Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda & 0 & 0 \\
0 & 2-\lambda & 0 \\
0 & 0 & 2 -\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
2-\lambda & 0 & 0 \\
0 & 2-\lambda & 0 \\
0 & 0 & 2 -\lambda \\
\end{bmatrix}=0$
$\left (2- \lambda \right ) (2- \lambda)(2-\lambda)=0$
$(2-\lambda)^3=0$
$\lambda_1= \lambda_2=\lambda_3=2$
2. Find eigenvectors:
For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
2-\lambda & 0 & 0 \\
0 & 2-\lambda & 0 \\
0 & 0 & 2 -\lambda \\
\end{bmatrix}=\begin{bmatrix}
0& 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Let $r,s$ and $t$ be free variables.
$\rightarrow \vec{V}=r(1,0,0)+s(0,1,0)+t(0,0,1) \\
E_2=\{(1,0,0),(0,1,0),(0,0,1)\}
\rightarrow dim(E_2)=3 $
Hence, matrix $A$ is non-defective.
