Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
3-\lambda & 1 & 0 \\
-1 & 5-\lambda & 0 \\
0 & 0 & 4 -\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
3-\lambda & 1 & 0 \\
-1 & 5-\lambda & 0 \\
0 & 0 & 4 -\lambda \\
\end{bmatrix}=0$
$\left (3- \lambda \right ) (5- \lambda)(4-\lambda)=0$
$(\lambda -4)^3=0$
$\lambda_1= \lambda_2=\lambda_3=4$
2. Find eigenvectors:
For $\lambda=4$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
3-\lambda & 1 & 0 \\
-1 & 5-\lambda & 0 \\
0 & 0 & 4 -\lambda \\
\end{bmatrix}=\begin{bmatrix}
-1 & 1 & 0 \\
-1 & 1 & 0 \\
0 & 0 & 0\\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$\begin{bmatrix}
1 & -1 & 0 | 0 \\
0 & 0 & 0 | 0\\
0 & 0 & 0 | 0\\
\end{bmatrix}$
Let $r$ and $s$ be free variables.
$\vec{V}=r(1,1,0) +s(0,0,1) \\
E_2=\{r(1,1,0) +s(0,0,1)\}
\rightarrow dim(E_2)=2$
Hence, matrix $A$ is defective because it does not have complete set eigenvectors.
