Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda & 2 & -1 \\
2 & 1-\lambda & -1 \\
2 & 3 & -1-\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
2-\lambda & 2 & -1 \\
2 & 1-\lambda & -1 \\
2 & 3 & -1-\lambda \\
\end{bmatrix}=0$
$\left (2- \lambda \right ) (1- \lambda)(-1-\lambda)=0$
$(\lambda-2).\lambda^2=0$
$\lambda_1= \lambda_2=2,\lambda_3=0$
2. Find eigenvectors:
For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 2 & -1 \\
2 & 1-\lambda & -1 \\
2 & 3 & -1-\lambda \\
\end{bmatrix}=\begin{bmatrix}
0 & 2 & -1 \\
2 & -1 & -1 \\
2 & 3 & -3 \\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
We obtain reduced row echelon form of B:
$\begin{bmatrix}
0 & 2 & -1 |0 \\
2 & -1 & -1|0 \\
2 & 3 & -3 |0\\
\end{bmatrix}\approx \begin{bmatrix}
4 & 0 & -3 | 0 \\
0 & 2 & -1 | 0\\
0 & 0 & 0 | 0\\
\end{bmatrix}$
Let $r$ be a free variable.
$\rightarrow \vec{V}=r(3,2,4) \\
E_2=\{(3,2,4)\}
\rightarrow dim(E_2)=1 $
For $\lambda=0$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 2 & -1 \\
2 & 1-\lambda & -1 \\
2 & 3 & -1-\lambda \\
\end{bmatrix}=\begin{bmatrix}
2 & 2 & -1 \\
2 & 1 & -1 \\
2 & 3 & -1 \\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
We obtain reduced row echelon form of B:
$\begin{bmatrix}
2 & 2 & -1 |0 \\
2 & 1 & -1|0 \\
2 & 3 & -1 |0\\
\end{bmatrix}\approx \begin{bmatrix}
2 & 0 & -1 | 0 \\
0 & 1 & 0 | 0\\
0 & 0 & 0 | 0\\
\end{bmatrix}$
Let $s$ be a free variable.
$\rightarrow \vec{V}=s(1,0,2) \\
E_2=\{(1,0,2)\}
\rightarrow dim(E_2)=1$
Hence, matrix $A$ is defective.
