Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
3-\lambda & -4 & -1 \\
0 & -1-\lambda & -1 \\
0 & -4 & 2 -\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
3-\lambda & -4 & -1 \\
0 & -1-\lambda & -1 \\
0 & -4 & 2 -\lambda \\
\end{bmatrix}=0$
$\left (3- \lambda \right ) (-1- \lambda)(2-\lambda)=0$
$(\lambda +2)(\lambda -3)^2=0$
$\lambda_1=-2, \lambda_2=\lambda_3=3$
2. Find eigenvectors:
For $\lambda_1=-2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
3-\lambda & -4 & -1 \\
0 & -1-\lambda & -1 \\
0 & -4 & 2 -\lambda \\
\end{bmatrix}=\begin{bmatrix}
5 & -4 & -1 \\
0 & 1 & -1 \\
0 & -4 & 4\\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$\begin{bmatrix}
1 & 0 & -1 | 0 \\
0 & 1 & -1 | 0\\
0 & 0 & 1 | 0\\
\end{bmatrix}$
Let $r$ be a free variable.
$\vec{V}=r(1,1,1) \\
E_1=\{(1,1,1)\}
\rightarrow dim(E_1)=1$
For $\lambda_1=3$
let $C=A-\lambda_1I$
$C=\begin{bmatrix}
3-\lambda & -4 & -1 \\
0 & -1-\lambda & -1 \\
0 & -4 & 2 -\lambda \\
\end{bmatrix}=\begin{bmatrix}
0 & -4 & -1 \\
0 & -4 & -1 \\
0 & -4 & -1\\
\end{bmatrix}$
Then,
$C\vec{V}$=$\vec{0}$
Use reduced row echelon form
$\begin{bmatrix}
0 & 4 & 1 | 0 \\
0 & 0 & 0 | 0\\
0 & 0 & 0 | 0\\
\end{bmatrix}$
Let $r$ and $s$ be free variables.
$\vec{V}=r(1,1,0) +s(0,1,-4) \\
E_2=\{r(1,1,0) +s(0,1,-4)\}
\rightarrow dim(E_2)=2$
Hence, matrix $A$ is nondefective.
