Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & -8\\
2 & -7-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & -8\\
2 & -7-\lambda
\end{bmatrix}=0$
$(-7- \lambda)(1-\lambda)=0$
$\lambda^2+6\lambda +9=0$
$\lambda_1=\lambda_2=-3$
2. Find eigenvectors:
For $\lambda=-3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1-\lambda & -8\\
2 & -7-\lambda
\end{bmatrix}=\begin{bmatrix}
-4 & -8 \\
2 & 4
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow v_1-2v_2=0$
Let $r$ be a free variable.
$\vec{V}=r(2,1) \\
E_2=\{(2,1)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{2,1)\}$ in $R$
Since there is just only one independent eigenvector exists, $A$ is not diagonalizable.
