eigenvector of 2x2 matrix

Question

A = $ \begin{pmatrix} -1 & 1 \\ -1 & -1\end{pmatrix} $

eigenvalues are $-1 + i$ and $-1-i$

I am a bit confused working out the eigenvectors for each eigenvalue, for example

for the second eigenvalue $ \lambda_{2}= -1-i$

I get

$ \begin{pmatrix} i & 1 \\ -1 & i \end{pmatrix} $ $ \begin{pmatrix} x_1 \\ -x_2 \end{pmatrix} $ = $ \begin{pmatrix} 0 \\ 0 \end{pmatrix} $

so first equation I get,

$ ix_1 + x_2 = 0 $

$ -x_1+ix_2 = 0 $

from the first equation

$ ix_1 = -x_2 $

$ x_1 = -x_2/i $

so setting $x_2 = -i, x_1 = 1$

so $ x_{2} = \begin{pmatrix} 1 \\ -i \end{pmatrix} $

but from the second equation I get a differenct eigenvector

$ -x_1 = ix_2 $

if I let $x_2 = 1$ then $x_1=-i$

so I get

$ x_2$ = $ \begin{pmatrix} -i \\ 1 \end{pmatrix} $

Which eigenvector is correct ?

Answer 1

$x_2$ is a complex number, not a vector. Subsitituting $x_2 = -ix_1$ in the second equation gives $-x_1 -i^2x_1 = 0$, which is always true. So the only constraint is $x_2 = -x_1i$, so any eigenvector is of the form $(x_1, -x_1 i)^T$ for any $x_1\in \mathbb{C}$.

Both vectors you gave (and mislabeled) $x_2$ are eigenvectors. Eigenvectors are not unique.