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Let $\mathbf{V}$ be a $2n-1$ by $2n-1$ [symmetric positive definite] matrix with a known inverse and define $\mathbf{A}=[[\mathbf{D},\mathbf{0}]',\mathbf{I}]$ where $\mathbf{D}$ is a diagonal matrix of size $n-1$, $\mathbf{0}$ is a column vector of zeros and $\mathbf{I}$ is an identity matrix of size $n$. Given that the inverse of $\mathbf{V}$ is known and $\mathbf{A}$ has a special structure with lots of zeros, can we find a simple closed form for $(\mathbf{AVA}')^{-1}$?

Thanks

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This matrix is not invertible in general: $$ A= \pmatrix{1 & 0 & 1}, \ V = \pmatrix{1 & 0 & 0 \\ 0&1&0 \\0 & 0 & -1} $$ Then $$ AVA^T = \pmatrix{1 & 0 & -1}A^T=0. $$

If $V$ is symmetric and positive definite in addition, then $AVA^T$ is at least invertible. Partition $$ V=\pmatrix{ V_{11} & * & V_{12} \\ * & * & * \\ V_{12}^T & * & V_{22}} $$ with $(n-1) \times (n-1)$ matrices $V_{ij}$. The submatrices $V_{11}$ and $V_{22}$ are symmetric positive definit as well. And it holds $$ AVA^T = DV_{11}D + DV_{12}+V_{12}D+V_{22}. $$ I am afraid that there is no nice formula to get the inverse of a sum of two matrices.

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  • $\begingroup$ Thanks, my $\mathbf{V}$ is in fact symmetric positive definite. Direct multiplication does not seem to be helpful, but it may still be possible to find some nice formula if the information on the inverse of $\mathbf{V}$ can somehow be used. I have a feeling (from my related works on this) that there might be a not too difficult solution to this inverse [basically, this multiplied by some other matrices seems to have a closed form solution]. $\endgroup$ – user41838 Jul 16 '15 at 6:38

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