Assume I have an $n\times n$ block matrix of the form $M=\begin{pmatrix}A & B \\ -B & A\end{pmatrix}$ and it has $n$ independent eigenvectors. Note here that $A$ and $B$ are square matrices (of the same size).
I can see that $\begin{pmatrix}\vec{v} \\ i\vec{v}\end{pmatrix}$ is an eigenvector when $\vec{v}$ is an eigenvector of $A+iB.$ I can also see that $\begin{pmatrix}\vec{w} \\ -i\vec{w}\end{pmatrix}$ is an eigenvector when $\vec{w}$ is an eigenvector of $A-iB.$
However, these two vectors are only orthogonal when $\vec{v}$ and $\vec{w}$ are orthogonal.
So if I find these $n$ eigenvectors, under what conditions will they be independent?
Thanks :)
I just realized the answer to this question. If you normalize these vectors and put them in as columns to some matrix $U,$ this matrix will be unitary. Further, because the matrices that I end up with are Hermitian (a property that happens to be true of the problem I'm working on), all of the vectors $\vec{v}$ and $\vec{w}$ span and thus we have linear independence.