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 $(\mathbf{AVA}')^{-1}$?



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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.