Prove that the transpose of a symplectic matrix is also symplectic

Question

I worked out with symplectic matrices that the transpose is also symplectic for the $2\times 2$ case since the algebra was easy and the determinant of the matrix just needed to equal $1$. [The expression for determinant is easy for $2\times 2$ matrices.] and the transpose has equal determinant. But what about the $n\times n$ case. Is it also true?

Answer 1

If $S$ is symplectic of size $2n\times 2n$, i.e., satisfies $S^TJS=J$, so is $S^T$. To see this, take the inverse of the equality $(S^{−1})^T JS^{-1}= J$ (because if $S$ is symplectic, then $S^{-1}$ is symplectic too).