It is well known what happens if two real symmetric matrices commute, i.e. if we have two matrices $A$ and $B$ such that $A=A^T$, $B=B^T$ and $AB=BA$. The answer is given in terms of diagonalization: there is a unitary matrix $M$ such that $A$ and $B$ are transformed into $A'=M^TAM$ and $B'=M^TBM$, and both $A'$ and $B'$ are diagonal.

Here I'm asking if any analogous property holds in the following case.

$A$ and $B$ are symmetric, i.e. $A=A^T$ and $B=B^T$. The following property holds:

$$A\Omega B=B\Omega A$$ (1)

where $\Omega$ is the matrix defining the symplectic bilinear form (skew-symmetric, nonsingular, and hollow), e.g.:

$$\Omega = \begin{bmatrix}0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}$$

The allowed transformations are the symplectic matrices $M$, i.e. matrices for which the following holds:

$$M^T\Omega M=\Omega$$

The transformed matrices are $A'=M^TAM$ and $B'=M^TBM$.

My question is if there is a form into which $A'$ and $B'$ can be put, by means of a suitable $M$, provided that Eq.1 holds.

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    $\begingroup$ No idea, but a comment, a bit unrelated: for many computations with such things it's better to write $\Omega$ as $diag(M,\ldots,M)$ where $M=(^0_{-1}{\ }^1_0)$. Of course, in practice, it doesn't make that much of a difference. $\endgroup$ – Richard Apr 7 at 17:24
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    $\begingroup$ For $n=2$, there's the formula $\Omega A\Omega = (\det A) A^{-1T}$ ($=(\det A) A^{-1}$ here), so if $\det A=\det B$, then (1) says that $A^{-1}B=B^{-1}A$, and in general, there's an extra constant. Of course, all this is a far cry from the full question, but it might give a hint. $\endgroup$ – Christian Remling Apr 7 at 18:09
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    $\begingroup$ If $A$ and $B$ are complex matrices for which $\Omega A$ and $\Omega B$ are (anti)symmetric and commute (the latter is equivalent to condition $(1)$), then there's a symplectic $S$ such that $S^{-1}\Omega AS=D$ and $S^{-1}\Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^\top AS=-\Omega D$ and $S^\top BS=-\Omega E$. $\endgroup$ – MTyson Apr 8 at 0:18
  • $\begingroup$ Interesting! Could you please make this an answer, and maybe extend it a bit? For example, I still do not understand if this applies when $A$ is symmetric, i.e. $A=A^T$. $\endgroup$ – Doriano Brogioli Apr 8 at 9:52
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    $\begingroup$ @DorianoBrogioli --- the symmetry of $A$ ensures that $\Omega A$ is Hamiltonian (symplectic skew-symmetric), which is the condition needed for Lemma 17 mentioned by MTyson to apply. $\endgroup$ – Carlo Beenakker Apr 8 at 10:09

Let $A$ and $B$ be complex symmetric matrices of even dimension which satisfy condition $(1)$ and for which $\Omega A$ and $\Omega B$ are diagonalizable. Then $\Omega A$ and $\Omega B$ are Hamiltonian (not (anti)symmetric as my comment said): $$\Omega^\top (\Omega A)^\top \Omega=-\Omega A^\top\Omega^\top\Omega=-\Omega A.$$ Condition $(1)$ means that $\Omega A$ and $\Omega B$ commute. By Lemma 17 in the paper in Carlo's answer, there's a symplectic $S$ such that $S^{-1}\Omega AS=D$ and $S^{-1}\Omega BS=E$ are diagonal. Hence $S^\top AS=-\Omega D$ and $S^\top BS=-\Omega E$ are of the same form.

  • $\begingroup$ Very useful and complete. Could you please comment on the requisite that $\Omega A$ and $\Omega B$ are diagonalizable? Can be expressed as a more explicit condition on $A$ and $B$? And what happens if they are not diagonalizable? $\endgroup$ – Doriano Brogioli Apr 8 at 20:59
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    $\begingroup$ @DorianoBrogioli If $S^\top AS=-\Omega D$ then $S^{-1}\Omega AS=D$, so this is a necessary condition. I don't know any equivalent conditions on $A$ nor if there's a Jordan-like form that the matrices can always be put into. $\endgroup$ – MTyson Apr 8 at 21:58
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    $\begingroup$ There is a symplectic version of the Jordan normal form (of a single matrix). Have a look at the paper Conjugacy classes in linear groups. $\endgroup$ – Tobias Diez Apr 10 at 9:49

The symplectic counterpart of the fact that a family of commuting diagonalizable matrices is simultaneously diagonalizable is discussed in section 3.1 of On the diagonalizability of a matrix by a symplectic equivalence, similarity or congruence transformation

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    $\begingroup$ I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting. $\endgroup$ – Christian Remling Apr 7 at 20:27
  • $\begingroup$ I agree. As far as I can understand, the paper has to do with the topic but does not directly contain the answer, and further elaboration and work is needed. Maybe the comment of MTyson (to the question) can help. $\endgroup$ – Doriano Brogioli Apr 8 at 9:54

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