This is a follow-up question to the one asked here:

Given a complex symmetric $2n\times2n$-matrix $A$, i.e., $A\in \mathbb{C}^{2n\times2n}$ with $A = A^T$. Is it possible, to block-diagonalize $A$ using a (complex) symplectic matrix $R$, i.e., is there always a matrix $R \in \mathbb{C}^{2n\times2n} $ with

$$ R^T J R = J \tag{1}$$ such that $$ R^T A R = \begin{pmatrix} 0 & D \\ D & 0 \end{pmatrix} \tag{1},$$ where $$ J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$ and $D$ a diagonal matrix. If yes, what would be a good procedure to obtain $D$ and $R$.

some remarks:

- so different from the former question, I assume a complex matrix $A$ and drop the requirement that the matrix is Hamiltonian
- numerical tests on random matrices seem to suggest that this is always possible
- I hope the solution to this problem is not too trivial. I did check some linear-algebra books and did not find the required result