Symplectic block-diagonalization of a complex symmetric matrix 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

 A: This fails even for $n=1$.  In this case, the matrix
$$
A = \begin{pmatrix} 1&0\\0&0\end{pmatrix}
$$
can't be diagonalized in the form that you want because it's not zero, yet its determinant vanishes.
There is a test for when this can be done, though.  It's enough to have $JA$ be semi-simple (i.e., diagonalizable).  In this case, the eigenvalues of $JA$ come in opposite pairs, and, if $\lambda_1,\ldots,\lambda_n,-\lambda_1,\ldots,-\lambda_n$ is a list of the eigenvalues of $JA$, then you can let $D$ be the diagonal matrix whose entries are $\lambda_1,\ldots,\lambda_n$.
The thing you want to think about is that the Lie algebra of the symplectic group is exactly the set of matrices of the form $JA$ where $A$ is symmetric, and you are trying to conjugate $JA$ into the Cartan subalgebra consisting of the diagonal matrices of this form via the adjoint representation of the symplectic group on its Lie algebra.   
(It's a general fact that you can conjugate the semi-simple elements of the Lie algebra into a Cartan subalgebra via the adjoint action, but you can't conjugate the nilpotent elements into a Cartan subalgebra.)
