I have this matrix
\begin{pmatrix} -3 & 1 \\ -2 & -1 \end{pmatrix}
and I found the eigenvalues, which are $\lambda_{1} = -2+i \ $ and $\ \lambda_{2}= -2-i$,
and now I have to find the eigenvectors, but they are with complex values and I don't know the steps for that.
Edit: So for the first eigenvalue $\lambda_{1}$= -2 + i, I tried:
\begin{pmatrix} -3-(-2+i) & 1 \\ -2 & -1-(-2+i) \end{pmatrix}
and then I've got:
(-1-i)x + y = 0
-2x + (1-i)y = 0
What should I do next?
Here's a solution of an analogous problem as a starting point for you:
Here's a matrix with eigenvalues $\pm i$: $$ A = \pmatrix{0 & 1 \\ -1 & 0} $$ I'll find the eigenvector for $+i$. I want to find a vector $v$ with $$ Av = i ~v, $$ right? I convert that to $$ Av = iI ~v\\ (A-iI)v = 0 $$ So now I want to find the kernel of $$ \pmatrix{-i & 1 \\ -1 & -i}, $$ so I need to solve $$ \pmatrix{-i & 1 \\ -1 & -i} \cdot \pmatrix{x \\ y} = \pmatrix{0 \\ 0} $$ which becomes $$ -ix + y = 0 \\ -x - iy = 0 $$ we see that the second equation is just $-i$ times the first one, so we can simply solve the first equation. Picking $x = 1$ (because it's easy!), we get $y = i$, so one eigenvector is $$ \pmatrix{1\\i}. $$ Any (complex) nonzero multiple of this is also an eigenvector, so $$ \pmatrix{i\\-1}, \pmatrix{2\\2i}, \pmatrix{1+i\\-1+i} $$ are all eigenvectors as well. (If we'd chosen $x = i, x = 2, x = 1 + i$, they're what we'd have found when we solved for $y$.)