# Show that $A^{-1} + B^{-1}$ is invertible when $A,B$ and $A+B$ are invertible

I have the following issue: $$A,B\in\mathbb C^{n\times n}$$ invertible, such that also $$A + B$$ is invertible. How is it shown that $$A^{-1} + B^{-1}$$ is invertible?

• – hardmath Mar 27 at 15:52
• Can we fix the title ? – T. Fo Mar 27 at 19:57
• @Andreu Gooz Biel: If your question has been answered below, please, accept an answer. Otherwise your question remains open indefinitely. Thank you! – Moritz Apr 1 at 20:59

$$A^{-1}+B^{-1} = A^{-1}(A+B)B^{-1}$$ By the way: (Spanish) demonstración --> (English) proof. ;-)

Here is a more pedantic approach to @amsmath's slick approach:

Suppose we want to solve $$(A^{-1} + B^{-1}) x = A^{-1}x + B^{-1} x = y$$.

Then $$Ay=x + AB^{-1} x$$, and letting $$x'=B^{-1} x$$ we get $$Ay = B x' + A x' = (A+B)x'$$ and so $$x'= (A+B)^{-1} Ay$$ and finally $$x=Bx' = B(A+B)^{-1} Ay$$.

Hence $$(A^{-1} + B^{-1})^{-1} = B(A+B)^{-1} A$$. Of course you know that a square matrix has a two sided inverse if you have found an inverse from one side.