Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
4-\lambda & 1 & 6 \\
-4 & 0-\lambda & 7 \\
0 & 0 & -3 -\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
4-\lambda & 1 & 6 \\
-4 & -\lambda & 7 \\
0 & 0 & -3 -\lambda \\
\end{bmatrix}=0$
$\left (4- \lambda \right ) (- \lambda)(-3-\lambda)=0$
$(\lambda+3)(\lambda -2)^2=0$
$\lambda_1=-3, \lambda_2=\lambda_3=2$
2. Find eigenvectors:
For $\lambda=-3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
4-\lambda & 1 & 6 \\
-4 & 0-\lambda & 7 \\
0 & 0 & -3 -\lambda \\
\end{bmatrix}=\begin{bmatrix}
7 & 1 & 6 \\
-4 & 3 & 7 \\
0 & 0 & 0\\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Let $r$ be a free variable.
$\vec{V}=r(-11,1,1) \\
E_2=\{r(-1,1,1)\}
\rightarrow dim(E_2)=1$
For $\lambda=-2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
4-\lambda & 1 & 6 \\
-4 & 0-\lambda & 7 \\
0 & 0 & -3 -\lambda \\
\end{bmatrix}=\begin{bmatrix}
2 & 1 & 6 \\
-4 & -2 & 7 \\
0 & 0 & -5\\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$\begin{bmatrix}
2 & 1 & 0 | 0 \\
0 & 0 & 1 | 0\\
0 & 0 & 0 | 0\\
\end{bmatrix}$
Let $s$ be a free variable.
$\vec{V}=s(-1,2,0) \\
E_2=\{s(-1,2,0) \}
\rightarrow dim(E_2)=1 \ne 2$
Hence, matrix $A$ is defective.
