Compute inverse of $ABA^{T}$ by its component pseudo-inverse?

Asked Nov 12 '20 at 21:30

Active Nov 12 '20 at 22:07

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Is there a way to compute $(ABA^{T})^{-1}$, for $A \in R^{m,n}, B \in R^{n,n}$ and $m<n$. $B$ is PD, and each row of $A$ contain only one $1$ element, rest are zeros, rows are distinct.

I tried $(A^{T})^{\dagger}B^{-1}A^{\dagger}$, but it is not true, guess it's not correct for Moore-Penrose pseudoinverse. $A$ has above structure, so $A^{\dagger} = A^{T}$.

The reason for the question is if I can compute inverse of $B$ only one time, then I can get some speedup to compute $(ABA^{T})^{-1}$ with different $A$, because $\text{pinv}(A)$ is just its tranpose.

edited Nov 12 '20 at 22:07

Mrcrg

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asked Nov 12 '20 at 21:30

anormcat

So in other words, $ABA^T$ is a submatrix of $B$. As far as I know there no nice way to compute the inverse of a submatrix using the overall inverse – Ben Grossmann Nov 12 '20 at 21:43
yes it's true $ABA^{T}$ is a submatrix of B. the fact that i have to compute sub-matrices inverse thousands of times lead me to this question – anormcat Nov 12 '20 at 21:46
You could try using Sherman Morrison as in this post – Ben Grossmann Nov 12 '20 at 21:48
thanks. I'll take a look :) – anormcat Nov 12 '20 at 21:50
There are some more ideas on this post – Ben Grossmann Nov 12 '20 at 21:51

Compute inverse of $ABA^{T}$ by its component pseudo-inverse?

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