Symplectic equivalent of commuting matrices

Question

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.

Answer 1

It is not too difficult to prove (if I made no mistakes) that, under the further hypothesis added by Doriano in his answer, i.e. that one of the two matrices, say $A$, is positive definite (the invertibility of $B$ is not necessary), then the two real simmetric matrices $A$ and $B$ satisfying (1) can actually be simultaneously diagonalized by means of a real symplectic matrix $M$. This can be so chosen that the transformed matrices have the form $A'=\text{diag} (a_1, a_1, a_2, a_2, \dots, a_n, a_n)$, $B'=\text{diag} (b_1, b_1, b_2, b_2, \dots, b_n, b_n)$, with $a_i>0\ \forall\ i=1,\dots, n$.

Answer 2

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.

Answer 3

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

Answer 4

I see that the answer of MTyson requires making an additional assumption, i.e., that $A\Omega$ and $B\Omega$ are diagonalizable, in addition to the conditions described in the question. It is possible that the conditions described in the question are too generic, hence they does not lead to any interesting form for $A$ and $B$. Here, I discuss a different additional condition, that is more relevant for my interests. However, I thank the other users who posted interesting answers and suggestions!

The additional condition is that one of the two matrices, $A$ or $B$, is positive definite. Moreover, I will assume that both $A$ and $B$ are invertible; actually, the case of singular matrices can be easily handled in a similar (but longer) way and its discussion does not add too much.

First, I follow the suggestion of Richard and I write $\Omega$ as:

$$ \Omega = \begin{bmatrix} 0 & 1 & & & \cdots \\ -1 & 0 & & \\ & & 0 & 1 &\\ & & -1 & 0 & \\ \vdots & & & & \ddots \end{bmatrix}. $$

I claim that the two matrices $A$ and $B$ can be written as diagonal block matrices: \begin{gather*} A = \begin{bmatrix} \lambda_1 C_1 & & & \\ & \lambda_2 C_2 & & \\ & & \lambda_3 C_3 & \\ & & & \ddots \end{bmatrix} \\ B = \begin{bmatrix} C_1 & & & \\ & C_2 & & \\ & & C_3 & \\ & & & \ddots \end{bmatrix}, \end{gather*} where $C_i$ are symmetric square $2n_i\times 2n_i$ matrices, and $\lambda_i$ and $2n_i$ are the eigenvalues and their multiplicities, respectively, of the matrix $P=B^{-1}A$. Btw, it is important to notice that $P$ is diagonalizable and all the eigenvalues have even (multiple of 2) multiplicity.

Proof

I call $P=B^{-1}A$. From the fact that one of the two $A$ and $B$ is positive definite, $P$ is diagonalizable.

I define $Q=-B^{-1}A\Omega$. It can be easily shown that $Q$ is antisymmetric, $Q=-Q^T$. Thus: $$ P = Q\Omega = B^{-1}A $$ is both the product of two antisymmetric matrices ($Q$ and $\Omega$) and two symmetric matrices ($B^{-1}$ and $A$). This ensures that the eigenvalues, $\lambda_i$, have even multiplicity, $2n_i$.

It can easily be shown that: $$ P \Omega P^T \Omega = \Omega P^T \Omega P $$ This condition is called $\phi_J$-normality in the paper "On the diagonalizability of a matrix by a symplectic equivalence, similarity or congruence transformation" cited by Carlo Beenakker. Together with the condition that $P$ is diagonalizable, from the same paper we conclude that $P$ is symplectically similar to a diagonal matrix: $$ P = M^{-1} D M $$ where $M$ is symplectic.

This can be rewritten as: $$ \tilde{B}^{-1}\tilde{A} = D $$ where $\tilde{A}=M^T A M$ and $\tilde{B}=M^T B M$.

$$ \tilde{A} = \tilde{B} D.$$

It can be noticed that $\tilde{A}$ and $\tilde{B}$ are symmetric. Thus:

$$ \tilde{B} D = D \tilde{B}.$$

Since $\tilde{B}$ commutes with a diagonal matrix, and is symmetric, it has the block structure of the thesis. $\tilde{A}$ is then calculated as $ \tilde{A} = \tilde{B} D$.