# Proving an equation with invertible matrices

I need to prove that if $A$, $B$ and $(A + B)$ are invertible then $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$

I'm a bit lost with this one,
I can't find a way to make any assumptions about $(A^{-1} + B^{-1})$,
Neither by using $A^{-1}$, $B^{-1}$ and $(A + B)^{-1}$.

If someone could clue me in I'll be grateful.

UPDATE:

Thanks for all of your input - it really helped,
I got a solution, I'd like to know if my way of proving it is valid:

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

multiplying both sides by $A^{-1}$ and $B^{-1}$

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

now multiplying both sides by $(A + B)$, I get to

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

Lets call $(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})$ -> $C$

Now I determine that $C$ is $(A+B)^{-1}$,
because $(A+B)C$ equals the Identity Matrix.

So lastly to verify that I place $C$ in the original equation:

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

and from this I get:

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

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updated my solution – res Nov 20 '12 at 14:13

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

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

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Hint: To show that $A^{-1}=C$, compute $AC$ and show that $AC=I$.
In your case, you just need to show that $(A^{-1} + B^{-1})\cdot(A(A + B)^{-1}B)=I$, where $I$ is the identity matrix.

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You can do it the way as other users have mentioned or observe that $M(N^{-1})=(NM^{-1})^{-1}$. See how the matrix $M$ came inside the inverse. Also $M(N^{-1})O=(O^{-1}NM^{-1})^{-1}$. It is just a generalization of the first statement. Now put $M=A$, $N=A+B$ and $O=B$. You will have your answer.

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How about a purposedly complicated proof, but without matrix inversion?

Let $X = A(A + B)^{-1}B$. By writing the rightmost $B$ as $(A+B)-B$, we get $$X = A(A + B)^{-1}[(A+B)-A] = \underbrace{A - A(A + B)^{-1}A}_{X_1}.$$ Similarly, if we write the leftmost $A$ as $(A+B)-B$ instead, we get $$X = [(A+B)-B](A + B)^{-1}B = \underbrace{B - B(A + B)^{-1}B}_{X_2}.$$ Let $Y=A^{-1} + B^{-1}$. Then \begin{align} XY &= X(A^{-1} + B^{-1})\\ &= X_1 A^{-1} + X_2B^{-1}\\ &= [I - A(A + B)^{-1}] + [I - B(A + B)^{-1}]\\ &= 2I - (A+B)(A + B)^{-1} = I. \end{align} So $XY=I$ and we are done.

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