Answer
$\lambda_1=-2$
$\vec{V}=(-1,1,0)s+(-1, 0 ,1)r$
$\lambda_2=4$
$\vec{V}=(1,1,1)t$
Work Step by Step
1. eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
0-\lambda &2 &2 \\
2& 0-\lambda&2\\
2&2 &0-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2\\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
0-\lambda &2 &2 \\
2& 0-\lambda&2\\
2&2 &0-\lambda
\end{vmatrix}=0$
$\lambda_1=-2$
$\lambda_2=4$
===============================
2. eigenvectors
$\lambda_1=-2$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
0-\lambda_1 &2 &2 \\
2& 0-\lambda_1&2\\
2&2 &0-\lambda_1
\end{bmatrix}$
= $\begin{bmatrix}
2&2 &2 \\
2&2&2\\
2&2 &2
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 1 & 1& 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $r,s$ is a free variable.
$\vec{V}=(-1,1,0)s+(-1, 0 ,1)s$
=============================
$\lambda_2=4$
let $B=A-\lambda_2I$
$B= \begin{bmatrix}
0-\lambda_2 &2 &2 \\
2& 0-\lambda_2&2\\
2&2 &0-\lambda_2
\end{bmatrix}$
= $\begin{bmatrix}
-4&2 &2 \\
2& -4&2\\
2&2 &-4
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 0 & -1 & 0 \\
0 & 1 & -1& 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $t$ is a free variable.
$\vec{V}=(1,1,1)t$
