Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
5-\lambda & 5 \\
-2 & -1-\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
5-\lambda & 5 \\
-2 & -1-\lambda \\
\end{bmatrix}=0$
$\left (5- \lambda \right ) (-1- \lambda)=0$
$\lambda^2-4\lambda+5=0$
$\lambda_1=2+i, \lambda_2=2-i$
2. Find eigenvectors:
For $\lambda_1=2+i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
5-\lambda & 5 \\
-2 & -1-\lambda \\
\end{bmatrix}=\begin{bmatrix}
3-i & 5\\
-2 & -3-i\\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
2 & 3 + i & 0 \\
0 & 0& 0
\end{array}\right)
\]
Let $r$ be a free variable.
$\vec{V}=r(-3-i,2) \\
E_1=\{(-3-i,2)\}
\rightarrow dim(E_1)=1$
For $\lambda_1=2-i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
5-\lambda & 5 \\
-2 & -1-\lambda \\
\end{bmatrix}=\begin{bmatrix}
3+i & 5\\
-2 & -3+i\\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
2 & 3 - i & 0 \\
0 & 0& 0
\end{array}\right)
\]
Let $s$ be a free variable.
$\vec{V}=r(-3+i,2) \\
E_2=\{(-3+i,2)\}
\rightarrow dim(E_2)=1$
Hence, matrix $A$ is nondefective.
