Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 6 \\
2 & -3-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 6 \\
2 & -3-\lambda
\end{bmatrix}=0$
$\left (1- \lambda \right )\left (- \lambda -3 \right )+12=0$
$\lambda^2+2\lambda -15=0$
$\lambda_1=3$ and $\lambda_2=-5$
2. Find eigenvectors:
For $\lambda_1=3$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
1-\lambda_1 & 6 \\
2 & -3-\lambda_1
\end{bmatrix}= \begin{bmatrix}
-2 & 6 \\
2 &-6
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & -3 & 0 \\
0 & 0& 0 \\
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1=3x_2 = 3r$
$\vec{V}=(3,1)r$
For $\lambda_1=-5$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
1-\lambda_1 & 6 \\
2 & -3-\lambda_1
\end{bmatrix}= \begin{bmatrix}
6 & 6 \\
2 &2
\end{bmatrix}$
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 $r$ is a free variable.
Then $x_1=-x_2 = -r$
$\vec{V}=(-1,1)r$
