Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
3-\lambda & -2 \\
4 & -1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
3-\lambda & -2 \\
4 & -1-\lambda
\end{bmatrix}=0$
$\left (3- \lambda \right )\left (- \lambda -1 \right )+8=0$
$\lambda^2-2\lambda +5=0$
$\lambda_1=1+2i$ and $\lambda_2=1-2i$
2. Find eigenvectors:
For $\lambda_1=1+2i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
3-\lambda & -2 \\
4 & -1-\lambda
\end{bmatrix}= \begin{bmatrix}
2-2i & -2 \\
4 & -2-2i
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1-i & 1 & 0 \\
0 & 0& 0 \\
\end{array}\right)
\]
let $r$ is a free variable.
Then $(1-i)x_1=x_2 = r$
$\vec{V}=(1,1-i)r$
For $\lambda_1=1-2i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
3-\lambda & -2 \\
4 & -1-\lambda
\end{bmatrix}= \begin{bmatrix}
2+2i & -2 \\
4 & -2+2i
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1+i & 1 & 0 \\
0 & 0& 0 \\
\end{array}\right)
\]
let $r$ is a free variable.
Then $(1+i)x_1=x_2 = r$
$\vec{V}=(1,1+i)r$
