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