# Inverse of sum of inverses of matrices [duplicate]

Is it somehow possible to reformulate the following exuation into something easier to calculate:

$$(A^{-1}+ B^{-1})^{-1}$$

A and B are both square real matrices: $$A, B \in \mathbb{R}^{n \times n}$$, and are positive definite and therefore invertible.

• What’s a quadratic matrix? Is that the same as symmetric? – Joe Jul 15 at 10:55
• I don't think so as the sum is not neccessarily invertible (take A = Id = A^-1 and B = -Id = B^-1) – Stockfish Jul 15 at 10:58
• By quadratic matrix did you mean square matrix? – J. W. Tanner Jul 15 at 10:59
• @Joe More likely it's that "quadrata" is the word for square in Italian – Ben Grossmann Jul 15 at 11:52
• Related post – Ben Grossmann Jul 15 at 12:02

Note that $$A^{-1}(A + B)B^{-1} = A^{-1}AB^{-1} + A^{-1}BB^{-1} = B^{-1} + A^{-1}.$$ That is, we have $$A^{-1} + B^{-1} = A^{-1}(A + B)B^{-1} \implies (A^{-1} + B^{-1})^{-1} = B(A + B)^{-1}A.$$ If you prefer, this can also equal to $$A(A + B)^{-1}B$$.
Note that because $$A,B$$ are positive definite, $$A + B$$ is also positive definite and therefore invertible.
• Furthermore, we can't expect to get any other expressions, because the special case for $1\times1$ matrices has to be $AB(A+B)^{-1}$, so the general case must use these three factors. +1. – J.G. Jul 15 at 11:57
• The first line should read $A^{-1}(A+B)B^{-1} = A^{-1}AB^{-1} + A^{-1}BB^{-1} = B^{-1} + A^{-1}$ – Kai Jul 15 at 21:24