This question will seem over-specific and obscure, but the motivation comes from a problem I am trying to solve in game theory. I hope someone can help, as it requires only linear algebra!
Let $H$ be a real invertible matrix, decomposed into symmetric and antisymmetric parts as $H = S+A$. Assume that $S$ is positive semi-definite, and that there exists a simultaneous eigenvector $u$ of $S$ and $A$ such that $Su = 0$ and $Au = \lambda u$ with non-zero (pure imaginary) $\lambda$. Prove or disprove that $u$ is also an eigenvector of $S_d$, the sub-matrix of $S$ consisting of its diagonal part only.
I can neither prove this nor find a counter-example. It is definitely true for $2 \times 2$ matrices, and I think also for $3 \times 3$. In the general case, notice that $u$ is also an eigenvalue of $H$ since $Hu = Au = \lambda u$. [The assumption that $\lambda \neq 0$ is superfluous since $H$ is assumed invertible.] I tried to use a criterion on the possibility of a matrix having pure imaginary eigenvalues, e.g. that there exists a rank-1 matrix $M$ such that $HM$ is antisymmetric (ref). I have also tried to use a relationship between eigenvectors and diagonal elements of a diagonalisable matrix (ref). I didn't get very far.
EDIT: Thanks to fedja's counter-example below, the answer to this question is no. Any such $u$ is not necessarily an eigenvector of $S_d$. The question I am really interested in, however, is the following. If any two such eigenvectors $u_i$ and $u_j$ exist with distinct eigenvalues, they must be orthogonal since $A$ is anti-symmetric. Can we also prove that $u_i$ and $S_d u_j$ are orthogonal, namely $$ u_i^* S_d u_j = 0 \ ? $$ As you can see, if we could have shown that $u_i$ is an eigenvector of $S_d$, we would be done. This is not true, but fedja's counter-example does not contradict this more restrictive claim.