4$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 3 \\
2 & 1-\lambda
\end{bmatrix}=\begin{bmatrix}
-2 & 3 \\
2 & -3
\end{bmatrix}$
We obtain reduced row echelon form:
$\begin{bmatrix}
-2 & 3 | 0 \\
2 & -3 | 0
\end{bmatrix} \approx \begin{bmatrix}
2 & -3 | 0 \\
0 &0 | 0
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Let $r$ be a free variable.
$\vec{V}=r(2,3) \\
E_2=\{r(2,3)\}
\rightarrow dim(E_2)=1$
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 3 \\
2 & 1-\lambda
\end{bmatrix}=\begin{bmatrix}
3 & 3 \\
2 & 2
\end{bmatrix}$
We obtain reduced row echelon form:
$\begin{bmatrix}
3 & 3 | 0 \\
2 & 2| 0
\end{bmatrix} \approx \begin{bmatrix}
1 & 1 | 0 \\
0 &0 | 0
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Let $r$ be a free variable.
$\vec{V}=r(1,1) \\
E_2=\{r(1,1)\}
\rightarrow dim(E_2)=1$
Hence, matrix $A$ is non-defective.
