Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
5-\lambda &-4 \\
8 & -7-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
5-\lambda &-4 \\
8 & -7-\lambda
\end{bmatrix}=0$
$\left (5- \lambda \right )\left (- \lambda -7 \right )+32=0$
$\lambda^2+2\lambda -3=0$
$\lambda_1=1$ and $\lambda_2=-3$
2. Find eigenvectors:
For $\lambda_1=1$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
5-\lambda_1 &-4 \\
8 & -7-\lambda_1
\end{bmatrix}= \begin{bmatrix}
4 & -4 \\
8 &-8
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
\frac{1}{2} & 1 & 0 \\
1 & 1 & 0 \\
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1=-2x_2 = -2 r$
$\vec{V}=(-2,1)r$
