Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-7-\lambda & 0\\
-3 & -7-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
-7-\lambda & 0\\
-3 & -7-\lambda
\end{bmatrix}=0$
$\left (-7- \lambda \right ) (-7- \lambda)=0$
$(-7\lambda )^2=0$
$\lambda_1=\lambda_2=-7$
2. Find eigenvectors:
For $\lambda=-7$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-7-\lambda & 0\\
-3 & -7-\lambda
\end{bmatrix}=\begin{bmatrix}
0 & 6 \\
-3 & 0
\end{bmatrix}$
Reduce row echelon form:
$\begin{bmatrix}
0 & 6 | 0\\
-3 & 0 | 0
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 |0 \\
0 & 1 | 0
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Let $r$ be a free variable.
$\vec{V}=r(0,1) \\
E_2=\{(0,1)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{0,1)\}$ in $R^2$
