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Given the left hand side of the equation below, how to show the equality with the right hand side? I checked it numerically, but not sure how to prove it.

\begin{align} A - AC^T(CAC^T +R)^{-1} CA = (A^{-1}+ C^T R^{-1} C)^{-1} \end{align}

$A$ is an invertable square matrix.

$C$ is a rectangular matrix with compatible dimension w.r.t $A$ and $R$

$R$ is an invertable ssquare matrix.

user208428
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  • Is this really all the information you have? I'm not entirely convinced that $CAC^T+R$ or $A^{-1}+C^TR^{-1}C$ are even invertible. ($C$ could be the identity matrix and $A+R$ is not necessarily invertible even if $A$ and $R$ are. Even then a direct computation might get complicated - so you'll definately need to get rid of those in another way.) – Piwi Oct 11 '15 at 17:22
  • The matrix $A$, the matrix $R$ and the matrix $(CAC^T + R)$ are all co-variance matrices of Gaussian random vectors. – user208428 Oct 12 '15 at 09:39
  • However this doesn't imply that those inverses exist. So you can assume the special case in which those inverses exist. – user208428 Oct 12 '15 at 09:50

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