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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.

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  • $\begingroup$ Do you mean that $S_d$ has zeroes outside the diagonal and coincides with $S$ otherwise? $\endgroup$ – Arnaud Mortier Jun 11 '18 at 15:52
  • $\begingroup$ A eigenvector for a diagonal matrix has a very specific form. Do you know what that is? $\endgroup$ – Paul Jun 11 '18 at 15:52
  • $\begingroup$ Yes to both of your comments. Paul: the eigenvectors of $S_d$ are indeed very simple, but any linear combination of those with the same eigenvalue is also an eigenvector. So the question is whether $u$ is indeed such a vector. $\endgroup$ – Nao Jun 11 '18 at 15:56
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$H=\begin{bmatrix}4&-2&-1&-2\\ -2&1&-2&-4\\1&2&4& -2\\ 2&4&-2&1\end{bmatrix}$, $\lambda=-5i$, $u=\begin{bmatrix}1\\2\\i\\2i\end{bmatrix}$

For the new version take

$H=\begin{bmatrix}4&-2&-1&-2\\ -2&1&-2&-4\\1&2&8&-4\\ 2&4&-4&2\end{bmatrix}$, $\lambda_1=-5i$, $u_1=\begin{bmatrix}1\\2\\i\\2i\end{bmatrix}$, $\lambda_2=+5i$, $u_2=\begin{bmatrix}1\\2\\-i\\-2i\end{bmatrix}$,

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  • $\begingroup$ Thank you fedja for this counter-example! We now know that $u$ is not necessarily an eigenvector of $S_d$. I have edited my question to ask further about the behaviour of $S_d$, which would have been facilitated if all such $u$ were indeed eigenvectors. $\endgroup$ – Nao Jun 20 '18 at 7:08
  • $\begingroup$ @Nao The answer is still "no". Let $H=S+A$, $u$ be as in my example. Consider the block matrix $\begin{bmatrix}S+A&0\\0&-S+A\end{bmatrix}$ and the vectors $\begin{bmatrix}u\\u\end{bmatrix}$, $\begin{bmatrix}u\\-u\end{bmatrix}$. $\endgroup$ – fedja Jun 20 '18 at 13:21
  • $\begingroup$ I don't think your example works, since the given block matrix has symmetric part $\begin{bmatrix} S & 0 \\ 0 & -S \end{bmatrix}$ which is not positive semi-definite. $\endgroup$ – Nao Jun 21 '18 at 17:33
  • $\begingroup$ @Nao OK, then change $-S$ to $2S$ The effect will be the same. $\endgroup$ – fedja Jun 21 '18 at 23:55
  • $\begingroup$ There is an issue with either example. I required that the two eigenvectors we are considering have distinct eigenvalues, whereas $\begin{bmatrix} u \\ u \end{bmatrix}$ and $\begin{bmatrix} u \\ 2u \end{bmatrix}$ have the same. $\endgroup$ – Nao Jun 22 '18 at 10:48

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