Order of Symplectic Matrices

Question

I'm going through Appendix 3 of Lax's Linear Algebra, and I'm not entirely sure why symplectic matrices must be of order $2n$. He defines symplectic matrices as "linear maps that preserve a nonsingular bilinear alternating function of the form $(x,Ay)$, where $A$ is a real anti-self-adjoint matrix and $det A\ne0$. It then follows that $A$ must be of even order $2n$". However, I can't figure out why it must be of order $2n$, and as far as I can find online it's always just chalked up to "by definition". My intuition is that it has something to do with pairing off basis vectors (i.e. pairing off $(e_1,Ae_2)$ etc), but I haven't been able to find any contradictions for the $n=3$ case (which we would expect if it had to be of even order). It would seem to me that functions that are nonsingular and alternating must have even order, but I can't find out why. What is the reason for this?

Answer 1

From the statement you quote, $A$ must be a skew-symmetric matrix with nonzero determinant.

But skew-symmetric matrices of odd order have always determinant 0, because

$$\mathrm{det}(A)=\mathrm{det}(A^T)=\mathrm{det}(−A)=(−1)^n\mathrm{det}(A),$$

$n$ being the order of $A$.