How to prove the both identity (matrix)

Question

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 $(I+AB)^{-1}$ ?

Answer 1

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