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