# Prove that the transpose of a symplectic matrix is also symplectic

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?

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).
• I mean, taking the inverse on both sides, e.g., if the equation is $C=DE$ then taking the inverse gives $C^{-1}=E^{-1}D^{-1}$, right ? Dec 7, 2015 at 14:09
• Here $J$ is supposed to be the matrix of some non-degenerate skew-symmetric bilinear form with respect to some basis of a $2n$-dimensional real vector space. Then how can you even conclude that $J^{-1} = J^{\top}\$? I have no idea. Mar 19 at 6:40