Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-1-\lambda & -2\\
-2 & 2-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
-1-\lambda & -2\\
-2 & 2-\lambda
\end{bmatrix}=0$
$(-1- \lambda)(2-\lambda)=0$
$\lambda^2-\lambda -6=0$
$\lambda_1=3,\lambda_2=-2$
2. Find eigenvectors:
For $\lambda=3$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-1-\lambda & -2\\
-2 & 2-\lambda
\end{bmatrix}=\begin{bmatrix}
-4 & -2 \\
-2 & -1
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow 2v_1+v_2=0$
Let $r$ be a free variable.
$\vec{V}=r(-1,2) \\
E_2=\{(-1,2)\}
\rightarrow dim(E_2)=1$
The eigenvectors span $\{-1,2)\}$ in $R$
For $\lambda=-2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-1-\lambda & -2\\
-2 & 2-\lambda
\end{bmatrix}=\begin{bmatrix}
1 & -2 \\
-2 & 4
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} \\
\rightarrow w_1-2w_2=0$
Let $s$ be a free variable.
$\vec{V}=s(2,1) \\
E_1=\{(2,1)\}
\rightarrow dim(E_1)=1$
The eigenvectors span $\{2,1)\}$ in $R$
Hence, $A$ is diagonalizable
We have $S=\begin{bmatrix}
-1 & 2 \\
2 & 1
\end{bmatrix}$
such that $S^{-1}AS=diag(3,-2)$
