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