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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)

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    $\begingroup$ There is a property of the transpose that says $(A+B)^{\mathbb{T}} =A^{\mathbb{T}}+B^{\mathbb{T}}$ Seems like it could be useful? $\endgroup$
    – graydad
    Commented Sep 6, 2014 at 17:56
  • $\begingroup$ @graydad it is useful, but insufficient on its own. Note also that $(AB)^T = B^TA^T$ $\endgroup$ Commented Sep 6, 2014 at 18:13
  • $\begingroup$ @Omnomnomnom ah I see why that would ruin what I had planned. Now my gut is telling me we need to show that $AD^T$ and $BC^T$ must be symmetric. If that were true, then the two results we listed above would get us the result. $\endgroup$
    – graydad
    Commented Sep 6, 2014 at 18:20
  • $\begingroup$ @graydad how so? $\endgroup$ Commented Sep 6, 2014 at 18:22
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    $\begingroup$ I think this is as good as it gets (and is very neat IMO). $\endgroup$
    – Matt Rigby
    Commented Sep 14, 2014 at 14:18

1 Answer 1

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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.

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    $\begingroup$ The bounty is yours; consider it a reward for an interesting question $\endgroup$ Commented Sep 16, 2014 at 15:48

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