Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
7-\lambda & -8 & 6 \\
8 & -9-\lambda & 6\\
0 & 0 & -1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
7-\lambda & -8 & 6 \\
8 & -9-\lambda & 6\\
0 & 0 & -1-\lambda
\end{bmatrix}=0$
$\left (7- \lambda \right )\left (- \lambda -9 \right )(-1-\lambda)=0$
$\lambda^3+3\lambda^2+3\lambda +1=0$
$\lambda_1=\lambda_2=\lambda_3=-1$
2. Find eigenvectors:
For $\lambda_1=-1$
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,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & -1 & \frac{3}{4} & 0 \\
0 & 0& 0 \\
0 & 0 & 0
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1+x_2 +\frac{3}{4}x_3= 0 \\
$
$\vec{V}=(1,1,0)r +s(-3,0,4)$
