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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

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  • $\begingroup$ @enzotib here it is not given that A,I+AB invertible matrices, so the solution's may differ $\endgroup$ – avivgood2 Aug 2 '20 at 13:43
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Hint : $$A^{-1}(I+AB)A = I + BA$$

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

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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.

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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}

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