A $2n\times 2n$ dimensional Hermitian matrix that can be diagonalized by a symplectic transformation can be viewed as an $n\times n$ matrix with elements consisting of $2\times 2$ blocks of the quaternion real form
${\bar{z}\;-\bar{w}}\choose{w\;\; z}$
so if you choose real $z$ and $w$ you have constructed a real symmetric matrix $M$ that can be diagonalized by a symplectic $S$; is S$. @Federico: this what you need?is the general form for matrices that commute,$MT=TM$, with$T=1_{N}\otimes{0\; 1}\choose{-1\; 0}K$($K$is the operator of complex conjugation); alternatively, one can take matrices that anticommute,$MT=-TM$; then the$2\times 2$blocks have the form${\bar{z}\;\bar{w}}\choose{w\;\; -z}$and again, for a real$M$one would choose real$w,z$. these two choices exhaust the possibilities. In applications to physical systems, the matrix$M$is a Hamiltonian and$T$is the operator of time reversal. Then only commuting matrices,$MT=TM$, are permitted. For a discussion in the physics context, see Section 1.4.2 of Forrester's book, online here: http://www.ms.unimelb.edu.au/~matpjf/b1.ps Post Undeleted by Carlo Beenakker 2 new attempt at an answer The condition is A$2n\times 2n$dimensional Hermitian matrix that the 2n eigenvalues can be diagonalized by a symplectic transformation can be viewed as an$n\times n$matrix with elements consisting of M should come in n twofold degenerate pairs. Then$2\times 2$blocks of the matrix S is both orthogonal quaternion real form${\bar{z}\;-\bar{w}}\choose{w\;\; z}$so if you choose real$z$and$w$you have constructed a real symmetric matrix$M$that can be diagonalized by a symplectic .$S\$; is this what you need?