Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 2 \\
-2 & 5-\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 2 \\
-2 & 5-\lambda \\
\end{bmatrix}=0$
$\left (1- \lambda \right ) (5- \lambda)=0$
$(\lambda-3)^2=0$
$\lambda_1=3, \lambda_2=3$
2. Find eigenvectors:
For $\lambda_1=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1-\lambda & 2 \\
-2 & 5-\lambda \\
\end{bmatrix}=\begin{bmatrix}
-2 & 2\\
-2 & 2\\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & -1 & 0 \\
0 & 0& 0
\end{array}\right)
\]
Let $r$ be a free variable.
$\vec{V}=r(1,1) \\
E_1=\{(1,1)\}
\rightarrow dim(E_1)=1$
Since $A$ doesn't have complete set of eigenvectors, the matrix $A$ is defective.
