Answer
$\lambda_1=1$
$\vec{V}=(1,0,-1)r$
$\lambda_2 =2$
$\vec{V}=(-0.5,1,0)v + (0.5,0,1)w$
Work Step by Step
1. eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
0-\lambda &1 &-1 \\
0& 2-\lambda&0\\
2&-1 &3-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2\\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
0-\lambda &1 &-1 \\
0& 2-\lambda&0\\
2&-1 &3-\lambda
\end{vmatrix}=0$
$\lambda_1=1$
$\lambda_2 =2$
$\lambda_3 =2$
===============================
2. eigenvectors
$\lambda_1=1$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
0-\lambda_1 &1 &-1 \\
0& 2-\lambda_1&0\\
2&-1 &3-\lambda_1
\end{bmatrix}$
= $\begin{bmatrix}
-1&1&-1 \\
0 &1 &0\\
2&-1 &2
\end{bmatrix}$
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.
$\vec{V}=(1,0,-1)r$
=============================
$\lambda_2=2$
let $B=A-\lambda_2I$
$B= \begin{bmatrix}
0-\lambda_2 &1 &-1 \\
0& 2-\lambda_2&0\\
2&-1 &3-\lambda_2
\end{bmatrix}$
= $\begin{bmatrix}
-2&1&-1 \\
0 &0 &0\\
2&-1 &1
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & -0.5 & 0.5 & 0 \\
0 & 0 & 0& 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $v$and $w$ are a free variable.
$\vec{V}=(-0.5,1,0)v + (0.5,0,1)w$
