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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.

krishna
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    You are given the supposed inverse of $I - BA$ explicitly so all you need to do is multiply. What did you try? – levap Apr 19 '17 at 14:42
  • 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. – krishna Apr 19 '17 at 14:52
  • 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 $(I - BA)^{-1} = I + B(I - AB)^{-1}A$. – levap Apr 19 '17 at 14:55
  • @krishna: So, you could update your question adding your own effort. That is the best way to get help here. – Arnaldo Apr 19 '17 at 14:56
  • Thanks for the advice. I have updated the question with all the details. – krishna Apr 19 '17 at 15:05
  • @levap: You are just verifying the given expression not proving it and also I strongly feel we don't need to use the given expression in any way to just prove that (I-BA) is invertible. – krishna Apr 19 '17 at 15:31
  • @krishna: That is just wrong. Even if you manage to show "abstractly" that $I - BA$ is invertible, how would you know what is $(I - BA)^{-1}$? If you are asked to show that something is invertible then you can do it abstractly but if you are asked to show that something is invertible with a specific inverse, there's nothing to show - you just multiply and verify that the result is $I$. There's a reason the inverse was provided to you and that reason is to make the exercise easier... – levap Apr 19 '17 at 15:37
  • I agree that the author might have intended to verify the expression as opposed to somehow constructing it. But you used that expression to prove that (I-BA) is invertible which completely defeats the point of the question and as stated in my question I am looking for a method which proves (I-BA) is invertible and which can be extended to prove that AB and BA have same eigen values. – krishna Apr 19 '17 at 16:15

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