Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-\lambda & 2 & -1\\
-2 & -\lambda & -2\\
1 & 2 & -\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
-\lambda & 2 & -1\\
-2 & -\lambda & -2\\
1 & 2 & -\lambda
\end{bmatrix}=0$
$(\lambda^2+9)\lambda=0$
$\lambda_1=0,\lambda_2=3i,\lambda_3=-3i$
2. Find eigenvectors:
For $\lambda=0$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
0 & 2 & -1\\
-2 & 0 & -2\\
1 & 2 & 0
\end{bmatrix}=\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
Let $r$ be a free variable.
$\vec{V}=r(-2,1,2) \\
E_1=\{(-2,1,2)\}
\rightarrow dim(E_1)=1$
The eigenvectors span $\{(-2,1,2)\}$ in $R$
For $\lambda=3i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-3i & 0 & 0\\
3 & -3i & -1\\
0 & 2 & -3i
\end{bmatrix}=\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
Let $s$ be a free variable.
$\vec{V}=s(4+3i,6i-2,5) \\
E_2=\{(4+3i,6i-2,5) \}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(4+3i,6i-2,5) \}$ in $R$
For $\lambda=-3i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
3i & 0 & 0\\
3 & 3i & -1\\
0 & 2 & 3i
\end{bmatrix}=\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
Let $t$ be a free variable.
$\vec{V}=t(4-3i,-2-6i,5) \\
E_3=\{(4-3i,-2-6i,5)\}
\rightarrow dim(E_3)=1$
The eigenvectors span $\{(4-3i,-2-6i,5)\}$ in $R$
Since there are three independent eigenvectors exist, $A$ is diagonalizable.
Obtain $B=\begin{bmatrix}
-2 & 3i+4 & 4-3i\\
1 & 6i-2 & -2-6i\\
2 & 5 & 5
\end{bmatrix} $
