Answer
$\lambda_1=-2$
$\vec{V}=(-1,0,1)r$
$\lambda_2=-2\pm i$
$\vec{V}=(1,0,0)w+ (0, 1,0)x + (0,0,1)y$
Work Step by Step
1. eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-2-\lambda &1 &0 \\
1& -1-\lambda&-1\\
1&3 &-3-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2\\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
-2-\lambda &1 &0 \\
1& -1-\lambda&-1\\
1&3 &-3-\lambda
\end{vmatrix}=0$
$\lambda_1=-2$ and
$\lambda_2=-2\pm i$
===============================
2. eigenvectors
$\lambda_1=3$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
-2-\lambda_1 &1 &0 \\
1& -1-\lambda_1&-1\\
1&3 &-3-\lambda_1
\end{bmatrix}$
= $\begin{bmatrix}
0&1&0 \\
1 &1 &-1\\
1&3 &-1
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $r$ is a free variable.
$\vec{V}=(-1,0,1)r$
=============================
$\lambda_2=\pm i$
let $B=A-\lambda_2I$
$B= \begin{bmatrix}
-2-\lambda_2 &1 &0 \\
1& -1-\lambda_2&-1\\
1&3 &-3-\lambda_2
\end{bmatrix}$
= $\begin{bmatrix}
-2\pm i&1&0 \\
1 &-1\pm i&-1\\
1&3 &-3-\pm i
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array}\right)
\]
let $w$, $x$ , $y$ are a free variables.
$\vec{V}=(1,0,0)w+ (0, 1,0)x + (0,0,1)y$
