Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
7-\lambda & -8 & 6 \\
8 & -9-\lambda & 6 \\
0 & 0 & -1-\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
7-\lambda & -8 & 6 \\
8 & -9-\lambda & 6 \\
0 & 0 & -1-\lambda \\
\end{bmatrix}=0$
$\left (7- \lambda \right ) (-9- \lambda)(-1-\lambda)=0$
$(\lambda+1)^3=0$
$\lambda_1= \lambda_2=\lambda_3=-1$
2. Find eigenvectors:
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
7-\lambda & -8 & 6 \\
8 & -9-\lambda & 6 \\
0 & 0 & -1-\lambda \\
\end{bmatrix}=\begin{bmatrix}
8& -8 & 6 \\
8 & -8 & 6 \\
0 & 0 & 0 \\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
We obtain reduced row echelon form of B:
$\begin{bmatrix}
8 & -8 & 6 |0 \\
8 & -8 & 6 | 0\\
0 & 0 & 0 | 0\\
\end{bmatrix} \approx \begin{bmatrix}
4 & -4 & 3 | 0 \\
0 & 0 & 0 | 0\\
0 & 0 & 0 | 0\\
\end{bmatrix}$
Let $r,s$ be free variables.
$\rightarrow \vec{V}=r(1,0,0)+s(-3,0,4) \\
E_2=\{(1,0,0),(-3,0,4)\}
\rightarrow dim(E_2)=2 \ne 3$
Hence, matrix $A$ is defective.
