# proof that $(I+AB)^{-1}A = A(I+BA)^{-1}$ [duplicate]

Given $$A, I+AB\:$$ invertible matrices, prove that $$I+BA$$ is invertible and that $$(I+AB)^{-1}A = A(I+BA)^{-1}.$$

How should I approach this? The question seems similar to Proof if $I+AB$ invertible then $I+BA$ invertible and $(I+BA)^{-1}=I-B(I+AB)^{-1}A$ but I can't make the connection

• @enzotib here it is not given that A,I+AB invertible matrices, so the solution's may differ – avivgood2 Aug 2 '20 at 13:43

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

• Why is this true? – avivgood2 Aug 2 '20 at 13:06
• Distributive property... – ECL Aug 2 '20 at 13:07
• @avivgood2 Since $$\left(A^{-1}(I+AB)\right)A = (A^{-1} + B)A=I+BA$$ – Luxerhia Aug 2 '20 at 13:07

Start with $$A(I+BA)=(I+AB)A$$ If $$A$$, $$I+AB$$ are invertible, then $$(I+BA)=A^{-1}(I+AB)A,$$ which makes $$I+BA$$ invertible, with $$(I+BA)^{-1}=A^{-1}(I+AB)^{-1}A$$ Multiplying on the left by $$A$$ gives the desired result: $$A(I+BA)^{-1}=(I+AB)^{-1}A.$$

Well... if you follow the answer to that question you get that $$I+BA$$ is invertible. Furthermore,

\begin{align} A(I+BA)^{-1}&=A-AB(I+AB)^{-1}A=(I-AB(I+AB)^{-1})A\\ &=(I-(I+AB)(I+AB)^{-1}+(I+AB)^{-1})A\\ &= (I-I+(I+AB)^{-1})A\\ &=(I+AB)^{-1}A, \end{align} and there you go.

If you accept that $$I+AB$$ invertible implies $$I+BA$$ invertible and $$(I+BA)^{-1} =I-B(I+AB)^{-1}A$$ then \begin{align} A(I+BA)^{-1}&=A-AB(I+AB)^{-1}A=A+(I-(I+AB))(I+AB)^{-1}A\\ &=A+(I+AB)^{-1}A-A=(I+AB)^{-1}A. \end{align}