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I need to compute $ (A^{-1} + B^{-1})^{-1} $. Both $A$ and $B$ are symmetric and $A$ is invertible and PSD. I already know $B^{-1}$ and $A$, but I don't have $A^{-1}$ and $B$. Is there a formula to compute that avoids two inversions?

EDIT:

Another related question is if there are situations (e.g. $A$ and $B$ have additional properties) where it is possible to avoid both the inversions and under which conditions.

As additional properties consider that I know also $B$ together with $A$ and $B^{-1}$ and that $B$ is invertible and PSD as $A$. Furthermore, both $A$ and $B$ are square matrices. With these properties, is there a formual (or even an approximation) to avoid both the inversions?

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You have $(A^{-1}+B^{-1})^{-1} = ((I+B^{-1}A)(A^{-1}))^{-1} = A (I+B^{-1}A)^{-1}$, so if you already have $B^{-1}$, you only need two more multiplications, an addition and an inversion.

This only requires $A,B, A^{-1}+B^{-1}$ to be invertible.

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  • $\begingroup$ Thanks a lot, do you have any ideas about the situations I mentioned in the edit of the question? $\endgroup$ – gcswoosh Apr 10 '15 at 17:08
  • $\begingroup$ What sort of additional properties were you thinking of? $\endgroup$ – copper.hat Apr 10 '15 at 17:25
  • $\begingroup$ For example, knowing also $B$ (i.e., I know $A$, $B$ and $B^{-1}$) and knowing that $B$ is invertible and PSD can be useful to avoid the inversion? $\endgroup$ – gcswoosh Apr 11 '15 at 10:04

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