Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} -\lambda & 1 & -2 \\ -1 & -\lambda & 2\\ 2& -2 & -\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$
$\begin{bmatrix} -\lambda & 1 & -2 \\ -1 & -\lambda & 2\\ 2& -2 & -\lambda \end{bmatrix}=0$
$\lambda_1=0,\lambda_2=3i, \lambda_3=-3i$
2. Find eigenvectors:
For $\lambda=0$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 2\\ 2& 0\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} $
Let $r$ be free variables.
$\vec{V}=r(2,2,1)\\
E_1=\{(2,2,1)\} \\
\rightarrow dim(E_2)=1$
For $\lambda=3i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -3i & 1 & -2 \\ -1 & -3i & 2\\ 2& -2 & -3i \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} $
Let $s$ be free variables.
$\vec{V}=s(-4-3i,5,-2+6i)\\
E_2=\{(-4-3i,5,-2+6i)\} \\
\rightarrow dim(E_2)=1$
For $\lambda=-3i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 3i & 1 & -2 \\ -1 & 3i & 2\\ 2& -2 & 3i \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} $
Let $t$ be free variables.
$\vec{V}=t(-4+3i,5,-2-6i)\\
E_3=\{(-4+3i,5,-2-6i)\} \\
\rightarrow dim(E_2)=1$
