Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}6-\lambda & 5 \\
-5 & -4-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}6-\lambda & 5 \\
-5 & -4-\lambda
\end{bmatrix}=0$
$( \lambda -1)^2=0$
$\lambda_1=\lambda_2=1$
2. Find eigenvectors:
For $\lambda=1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix} 5 & 5 \\
-5 & -5
\end{bmatrix}$
Let $r$ be a free variable.
$\vec{V}=r(1,1) \\
E_1=\{r(1,1)\}
\rightarrow dim(E_2)=1\ne 2$
Hence, matrix $A$ is defective.
