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?
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 $(I - BA)$.