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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?

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  • $\begingroup$ See the introduction to posting mathematical notation. $\endgroup$ – hardmath Mar 27 at 15:52
  • $\begingroup$ Can we fix the title ? $\endgroup$ – T. Fo Mar 27 at 19:57
  • $\begingroup$ @Andreu Gooz Biel: If your question has been answered below, please, accept an answer. Otherwise your question remains open indefinitely. Thank you! $\endgroup$ – Moritz Apr 1 at 20:59
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$$ A^{-1}+B^{-1} = A^{-1}(A+B)B^{-1} $$ By the way: (Spanish) demonstración --> (English) proof. ;-)

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

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Of course you know that a square matrix has a two sided inverse if you have found an inverse from one side.

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