Answer
1.
$\lambda_1=1+ 3i
$ and $\lambda_2=1- 3i$
2.
$\vec{V}=(-(1-i),1)r$
$\vec{V}=(-(1+i),1)r$
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-2-\lambda &-6 \\
3 & 4-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
-2-\lambda &-6 \\
3 & 4-\lambda
\end{vmatrix}=0$
$\lambda_1=1+ 3i
$ and $\lambda_2=1- 3i$
2. $\lambda_1=1+ 3i$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
-2-\lambda_1 &-6 \\
3 & 4-\lambda_1
\end{bmatrix}$
= $\begin{bmatrix}
-3 - 3i& -6\\
3& 3 + 3i
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & 1-i & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1=-(1-i)x_2 = -(1-i) r$
$\vec{V}=(-(1-i),1)r$
=============================
$\lambda_2=1- 3i$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
-2-\lambda_2 &-6 \\
3 & 4-\lambda_2
\end{bmatrix}$
= $\begin{bmatrix}
-3 + 3i& -6\\
3& 3 + 3i
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & 1+i & 0 \\
0 & 0 & 0 \\
\end{array}\right)
\]
let $r$ be a free variable.
Then $x_1=-(1+i)x_2 = -(1+i) r$
$\vec{V}=(-(1+i),1)r$
