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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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    $\begingroup$ Smells like The Woodbury identity. $\endgroup$
    – Fabian
    Commented Jan 19, 2016 at 10:10
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    $\begingroup$ Don't we first need to assume that $A+B$ is invertible? $\endgroup$
    – Arpit Kansal
    Commented Jan 19, 2016 at 10:11
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    $\begingroup$ @ArpitKansal There is a $(A+B)^{-1}$ in the identity. Obviously you need to assume that $A+B$ is invertible... $\endgroup$
    – Najib Idrissi
    Commented Jan 19, 2016 at 10:15
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    $\begingroup$ Page 18, relation 157 of your linked pdf. $\endgroup$
    – N74
    Commented Jan 19, 2016 at 10:16
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    $\begingroup$ @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.. $\endgroup$
    – rschwieb
    Commented Jan 19, 2016 at 12:53

1 Answer 1

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\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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  • $\begingroup$ Dear @Urgje You've nailed it.bravo! $\endgroup$
    – Arpit Kansal
    Commented Jan 19, 2016 at 10:31

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