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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.

Widawensen
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smalldog
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  • Do you mean that $S_d$ has zeroes outside the diagonal and coincides with $S$ otherwise? – Arnaud Mortier Jun 11 '18 at 15:52
  • A eigenvector for a diagonal matrix has a very specific form. Do you know what that is? – Paul Jun 11 '18 at 15:52
  • 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. – smalldog Jun 11 '18 at 15:56

1 Answers1

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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

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

fedja
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  • 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. – smalldog Jun 20 '18 at 07:08
  • @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 $\begin{bmatrix}u\u\end{bmatrix}$, $\begin{bmatrix}u\-u\end{bmatrix}$. – fedja Jun 20 '18 at 13:21
  • 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. – smalldog Jun 21 '18 at 17:33
  • @Nao OK, then change $-S$ to $2S$ The effect will be the same. – fedja Jun 21 '18 at 23:55
  • 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. – smalldog Jun 22 '18 at 10:48
  • @Nao All right, missed that. Will try to fix it later. It is still quite likely that the counterexample is there because the game I'm playing is just that the conditions you impose are invariant under rotations while the diagonal part is not. – fedja Jun 22 '18 at 18:46
  • I see. I was wondering where the examples were coming from. If you find a working counter-example do let me know! – smalldog Jun 23 '18 at 08:02
  • @Nao OK, posted. – fedja Jun 25 '18 at 01:41
  • $H$ is required to be invertible, which is not the case here... – smalldog Jun 25 '18 at 13:55
  • @Nao OK, how about now? – fedja Jun 25 '18 at 15:19
  • Thank you for this, I've accepted your answer after lots of helpful work! I have one further question that would be worth exploring with you -- I don't know if there is a private message system on SE? – smalldog Jun 28 '18 at 13:18
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    I have just posted a new question (https://math.stackexchange.com/questions/2837698/stability-of-a-matrix-product) which would really help me out if you have a chance! – smalldog Jul 01 '18 at 15:53