We have for a $2 \times 2$ matrix $A$ that $A$ is symplectic if and only if $\det A =1$.
Is there any way to use this fact as the base for an inductive proof of the fact that the determinant of any symplectic matrix is $1$?
I tried writing it out in block matrix form but realized quickly that it was not as simple as I had thought, since:
- $a_{1,2} \dots a_{1,2n}$, and $a_{2,1}, \dots , a_{2n,1}$ don't have to be zero (although maybe this is possible to achieve via some similarity transformation that preserves the symplectic property?)
- The block matrix formed by deleting the first row and column doesn't have the correct dimensions to even be a symplectic matrix.
- It is unclear what the $2n-2 \times 2n-2$ matrix would be to use the inductive hypothesis for $n-1$ would be.
Even just showing that it is possible to extend the result from $2 \times 2$ symplectic matrices to $4 \times 4$ symplectic matrices seems like it would be an accomplishment (to me).