Answer
$\lambda_1=0$
$\vec{V}=(-0.6,1.8,1)r$
$\lambda_2=4$
$\vec{V}=(1,1,1)t$
$\lambda_3=2$
$\vec{V}=(1,3,1)s$
Work Step by Step
1. eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda &-1 &3 \\
3& 1-\lambda&0\\
2&-1 &3-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2\\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
2-\lambda &-1 &3 \\
3& 1-\lambda&0\\
2&-1 &3-\lambda
\end{vmatrix}=0$
$\lambda_1=0$
$\lambda_2=4$
$\lambda_3=2$
===============================
2. eigenvectors
$\lambda_1=0$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
2-\lambda_1 &-1 &3 \\
3& 1-\lambda_1&0\\
2&-1 &3-\lambda_1
\end{bmatrix}$
= $\begin{bmatrix}
2&-1 &3 \\
3& 1&0\\
2&-1 &3
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 0 & 0 .6& 0 \\
0 & 1 & -1.8 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $r$ is a free variable.
$\vec{V}=(-0.6,1.8,1)r$
=============================
$\lambda_2=4$
let $B=A-\lambda_2I$
$B= \begin{bmatrix}
2-\lambda_2 &-1 &3 \\
3& 1-\lambda_2&0\\
2&-1 &3-\lambda_2
\end{bmatrix}$
= $\begin{bmatrix}
-2&-1 &3 \\
3& -3&0\\
2&-1 &-1
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 0 & -1 & 0 \\
0 & 1 & -1& 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $t$ is a free variable.
$\vec{V}=(1,1,1)t$
=============================
$\lambda_3=2$
let $B=A-\lambda_3I$
$B= \begin{bmatrix}
2-\lambda_3 &-1 &3 \\
3& 1-\lambda_3&0\\
2&-1 &3-\lambda_3
\end{bmatrix}$
= $\begin{bmatrix}
0&-1 &3 \\
3& -1&0\\
2&-1 &1
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 0 & -1 & 0 \\
0 & 1 & -3& 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $s$ is a free variable.
$\vec{V}=(1,3,1)s$
