Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
3-\lambda & 0 \\
0 & 3-\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
3-\lambda & 0 \\
0 & 3-\lambda \\
\end{bmatrix}=0$
$\left (3- \lambda \right ) (3- \lambda)=0$
$(3-\lambda)^2=0$
$\lambda_1=3, \lambda_2=3$
2. Find eigenvectors:
For $\lambda_1=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
3-\lambda & 0 \\
0 & 3-\lambda \\
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0 & 0 \\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
0 & 0 & 0 \\
0 & 0& 0
\end{array}\right)
\]
let $r$ and $s$ be free variables.
$\vec{V}=r(1,0)+s(0,1) \\
E_1=\{(1,0);(0,1)\}
\rightarrow dim(E_1)=2$
Hence, the matrix $A$ is nondefective.
