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$\mathbf{A}$ is an $M\times N$ matrix with $M\leq N$ and $\mathbf{C}$ is an $N\times N$ diagonal matrix. $\mathbf{A}^{-1}$ does not exist, $(\mathbf{A}\mathbf{A}^H)^{-1}$ exists.

Matrix $\mathbf{A}\mathbf{C}\mathbf{A}^H$ is invertible. Is it possible to take out $\mathbf{C}^{-1}$ from $(\mathbf{A}\mathbf{C}\mathbf{A}^H)^{-1}$ , either using Kronecker operations or something else?

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    I presume $\mathbf A$ is known?2011-05-05
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    yes. $\mathbf{A}$ and and $(\mathbf{A}\mathbf{A}^H)^-1$ are know. I am looking some tricks where we can bring out inverse of C, either using vec operator or kroneckor or something else.2011-05-05
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    For the inverse of $I+A C A^H$ you could use the Binomial inverse theorem (http://en.wikipedia.org/wiki/Binomial_inverse_theorem).2011-05-05
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    Thanks, but I am not trying to invert something of that form2011-05-05
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    @Sun: your are welcome. I thought it would not help. I just wanted to share some of the matrix-identities which I like most ;-)2011-05-05
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    Is $\mathbf A^T$ full-rank? On the other hand, I'm not sure if what you propose is possible; if you substitute in the (economy) singular value decomposition $\mathbf A^T=\mathbf U\mathbf \Sigma\mathbf V^T$ in your expression, you end up with $(\mathbf U^T\mathbf C\mathbf U)^{-1}$ and it does not seem to me that you can recover $\mathbf C$ or its inverse from that matrix.2011-05-05

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Since the matrix $(ACA^H)^{-1}$ is a smaller matrix than $C^{-1}$, it should be impossible by a dimensional argument.

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    Quite right. Take a tiny example: $A=(1 2)$, $C=\pmatrix{a&0\cr0&b\cr}$. You want to find $C^{-1}$, knowing only the number $1/(a+4b)$. That's not even enough information to conclude that $C$ is invertible, much less to find its inverse.2011-05-05