Answer
1. $\lambda_1= 4 + 3i$ and $\lambda_2= 4 - 3i$
2.
$\vec{V}=(−(0.5+0.5i),1)r$
$\vec{V}=(−(0.5-0.5i),1)r$
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
7-\lambda &3 \\
-6& 1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
7-\lambda &3 \\
-6& 1-\lambda
\end{vmatrix}=0$
$\lambda_1= 4 + 3i$ and $\lambda_2= 4 - 3i$
2. $\lambda_1= 4 + 3i$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
7-\lambda_1 &3 \\
-6& 1-\lambda_1
\end{bmatrix}$
= $\begin{bmatrix}
3 - 3i &3 \\
-6&-3 - 3i
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & 0.5 + 0.5i
& 0 \\
0 & 0 & 0 \\
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1=-(0.5 + 0.5i)x_2 = -(0.5 + 0.5i) r$
$\vec{V}=(−(0.5+0.5i),1)r$
========================
$\lambda_2= 4 - 3i$
let $B=A-\lambda_2I$
$B= \begin{bmatrix}
7-\lambda_2 &3 \\
-6& 1-\lambda_2
\end{bmatrix}$
= $\begin{bmatrix}
3 + 3i &3 \\
-6&-3 + 3i
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & 0.5 - 0.5i
& 0 \\
0 & 0 & 0 \\
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1=-(0.5 - 0.5i)x_2 = -(0.5 - 0.5i) r$
$\vec{V}=(−(0.5-0.5i),1)r$
