Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
4-\lambda & 0 & 0\\
3 & -1-\lambda & -1\\
0 & 2 & 1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
4-\lambda & 0 & 0\\
3 & -1-\lambda & -1\\
0 & 2 & 1-\lambda
\end{bmatrix}=0$
$(\lambda^2+1)(\lambda-4)=0$
$\lambda_1=4,\lambda_2=i,\lambda_3=-i$
2. Find eigenvectors:
For $\lambda=4$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
0 & 0 & 0\\
3 & -5 & -1\\
0 & 2 & -3
\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.
$\vec{V}=r(17,9,6) \\
E_1=\{(17,9,6)\}
\rightarrow dim(E_1)=1$
The eigenvectors span $\{(17,9,6)\}$ in $R$
For $\lambda=i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
4-i & 0 & 0\\
3 & -1-i & -1\\
0 & 2 & 1-i
\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.
$\vec{V}=s(0,1-i,2) \\
E_2=\{(0,1-i,2)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(0,1-i,2)\}$ in $R$
For $\lambda=-i$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
4+i & 0 & 0\\
3 & -1+i & -1\\
0 & 2 & 1+i
\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.
$\vec{V}=t(0,1+i,2) \\
E_3=\{(0,1+i,2)\}
\rightarrow dim(E_3)=1$
The eigenvectors span $\{(0,1+i,2)\}$ in $R$
Since there are three independent eigenvectors exist, $A$ is diagonalizable.
Obtain $B=\begin{bmatrix}
17 & 0 & 0\\
9 & 1-i & 1+i\\
6 & 2 & 2
\end{bmatrix} $
