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Suppose $A,B\in GL(2,\mathbb Z)$, and $AB=BA$. If $A,B$ are hyperbolic, i.e. their eigenvalues do not have length $1$, show that $A^n=B^m$ for some integers $n,m$ ($n^2+m^2\ne 0$).

I have tried to prove it using basic linear algebra. Suppose $v$ is a common eigenvector and $$Av=\lambda v,\quad Bv=\mu v.$$ We only need to show that $\lambda^n=\mu^m$. I expand the two identities above but cannot find anything useful.

Should we take into some background knowledge of hyperbolic automorphisms of $\mathbb T^2$?

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    I think there must be something missing from your definition of "hyperbolic". If we take $A=2I$ and $B=3I$, then the statement clearly fails to hold. – 2017-01-01
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    they are not in $Gl(n,Z)$ – 2017-01-01
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    @Tsemo ah, good catch – 2017-01-01

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