Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 2 \\
2 & -2-\lambda\\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 2 \\
2 & -2-\lambda\\
\end{bmatrix}=0$
$\left (1- \lambda \right )\left (- \lambda -2\right)=0$
$\lambda^2+\lambda-6=0$
$\lambda_1=2, \lambda_2=-3$
2. Reduce A to row-echelor form
$A=\begin{bmatrix}
1 & 2 \\
2 & -2
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 \\
0 & -6
\end{bmatrix} \approx\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}=B$
Then, find eigenvalues:
(b-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 0 \\
0 & 1-\lambda\\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 0 \\
0 & 1-\lambda\\
\end{bmatrix}=0$
$\left (1- \lambda \right )\left (- \lambda +1\right)=0$
$\lambda^2-2\lambda+1=0$
$\lambda_1=1, \lambda_2=1$
Hence, matrix A and B don't have the same eigenvalues.
