TITLE: Matrices A+B=AB implies A commutes with B
QUESTION [5 upvotes]: $A$ and $B$ are $n\times n$ matrices and $A+B=AB$. I have an interesting proof that this implies $A$ commutes with $B$, but the proof only works when $||B|| \lt 1$. I'm looking for a way to salvage the proof so that it works for all $B$.
$$A=AB-B$$
$$A=(A-I)B$$
by repeatedly substituting $A$ with $(A-I)B$ on the RHS, $$A= (A-I)B^N - \sum_{i=1} ^{N-1} B^i$$
Since the space of $n\times n$ matrices is Banach (hence complete) under the standard operator norm, the limit $\sum_{i=1}^{\infty}B^i$ is well defined given $||B|| \lt 1$. Hence, taking the limit as $N \rightarrow \infty$, $$A =  -\sum_{i=1}^{\infty}B^i$$
Now the LHS commutes with B, hence so does the RHS.  
Is there some trick to extend the method to work with any $B$ where $||B|| \geq 1$? I know that the result is true due to another simple algebraic proof, but I'd like to see if this method can be salvaged.

REPLY [4 votes]: This is perhaps not quite what you’re looking for, but here is another algebraic method : I claim that $I-B$ must be invertible.
Otherwise, the kernel is nontrivial, and we have a nonzero $x$ such that $Bx=x$. But then $A+B=AB$ yields $Ax+Bx=ABx$, so $Bx=0$ which is absurd.
So $I-B$ is invertible. Let $\chi_B$ be the characteristic polynomial of $B$. We have just shown that $1$ is not a root of $\chi_B$, so $\chi_B$ is coprime to $X-1$. We then have a Bezout identity $Q(X)(X-1)+P(X)\chi_B(X)=1$. By the Cayley-Hamilton theorem, we deduce $Q(B)(B-I)=I$ so $(I-B)^{-1}=-Q(B)$. Then $A=-BQ(B)$ commutes with $B$.
The moral of the story : $(I-B)^{-1}$ (and therefore $A$) is always a polynomial in $B$.