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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
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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.)

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    $\begingroup$ Thank you for this simple counterexample (it is hard to sample a null set with random numbers). I still have one question: why is it that the eigenvalues of JA come in opposite pairs? (I guess this should be immediately obvious but sadly not for me) $\endgroup$
    – Fabian
    Commented Apr 23, 2014 at 18:29
  • $\begingroup$ @Fabian: One way to see it is to notice that $J^T = J^{-1}$, so $J(JA)^TJ^T$ has the same eigenvalues as $(JA)^T$, and hence the same eigenvalues as $JA$ itself. But if we expand things, we see that $J(JA)^TJ^T = JAJ^TJ^T = -JA$ (since $J^2 = -I$), so the spectrum of $JA$ and its negative coincide. $\endgroup$
    – Nathaniel Johnston
    Commented Apr 24, 2014 at 16:43

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