TITLE: When are the eigenvectors of an Hermitian matrix real?
QUESTION [5 upvotes]: Consider a Hermitian matrix $M=M^\dagger$. Clearly, its eigenvalues are real, but what is the condition for the eigenvectors to be real as well?
Edit 1: I consider the entries of $M$ to be complex numbers. Moreover, I call a vector real if all its components are real numbers.
Edit 2: As pointed out in the comments, if $v$ is a real eigenvector of $M$, then also is $i v$. For this reason I might need to emphasise that I do not care about ``global'' complex coefficients.

REPLY [2 votes]: Here is a necessary condition for an eigenvector to be real.
Let $v\in \mathbb R^n$ be a real eigenvector to the real eigenvalue $\lambda$. Split $M$ into real and imaginary part: $M=A+iB$, $A,B\in\mathbb R^{n,n}$.
Then
$$
Mx = (A+iB)x = Ax  +iBx= \lambda x.
$$
Splitting into real and imaginary parts again, it follows $Ax=\lambda x$ and $Bx=0$. That is, $v$ is in the null space of $B$, which is the imaginary part of $M$.
Of course, this is not sufficient to force an eigenvector to be real: the real matrix $\pmatrix{1 & 0\\0&1}$ has
has an orthonormal basis of eigenvectors that are not real
$\pmatrix{1\\i}$, $\pmatrix{1\\-i}$. Multiplication by scalars does not help.