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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? –  J. M. May 5 '11 at 11:44
    
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. –  Sun May 5 '11 at 11:46
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For the inverse of $I+A C A^H$ you could use the Binomial inverse theorem (en.wikipedia.org/wiki/Binomial_inverse_theorem). –  Fabian May 5 '11 at 11:51
    
Thanks, but I am not trying to invert something of that form –  Sun May 5 '11 at 11:54
    
@Sun: your are welcome. I thought it would not help. I just wanted to share some of the matrix-identities which I like most ;-) –  Fabian May 5 '11 at 11:58

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up vote 1 down vote accepted

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. –  Gerry Myerson May 5 '11 at 12:35

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