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I read a paper, and the paper use the following identities (that hold true in any ring)

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

Any way to prove this? How to open the term $(I+AB)^{-1}$ ?

sleeve chen
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    For the first: multiply by $(I + AB)$ on the left and $(I + BA)$ on the right. Expand both sides.

    For the second, use the fact that $A(I + BA)^{-1}$ is just $(I + AB)^{-1}A$. Move the term over to the other side and the identity reduces to $(I + AB)^{-1}(I + AB) = I$.

     –  Jan 04 '15 at 08:03
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    You can expand $(I+AB)^{-1}$ into a formal power series. See this answer to a closely related question. – user1551 Jan 04 '15 at 09:04

1 Answers1

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From the equality $A(I+BA)=(I+AB)A$, one obtains $(I+AB)^{-1}A=A(I+BA)^{-1}$, assuming $I+AB$ and $I+BA$ are both invertible.

From this, we get $I-A(I+BA)^{-1}B=I-(I+AB)^{-1}AB$. On the other hand, \begin{align*}\left(I-(I+AB)^{-1}AB\right)(I+AB)&=I+AB-(I+AB)^{-1}AB(I+AB)\\&=I+AB-(I+AB)^{-1}(I+AB)(AB)\\&=I+AB-AB\\&=I.\end{align*}

By uniqueness of inverses, we must have $(I+AB)^{-1}=I-A(I+BA)^{-1}B$.

Casteels
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