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If $A, B$ and $A + B$ are all $n × n$ invertible matrices. Prove that $A^{−1} + B^{−1}$ is invertible and the inverse is $A(A + B) ^{−1}B$.

I am afraid I am really stuck on this one, and I haven't really tried much because I don't know what to try.

Thanks for all the help guys, I understand now.

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    $\begingroup$ You need to show that $(A^{-1}+B^{-1})(A(A+B)^{-1}B)=I$. Multiply the LHS out and massage it a little bit. $\endgroup$
    – A.S.
    Commented Oct 30, 2015 at 21:54
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    $\begingroup$ To show $X$ is an inverse of $Y$, you must show $XY=I=YX$. (Note $XY=I$ and $YX=I$ are equivalent for finite square matrices, but are not equivalent in general.) Showing some $Y$ is an inverse of $X$ suffices to say that $X$ is invertible. $\endgroup$
    – anon
    Commented Oct 30, 2015 at 21:54
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    $\begingroup$ @A.S. I guess I wasn't so clear, I did try this, but I couldn't seem to get it to simplify so I though it would work. How do you get around the + in the second statement. No generic X + Y = I, and there are no negatives so nothing can cancel. I must be doing something wrong. $\endgroup$
    – Luke
    Commented Oct 30, 2015 at 22:08
  • $\begingroup$ Multiply both sides by two different matrices - one on the left, one on the right, so that LHS simplifies and RHS stays $I$. $\endgroup$
    – A.S.
    Commented Oct 30, 2015 at 22:16
  • $\begingroup$ math.stackexchange.com/q/3915572/321264, math.stackexchange.com/q/36301/321264 $\endgroup$ Commented Jan 28, 2022 at 13:32

3 Answers 3

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Here's a concise proof: $$ \begin{align} A^{-1} + B^{-1} &= B^{-1} + A^{-1} \\ &= B^{-1}AA^{-1} + B^{-1}BA^{-1}\\ &= B^{-1}(A + B)A^{-1} \\ \end{align} $$ So $A^{-1} + B^{-1}$ is the product of invertible matrices and thus is invertible, with inverse equal to $$ \begin{align} (A^{-1} + B^{-1})^{-1} &= (B^{-1}(A + B)A^{-1})^{-1} \\ &= A(A + B)^{-1}B \\ \end{align} $$

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Here's half of the solution. $$A(A+B)^{-1}B=((A+B)A^{-1})^{-1}B=(I+BA^{-1})^{-1}B$$ (Note that inversion reverses the order of products.) What happens if you do the same thing with $B$? We then get $$(B^{-1}(I+BA^{-1}))^{-1}=(B^{-1}+A^{-1})^{-1}$$ We didn't even have to multiply anything out!

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\begin{equation} (A^{-1} + B^{-1}) A (A+B)^{-1} B = A^{-1}A(A+B)^{-1}B + B^{-1} A (A+B)^{-1} B \end{equation} by the distributive law. Now here comes the trick: Write $A = (A+B) - B$ to get

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

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