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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 ( so $AB^T=A^TB$ and$CD^T=C^TD$ )
and $AD^T-BC^T=I$
Prove that: $A^TD-C^TB=I$

$T$ means transpose

$I$ is identity matrix

I need some advice, maybe there is some trick with trace of an matricx.

  • 1
    be careful, $(XY)^T=Y^TX^T$ – 2017-01-03
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    Isn't this false? Let A, C, and D be identity matrices and B be the zero matrix. – 2017-01-03
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    Also, for that matter, it seems to be answered here http://math.stackexchange.com/questions/463791/prove-that-atd-ctb-i?rq=1 – 2017-01-03

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