If $A^kB^l=B^lA^k$ for all $k,l>1$, is $AB=BA$?

Question

Let $R$ be a commutative ring with unity and $A, B\in\operatorname{Mat}(n, R)$ are invertible. Suppose $A^kB^l=B^lA^k$ for all natural numbers $k,l>1$. Is $AB=BA$?

I think that there can be useful that $AB=BA$ if and only if $A$ and $B$ have same eigenvalue.

Answer 1

We can conmute $a^n$ and $b^m$ if both $n$ and $m$ have norm greater than $1$.

It follows that

$a^3a^{-2}b^3b^{-2}a^{-3}a^{2}b^{-3}b^2=e$

This is equal to $aba^{-1}b^{-1}$ which clearly implies $ab=ba$.

Note that we only needed that the elements $a^2,a^3,b^2,b^3$ commute with each other.

Answer 2

The condition that $A$ and $B$ are invertible can be weakened to that $a=\det(A)$ and $b=\det(B)$ are not zero divisors.

By Cayley-Hamilton theorem, $Ap(A)=aI$ and $Bq(B)=bI$ for some polynomials $p$ and $q$. Therefore $$ abAB=(aA)(bB)=A^2p(A)B^2q(B)=B^2q(B)A^2p(A)=(bB)(aA)=abBA. $$ As $a$ and $b$ are not zero divisors, we must have $AB=BA$.