# 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. Apr 7, 2019 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. Apr 7, 2019 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 $S^\top BS=-\Omega E$. Apr 8, 2019 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. $A=A^T$. Apr 8, 2019 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. Apr 8, 2019 at 10:09

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 $$a_i>0\ \forall\ i=1,\dots, n$$.

• Thank you! I look forward to see a sketch of the proof here, or a complete discussion in a paper (in the second case, please put a link in the answer). Sep 2, 2019 at 11:44

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? Apr 8, 2019 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. Apr 8, 2019 at 21:58
• 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. Apr 10, 2019 at 9:49
• @TobiasDiez and MTyson: thanks to your help, I developed a solution to my own question, which has a quite unexpected form. If you are interested, have a look at my answer here! Jun 19, 2019 at 20:10

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

• I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting. Apr 7, 2019 at 20:27
• 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. Apr 8, 2019 at 9:54
• @ChristianRemling and Carlo Beenakker : thanks to your help, I developed a solution to my own question, which has a quite unexpected form. If you are interested, have a look at my answer here! Jun 19, 2019 at 20:09

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

• I'm not sure of the consensus, but I think it is better form to accept someone else's answer that inspired your own. Jun 20, 2019 at 21:46
• Thanks for the editing and for the comment. I would like to drive the attention of the readers to my answer, which is the final outcome of the discussion. That's why I "accepted" it. If there is a consensus against this, or an alternative suggestion, then I will change. Basically, the question asks what happens if two Hamiltonians are in involution, limiting to quadratic forms defined by $A$ and $B$ and to linear canonical transformations $M$. Thus requiring positive definitness is a more natural condition for me. But I acknowledge the contribution of the other participants. Jun 21, 2019 at 8:41