Given $n\times n$ real matrices $A,B,C,D$ such that:
$AB^T$ and $CD^T$ are symmetric
$AD^T-BC^T=I$
Prove that $A^TD-C^TB=I$
The solution I have come up with after a very long time is to consider:
$$\left( \begin{array}{cc} A & -B \\ -C & D \end{array} \right)\left( \begin{array}{cc} D^T & B^T \\ C^T & A^T \end{array} \right)=\left( \begin{array}{cc} I & 0 \\ 0 & I \end{array} \right)\rightarrow\left( \begin{array}{cc} D^T & B^T \\ C^T & A^T \end{array} \right)\left( \begin{array}{cc} A & -B \\ -C & D \end{array} \right)=\left( \begin{array}{cc} I & 0 \\ 0 & I \end{array} \right)$$ then $D^TA-B^TC=I$.
However this solution is apparently tricky. I would love to have a more natural solution (which may be applied to other problems as well)
I have found out that this $2n\times 2n$ matrix, indeed, has a name: symplectic matrix.
http://en.wikipedia.org/wiki/Symplectic_matrix
It seems like this property is a special property of a broader kind of matrix.