Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & -3 & 3\\
-2 & -4-\lambda & 6 \\
-2 & -6 & 8-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & -3 & 3\\
-2 & -4-\lambda & 6 \\
-2 & -6 & 8-\lambda
\end{bmatrix}=0$
$-\lambda^3+5\lambda^2-8\lambda +4=0$
$\lambda_1=\lambda_2=2, \lambda_3=1$
2. Find eigenvectors:
For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1-\lambda & -3 & 3\\
-2 & -4-\lambda & 6 \\
-2 & -6 & 8-\lambda
\end{bmatrix}=\begin{bmatrix}
-1 & -3 & 3\\
-2 & -6 & 6 \\
-2 & -6 & 6
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix} $
Let $r$ and $s$ be free variables.
$\vec{V}=r(-3,1,0)+s(3,0,1) \\
E_1=\{(-3,1,0)+(3,0,1)\}
\rightarrow dim(E_1)=2$
The eigenvectors span $\{(-3,1,0),(3,0,1)\}$ in $R$
For $\lambda=1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1-\lambda & -3 & 3\\
-2 & -4-\lambda & 6 \\
-2 & -6 & 8-\lambda
\end{bmatrix}=\begin{bmatrix}
0 & -3 & 3\\
-2 & -5 & 6 \\
-2 & -6 & 7
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix} $
Let $r$ be free variable.
$\vec{V}=r(1,2,2) \\
E_2=\{(1,2,2)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(1,2,2)\}$ in $R$
We obtain $S=\begin{bmatrix}
-3 & 3 & 1\\
1 & 0 & 2 \\
0 & 1 & 2
\end{bmatrix} \rightarrow
S^{-1}AS=diag(2,2,1)$
