# Diagonalization of elements of the symplectic algebra.

Let $A$ a symmetric positive definite real matrix of dimension $2n\times 2n$ and $J$ the standard symplectic matrix, with block representation \begin{gather} J= \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} \end{gather} with $I$ diagonal matrix of order $n \times n$. Consider the product $U=JA$, then there exists a basis that diagonalizes $U$ and have only imaginary eigenvalues.

How can I prove that?

• Why the algebra-precalculus tag? – user1551 Apr 6 '16 at 14:35

It's simply because $JA$ is similar to $A^{1/2}JA^{1/2}$, which is skew-Hermitian.