Answer
$\lambda_1=3$
$\vec{V}=(0,-1,1)r$
and
$\lambda_2=1$
$\vec{V}=(0,1,1)r$
Work Step by Step
1. eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
3-\lambda &0 &0 \\
0& 2-\lambda&-1\\
1&-1 &2-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2\\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
3-\lambda &0 &0 \\
0& 2-\lambda&-1\\
1&-1 &2-\lambda
\end{vmatrix}=0$
$\lambda_1=3$ and
$\lambda_2=1$
===============================
2. eigenvectors
$\lambda_1=3$
let $B=A-\lambda_1I$
$B= \begin{bmatrix}
3-\lambda_1 &0 &0 \\
0& 2-\lambda_1&-1\\
1&-1 &2-\lambda_1
\end{bmatrix}$
= $\begin{bmatrix}
0&0&0 \\
0 &-1 &-1\\
1&-1 &-1
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1=0$
$x_2=-x_3$
$\vec{V}=(0,-1,1)r$
=============================
$\lambda_2=1$
let $B=A-\lambda_2I$
$B= \begin{bmatrix}
3-\lambda_2 &0 &0 \\
0& 2-\lambda_2&-1\\
1&-1 &2-\lambda_2
\end{bmatrix}$
= $\begin{bmatrix}
2&0&0 \\
0 &1 &-1\\
1&-1 &1
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}ccc|c@{}}
1 & 0 & 0 & 0 \\
0 & 1 & -1& 0 \\
0 & 0 & 0 & 0
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1=0$
$x_2=x_3=r$
$\vec{V}=(0,1,1)r$
