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?


1 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).

  • $\begingroup$ What do you mean by take the inverse of the equality? $\endgroup$
    – Arcane1729
    Dec 7, 2015 at 13:12
  • $\begingroup$ 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 ? $\endgroup$ Dec 7, 2015 at 14:09
  • $\begingroup$ Ah I understand, thank you. $\endgroup$
    – Arcane1729
    Dec 7, 2015 at 23:24
  • $\begingroup$ Thank you! You helped me get unstuck today! $\endgroup$
    – mathreader
    May 13, 2019 at 21:05
  • 1
    $\begingroup$ 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. $\endgroup$
    – Anacardium
    Mar 19 at 6:40

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