Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda & 3 & 0 \\
-1 & -\lambda & 1 \\
-2 & -1 & 4-\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
2-\lambda & 3 & 0 \\
-1 & -\lambda & 1 \\
-2 & -1 & 4-\lambda \\
\end{bmatrix}=0$
$\left (2- \lambda \right ) (- \lambda)(4-\lambda)=0$
$(\lambda-2)^3=0$
$\lambda_1= \lambda_2=\lambda_3=2$
2. Find eigenvectors:
For $\lambda=2$
let $B=A-\lambda_1I$
$B=$\begin{bmatrix}
2-\lambda & 3 & 0 \\
-1 & -\lambda & 1 \\
-2 & -1 & 4-\lambda \\
\end{bmatrix}=\begin{bmatrix}
0 & 3 & 0 \\
-1 & -2 & 1 \\
-2 & -1 & 2 \\
\end{bmatrix}$
Then,
$B\vec{V}=\vec{0}$
We obtain reduced row echelon form of B:
$\begin{bmatrix}
0 & 3 & 0 |0 \\
-1 & -2 & 1 | 0\\
-2 & -1 & 2 | 0\\
\end{bmatrix} \approx\approx \begin{bmatrix}
1 & 0 & -1 | 0 \\
0 & 1 & 0 | 0\\
0 & 0 & 0 | 0\\
\end{bmatrix}$
Let $r$ be a free variable.
$\rightarrow \vec{V}=r(1,0,1) \\
E_1=\{(1,0,1)\}
\rightarrow dim(E_1)=1 \ne 3$
Hence, matrix $A$ is defective.
