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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?

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    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. – 2017-01-07
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    @Arthur Yea, I see it works. but my question is How to come up with this block matrix multiplication? – 2017-01-07
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    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. – 2017-01-07
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    @Arthur So to construct it i take $AD^T-BC^T=I$ and only this. So it has to give me Identity matrix? – 2017-01-07
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    "So it has to give me Identity matrix?" and if so, why? – 2017-01-07
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    By given condition I can fill in $A, -B, D^T, C^T$ and $I$. or anything else? – 2017-01-07
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    To answer your first question: if $B=0$ or $C=0$, the statement reduces to $AD^T=I\ \Rightarrow\ A^TD=I$; if $A=0$ or $D=0$, the statement reduces to $-BC^T=I\ \Rightarrow\ -C^TB=I$. So, both special cases follow from the equivalences of $XY=I,\ YX=I$ and $X^TY^T=I$, and it is natural to **try** to put the original problem in the context of matrix multiplications. This is where the intuition comes from. It's partly experience and partly epiphany. – 2017-01-07
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    @user1551 whether it uses determinants? – 2017-01-07
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    You don't need determinants. See [If $AB=I$ then $BA=I$](http://math.stackexchange.com/q/3852). – 2017-01-07
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    @user1551 Ok I think, I'm close. But how do I use that if (B=0 or C=0)? – 2017-01-07
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    $AD^T=I\ \Rightarrow D^TA=I\ \Rightarrow A^TD=(AD^T)^T=I^T=I$. – 2017-01-07
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    @user1551 I did it, thanks :) – 2017-01-07

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