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If $A$ is symmetric show that $(BA^{-1})^T(A^{-1}B^T)^{-1}=I$

I can see that:

$ (BA^{-1})^T(A^{-1}B^T)^{-1}\\ (A^{-1})^TB^T(B^T)^{-1}(A^{-1})^{-1}\\ A^{-1}B^T(B^T)^{-1}A\\ ...\\ A^{-1}...A=I $

I assume I would arrive at the last step, but I am confused how to get there.

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    $B^T$ is sitting right there next to its inverse, just cancel it!2012-09-21

1 Answers 1

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You just have to use

$ (X^T)^{-1}=(X^{-1})^T, \quad (XY)^T=Y^TX^T, \quad (XY)^{-1}=Y^{-1}X^{-1}$

for all quadratic matrices $X,Y$ (with existing inverse). Didn't you use these equations above?

$ (BA^{-1})^T(A^{-1}B^T)^{-1}=((A^{-1})^TB^T)((B^T)^{-1}(A^{-1})^{-1})\\ =(A^{-1})^T(B^T(B^T)^{-1})A\\ =(A^{-1})^TA\\ =(A^{-1})^TA^T \\ =I $

In the fourth equation you need, that $A$ is symmetric.

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    @Ben I think quadratic = square2012-09-21