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Let A and B be two $ n\times n$ matrcies over the field F.Prove that if (I-AB) is invertible, then (I-BA) is invertible and that $ (I-BA)^{-1}=I+B(I-AB)^{-1}A $

Using this result prove that AB and BA have precisely the same characteristic values in F.

My Attempt: For the first part I tried to prove the contrapositive ie I assumed (I-BA)X=0 has a non trivial solution and tried to show a non trivial solution for (I-AB)X=0. I could show this by making use of the fact that AB and BA have same eigen values but that method becomes circular in this case considering the follow up question mentioned at the end.Anyway even using that AB and BA have same eigen values I could not prove the expression for the inverse.I would request a solution which can then be used to prove that AB and BA have same eigen values and also proving the expression for the inverse.

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marked as duplicate by Brahadeesh, Leucippus, max_zorn, metamorphy, Lord Shark the Unknown Jan 4 at 4:43

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    $\begingroup$ You are given the supposed inverse of $I - BA$ explicitly so all you need to do is multiply. What did you try? $\endgroup$ – levap Apr 19 '17 at 14:42
  • $\begingroup$ I don't think verifying is the intention behind the question. We need to first prove that (I-BA) is invertible using that (I-AB) is invertible and then prove the given expression for inverse not verify it. For the first part I tried to prove the contrapositive ie I assumed (I-BA)X=0 has a non trivial solution and tried to get a non trivial solution for (I-AB)X=0 but I couldn't make much progress. $\endgroup$ – krishna Apr 19 '17 at 14:52
  • $\begingroup$ You are given that $I - AB$ is invertible so $(I - AB)^{-1}$ exists and you want to show that $I - BA$ is invertible. If you multiply $I - BA$ by $I + B(I - AB)^{-1} A$ and show that this is $I$ you'll prove that $I - BA$ is invertible and $(I - BA)^{-1} = I + B(I - AB)^{-1}A$. $\endgroup$ – levap Apr 19 '17 at 14:55
  • $\begingroup$ @krishna: So, you could update your question adding your own effort. That is the best way to get help here. $\endgroup$ – Arnaldo Apr 19 '17 at 14:56
  • $\begingroup$ Thanks for the advice. I have updated the question with all the details. $\endgroup$ – krishna Apr 19 '17 at 15:05

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