Let $A = \begin{pmatrix}0& 1\\ -1& 0\end{pmatrix}$. Then $ \begin{pmatrix}1\\-i\end{pmatrix}$
A. is an eigenvector with eigenvalue 1.
B. is an eigenvector with eigenvalue -1.
C. is an eigenvector with eigenvalue i.
D. is an eigenvector with eigenvalue -i.
E. Is not an eigenvector
The answer is D. How do I get to that answer?
I think I correctly get the eigenvalues i, -i, but when I try to get the associated eigenvectors I get something very different.
When I solve for $\lambda = -i$, I get the matrix $\begin{pmatrix}-i& -1\\ 1& -i\end{pmatrix}$. After computing the solution, I get $ \begin{pmatrix}i\\1\end{pmatrix}$.
EDIT: I figured it out. I needed to set the free parameter to $-i$, which makes the first value in the eigenvector a 1.
$$\begin{pmatrix}0& 1\\ -1& 0\end{pmatrix}\begin{pmatrix}1\\-i\end{pmatrix} = \begin{pmatrix}-i\\-1\end{pmatrix} =-i \begin{pmatrix}1\\-i\end{pmatrix} $$ Thus $-i$ is the eigenvalue.