# Invertibility of $I-AB$ [duplicate]

I got a question in linear algebra: 1) Let A and B be $n\times n$ matrices. If $I - AB$ is an invertible matrix, then prove that $I - BA$ is invertible.

Can someone tell me how to solve this question? I've no idea how to start. Thank you!

## marked as duplicate by Martin Sleziak, Vladimir, user91500, Cookie, Davide GiraudoJul 25 '14 at 9:26

This happens generally in any ring. Suppose $1-xy$ is invertible. Then we may do the informal
$$(1-yx)^{-1}=1+yx+yxyx+yxyxyx+\cdots=1+y(1+xy+xyxy+\cdots)x\\=1+y(1-xy)^{-1}x$$ It turns out this works. Some (unnecessary) jargon: if $1-x$ is invertible, we say $x$ is quasi-invertible, or quasi-regular. Thus, you're proving that $yx$ is quasi-regular iff $xy$ is.
• @Guess I just used the word "ring" but this is entirely elementary, and simple. The formal method is "multiply $(1-yx)$ by $1+y(1-xy)^{-1}x$ and see it gives $1$". – Pedro Tamaroff Jul 24 '14 at 5:02