Eigenvalue and Eigenvector with (-3,1;2,-2)

Question

$\begin{pmatrix}-3 & 1 \\ 2 & -2 \end{pmatrix}$

First of all, why can I never align my matrix??

Second I found the eigenvalues to be of multiplicity two at $\lambda = -2$

Which reduced to $\begin{pmatrix}-1 & 1 \\ 2 & 0 \end{pmatrix}$ = $\begin{pmatrix}0\\ 0\end{pmatrix}$, which gives me $x_1 = x_2 = 0$, is this correct? My one given vector is then $ k \begin{pmatrix}0\\ 0\end{pmatrix} = \vec0$?

Is this correct?

Answer 1

First of all, eigenvectors need to be non-zero.

Second, I don't think you found the eigenvalues correctly.

$|\lambda I - A| = \left|\begin{matrix}\lambda+3 & -1 \\ -2 & \lambda+2 \end{matrix}\right| = (\lambda+3)(\lambda+2)-(-1)(-2) = \lambda^2 + 5\lambda + 4 = 0$

Can you solve this quadratic? The solutions are not $\lambda = -2$ with multiplicity $2$.