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Is there a proof for the following identity?

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

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what happens if $a=i = -b?$ –  abel May 1 '11 at 23:25
    
Well, I generously assumed that everything is defined. –  Phira May 1 '11 at 23:27
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3 Answers 3

up vote 8 down vote accepted

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

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

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

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

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\begin{equation} (A^{-1} + B^{-1})^{-1} = \left(B^{-1}(A + B) A^{-1}\right)^{-1} \end{equation}

Since $(CD)^{-1} = D^{-1} C^{-1}$, we have

\begin{equation} \left(B^{-1}(A + B) A^{-1}\right)^{-1} = A(A+B)^{-1}B \end{equation}

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Write the matrices as $(A^{-1})^{-1}$ and repeatedly use $(M N)^{-1}=N^{-1}M^{-1}$.

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