Is it always true that matrix representation of a skew-symmetric non-degenerate bilinear form is orthogonal with respect to some basis?

I am going through the definition of symplectic matrices. While searching through various properties of those class of matrices I found that every symplectic matrix is necessarily invertible and the inverse of such matrices are again symplectic. Then out of curiosity I just ask myself a question $$:$$ Are the symplectic matrices closed under taking transpose? After several ineffectual attempts I have tried searching it in MSE and here's what I found. I have tried to use the same trick mentioned there earlier but I don't get the point how $$\Omega^{-1}$$ becomes $$\Omega$$ in this equation. I know that $$\Omega$$ is the matrix representation of some skew-symmetric non-degenerate form. So the second equation there clearly suggests that either $$\Omega$$ is involutary or orthogonal but I can't deduce any of these facts from the given condition. Could anyone please shed some light on it?

Thanks a bunch!

• I'm not sure what you don't understand in the answer you link. Just write the symplectic condition for $\Omega^{-1}$, as is done in the answer, and then take the inverse on both sides. This will give the symplectic condition for the transpose of $\Omega$. Mar 19 at 16:34
• @CaptainLama$:$ $\Omega$ is not symplectic. Rather, in the answer $S$ is symplectic. The author writes the condition of symplecticity for $S^{-1}$ and then told us to invert the condition. Then finally we ended up with $S \Omega^{-1}S^{\top}=\Omega^{-1}.$ But we want $S \Omega S^{\top}=\Omega.$ When can it happen? I am just trying to explore the possibilities that under what condition on $\Omega$ these two relations represent the same thing and I found that $\Omega=\lambda \Omega^{-1}$ i.e. $\Omega^2 =\lambda I$ for some scalar $\lambda.$ But what is this scalar actually in our case? contd... Mar 19 at 18:13
• ...This is what I intended to know. Hopefully my query is clear now. Mar 19 at 18:14
• @NicholasTodoroff$:$ Wait! Wait! Wait! Don't be in a hurry or pretend to become super smart. How do you get the second equation from the first? Could you add one more line of argument? Mar 19 at 18:34
• @Anacardium I deleted my comment because I was wrong. Mar 19 at 21:12

I think there is a small confusion. Let us say that is $$\Omega$$ is an invertible anti-symmetric matrix, then $$S$$ is $$\Omega$$-symplectic if it satifies $$S^t\Omega S=\Omega$$. Then the $$\Omega$$-symplectic matrices form a group, so if $$S$$ is $$\Omega$$-symplectic then so is $$S^{-1}$$.
On the other hand, $$S^t$$ is in general not $$\Omega$$-symplectic but $$\Omega^{-1}$$-symplectic, following the computation you did.
But if you use the classical $$\Omega=J$$, then $$J^{-1}=-J$$, so being $$\Omega^{-1}$$-symplectic is the same as being $$\Omega$$-symplectic.
• But it should be noted that there is a basis such that $\Omega$ takes on the form $J$, i.e. there is a change of basis matrix $A$ such that $J = A^T\Omega A$. What goes wrong is that $S$ transforms to $S' = A^{-1}SA$ and we know $(S')^T$ is symplectic wrt. $A^T\Omega A$, but transforming back what we get is $A(S')^TA^{-1} = (AA^T)S^T(AA^T)^{-1} \ne S^T$ is symplectic. Mar 19 at 22:32
• What is $J$ in your answer? If the answer I linked is confusing then how did it get $6$ upvotes and how does one accept it as an answer? Mar 20 at 17:32