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For invertible matrices A and B does the identity:

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

hold? My supervisor suggested that they are equal but I haven't been able to prove this and in the matrix cookbook (http://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf) there are separate identities for both sides of this equation, but they are not given as equal to each other.

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2  
Smells like The Woodbury identity. – Fabian Jan 19 at 10:10
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Don't we first need to assume that $A+B$ is invertible? – Arpit Kansal Jan 19 at 10:11
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@ArpitKansal There is a $(A+B)^{-1}$ in the identity. Obviously you need to assume that $A+B$ is invertible... – Najib Idrissi Jan 19 at 10:15
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Page 18, relation 157 of your linked pdf. – N74 Jan 19 at 10:16
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@Fabian It smells even more like Hua's identity, because it is just the second form listed at wikipedia with $a$ and $b$ exchanged with their respective inverses. I'm glad to learn of this more general thing, though.. – rschwieb Jan 19 at 12:53
up vote 16 down vote accepted

\begin{eqnarray*} A-A(A+B)^{-1}A &=&A-(A+B-B)(A+B)^{-1}A \\ &=&B(A+B)^{-1}A=[A^{-1}(A+B)B^{-1}]^{-1} \\ &=&(A^{-1}+B^{-1})^{-1} \end{eqnarray*}

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Dear @Urgje You've nailed it.bravo! – Arpit Kansal Jan 19 at 10:31

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