# 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.

• Hi, Pedro. Thank you for your answer. But is there any simpler and also formal method to prove this question? It is just an elementary linear algebra course, the concept of "ring" and "field" are still some distant away from us now. – Guess Jul 24 '14 at 4:53
• @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