# Symplectic equivalent of commuting matrices

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.

• 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. – Richard Apr 7 at 17:24
• 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. – Christian Remling Apr 7 at 18:09
• 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 $S^\top BS=-\Omega E$. – MTyson Apr 8 at 0:18
• 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. $A=A^T$. – Doriano Brogioli Apr 8 at 9:52
• @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. – 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.
• 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? – Doriano Brogioli Apr 8 at 20:59
• @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. – MTyson Apr 8 at 21:58