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A, B are two quadratic matrices and I-AB is invertible.

How do I prove that I-BA is also invertible and (I-BA)^-1 = I + B(I-AB)^-1 * A?

Not sure how to start with, but I have tried:

I-AB = C

⇔ B - BAB = BC

⇔ I - BA = BCB^-1

⇔ I - BA = B(I-CB^-1)

But this does not lead to anything and Im just guessing and I have no strategy. Any tips?

1 Answers1

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To show that $M^{-1}$ is the inverse of $M$, it suffices to show that $M^{-1}M = I$.

To that end, note that $$ [I + B(I - AB)^{-1}A](I - BA) = \\ I - BA + B(I - AB)^{-1}A - B(I - AB)^{-1}ABA =\\ I - BA + B[(I - AB)^{-1} - (I - AB)^{-1}AB]A =\\ I - BA + B[(I - AB)^{-1}(I - AB)]A =\\ I - BA + BA = I. $$ So indeed, $I + B(I - AB)^{-1}A$ is the inverse of $(I - BA)$.

Ben Grossmann
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  • It is to show also that $MM^{-1}=I$ or prove that it is enough wilth a left-inverse only, the latter being non-trivial. – A.Γ. Dec 22 '21 at 17:24
  • @A.Γ. I suppose it depends on the linear algebra course, but the sufficiency of left-inverses is a fact I'm used to assuming, non-trivial though it's proof may be. Here is a post about this proof, for reference. – Ben Grossmann Dec 22 '21 at 17:34