Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-1-\lambda & 1 & 0 & 0\\
0 & -1-\lambda & 0 & 0\\
0& 0 & -1-\lambda & 0\\
0 & 0 & 0 & 1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2 \\
v_3
\end{bmatrix}=\begin{bmatrix}
0\\
0 \\
0
\end{bmatrix}$
$\begin{bmatrix}
-1-\lambda & 1 & 0 & 0\\
0 & -1-\lambda & 0 & 0\\
0& 0 & -1-\lambda & 0\\
0 & 0 & 0 & 1-\lambda
\end{bmatrix}=0$
$(\lambda-1)^2(\lambda+1)^2=0$
$\lambda_1=\lambda_2=1,\lambda_3=\lambda_4=-1$
2. Find eigenvectors:
For $\lambda=-1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
0& 0 & 0 & 0\\
0 & 0 & 0 & 2
\end{bmatrix}=\begin{bmatrix}
v_1\\
v_2 \\
v_3 \\ v_4
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0 \\ 0
\end{bmatrix} $
Let $r$ be a free variable.
$\vec{V}=r(1,0,1,0) \\
E_1=\{(1,0,1,0)\}
\rightarrow dim(E_1)=2$
The eigenvectors span $\{(1,0,1,0)\}$ in $R$
For $\lambda=1$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-2 & 1 & 0 & 0\\
0 & -2 & 0 & 0\\
0& 0 & -2 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}=\begin{bmatrix}
v_1\\
v_2 \\
v_3 \\ v_4
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0 \\ 0
\end{bmatrix} $
Let $r$ be a free variable.
$\vec{V}=r(0,0,0,1)\\
E_2=\{(0,0,0,1)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{(0,0,0,1)\}$ in $R$
Since there are three independent eigenvectors exist, $A$ is not diagonalizable.
