Find the general solution:
$x'= \pmatrix{-2&-1\\2&0}x$
I got the eigenvalues to be $\lambda = -1 \pm i$.
Now finding the eigenvectors:
$(\lambda = -1 + i )= v_1 =\pmatrix{-2 -(-1 + i) &-1\\2&-(-1 + i)} =\pmatrix{2 &1-i\\2&1 -i}= \pmatrix{2 &1-i\\0&0} $
Thus the eigenvector is $v_1 = \pmatrix{1\\ -\frac{2}{1-i}}$
but the eigenvector in the solution set is $v_1 = \pmatrix{1\\ -1-i}$
how did they get that?