Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & -2 & 0\\
2 & -3-\lambda & 0\\
2 & -2 & -1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & -2 & 0\\
2 & -3-\lambda & 0\\
2 & -2 & -1-\lambda
\end{bmatrix}=0$
$(\lambda+1)^3=0$
$\lambda_1=\lambda_2=\lambda_3=-1$
2. Find eigenvectors:
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2 & -2 & 0\\
2 & -2 & 0\\
2 & -2 & 0
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} \\
\rightarrow v_1-2v_2=0$
Let $r,s$ be a free variable.
$\vec{V}=r(1,1,0)+s(0,0,1) \\
E_2=\{(1,1,0);(0,0,1)\}
\rightarrow dim(E_2)=2$
The eigenvectors span $\{(1,1,0);(0,0,1)\}$ in $R$
Since there are just two independent eigenvectors exist, $A$ is not diagonalizable.
