Answer
See below
Work Step by Step
1. Find eigenvalues
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix} -\lambda & 4 & -4 \\ -4 & -\lambda & -2\\ 4 & 2& -\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$
$\begin{bmatrix} -\lambda & 4 & -4 \\ -4 & -\lambda & -2\\ 4 & 2 & -\lambda \end{bmatrix}=0$
$\lambda^3+36\lambda=0$
$\lambda_1=0,\lambda_2=6i, \lambda_3=-6i$
2. Find eigenvectors:
For $\lambda=0$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 0 & 4 & -4 \\ -4 & 0 & -2\\ 4 & 2 & 0\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} $
Let $r$ be a free variable.
$\vec{V}=r(-1,2,2)\\
E_1=\{(-1,2,2)\} \\
\rightarrow dim(E_2)=1$
For $\lambda=6i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} -6i & 4 & -4 \\ -4 & -6i & -2\\ 4 & & -6i \end{bmatrix}$
Let $s$ be a free variable.
$\vec{V}=s(2+6i,-4+3i,5)\\
E_2=\{(2+6i,-4+3i,5)\} \\
\rightarrow dim(E_2)=1$
For $\lambda=-6i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 6i & 4 & -4 \\ -4 & 6i & -2\\ 4 & 2 & 6i \end{bmatrix}$
Let $t$ be a free variable.
$\vec{V}=t(2-6i,-4-3i,5)\\
E_3=\{(2-6i,-4-3i,5)\} \\
\rightarrow dim(E_3)=1$
