Answer
1. $\lambda_1=2$ and $\lambda_2=2$
2. No eigenvectors
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda &0 \\
0 & 2-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{vmatrix}
2-\lambda &0 \\
0 & 2-\lambda
\end{vmatrix}=0$
$(2-\lambda)(2-\lambda)=0$
$\lambda_1=2$ and $\lambda_2=2$
2. Let $B=A-\lambda_1I$
$B= \begin{bmatrix}
2-\lambda &0 \\
0 & 2-\lambda
\end{bmatrix}$
= $\begin{bmatrix}
0 &0 \\
0&0
\end{bmatrix}$
No eigenvectors.
