TITLE: $I-AB$ be invertible $\Leftrightarrow$ $I-BA$ is invertible
QUESTION [13 upvotes]: assume $A,B\in M_n(F)$ if  $I-AB$ be  invertible then  how  to prove $I-BA$ is invertible and how find inverse of $I-BA$ 
Thanks in advance

REPLY [16 votes]: Hint :$(I - BA)(I + B(I - AB)^{-1}A) = I$

REPLY [9 votes]: Claim: $\det(I-AB)=\det(I-BA)$.
Proof: 
Case 1: $A$ is invertible. Exercise!
Case 2: $A$ is not invertible. Consider $k$ the algebraic closure of $F$.
Let $P(X)=\det(I-(A-xI)B)- \det(I-B(A-xI))$. By part Case 1, $P(x)=0$ for all $\{ x \in k| x \mbox{ is not an eigenvalue of} A \}$. Since $P(x)$ is a polynomial of degree $n$ which has infinitely many roots, it follows that $P(X) \equiv 0$, and hence $P(0)=0$.