When are the eigenvectors of an Hermitian matrix real?

Question

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.

Answer 1

Let $A= \begin{bmatrix} 0 & i \\ -i & 0 \\ \end{bmatrix}$. Then $1$ is an eigenvalue of $A$ , but $ker(I-A)=span\{(i,1)^T\}$

Answer 2

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.