TITLE: $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.
QUESTION [29 upvotes]: The problem is from  Algebra by Artin, Chapter 1, Miscellaneous Problems, Exercise 8. I have been trying for a long time now to solve it but I am unsuccessful. 

Let $A,B$ be $m \times n$ and $n \times m$ matrices. Prove that $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible. 

Please provide with only a hint.

REPLY [4 votes]: We know that a square matrix $M$ is invertible if and only if
$$\{\vec x; \vec x M = \vec 0\}=\{\vec 0\}.$$
(The corresponding homogeneous linear system has only trivial solution. The corresponding linear map has trivial kernel.)
So the claim
$$I_m-AB\text{ is invertible }\qquad\implies\qquad I_n-BA\text{ is invertible}$$
is equivalent to
$$\{\vec x\in F^m; \vec x(I-AB)=\vec 0\}=\{\vec0\} \qquad\implies\qquad  \{\vec y\in F^n; \vec y(I-BA)=\vec 0\}=\{\vec0\}.$$
(Where $F$ denotes the base field.)
Proof. So let us assume that $I-AB$ is invertible, i.e., that $\vec x=\vec 0$ is the only vector such that $\vec x(I-AB)=\vec 0$.
Let $\vec y(I-BA)=\vec 0$.
Then we get 
$$\vec y=\vec yBA\tag{1}$$
and, consequently $\vec y B = \vec y BAB$. This is equivalent to
$$\vec yB(I-AB)= \vec 0.\tag{2}$$
This means that $\vec yB=\vec 0$, since $I-AB$ is invertible.
From (1) we now get that $\vec y=\vec yBA=\vec 0A=\vec 0$.
$\square$