# Prove matrix $A^TD-C^TB=I$

Let $A$, $B$, $C$ and $D$ be square matrices $n\times n$ over $\mathbb{R}$ .
Assume that $AB^T$ and $CD^T$ are symmetric .
and $AD^T-BC^T=I$
Prove that: $A^TD-C^TB=I$

I know this question has answer here: Prove that $A^TD-C^TB=I$
But I don't understand the accepted answer(Where that equation came from?):

Hint: The given condition says that $$\pmatrix{A&-B\\ -C&D}\pmatrix{D^T&B^T\\ C^T&A^T} = \pmatrix{I&0\\ 0&I}.$$ Now, note that $XY=I$ implies that $YX=I$ and in turn $X^TY^T=I$.

Is there any simpler way?
Do $A^TD$ and $C^TB$ must be symmetric? And if so, how to prove it?

• I would think that the given answer is very elegant. It rephrases the conditions as a block matrix multiplication, and you can easily read off the result. If you don't understand it, I suggest you try it out with pen and paper yourself. – Arthur Jan 7 '17 at 1:24
• @Arthur Yea, I see it works. but my question is How to come up with this block matrix multiplication? – UfmdFkiF Jan 7 '17 at 1:39
• Because you've seen the trick before, and try it to see whether it works this time. Most proofs go like that, except you don't see that part in writing very often. – Arthur Jan 7 '17 at 1:40
• @Arthur So to construct it i take $AD^T-BC^T=I$ and only this. So it has to give me Identity matrix? – UfmdFkiF Jan 7 '17 at 1:50
• "So it has to give me Identity matrix?" and if so, why? – UfmdFkiF Jan 7 '17 at 2:03