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If a matrix $A\in SL_2(\mathbb Q)$ has finite multiplicative order, then $\operatorname{tr}(A)\le 2$. Does anyone know a demonstration of this fact?

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Using the triangle inequality, if $|\lambda_1|,|\lambda_2|\le1$ then $|\mathrm{tr}(A)|=|\lambda_1+\lambda_2|\le|\lambda_1|+|\lambda_2|\le2$.

If $\lambda,v$ is an eigenpair* of a matrix $A$, and $A^n=I$, then $A^nv=\lambda^nv=Iv=v$, hence $\lambda^n=1$, which implies $|\lambda|=1$ (in particular it is an $n$th root of unity). This applies to all $\lambda$, so $|\mathrm{tr}(A)|\le2$.

Note this applies to all of $\mathrm{GL}_2(\Bbb C)$, not just $\mathrm{SL}_2(\Bbb Q)$, and bounds the trace in all of $\Bbb C$ instead of one side of the real line. More generally, if $A\in\mathrm{GL}_m(\Bbb C)$ satisfies $A^n=I$, then $|\mathrm{tr}(A)| except when $A=zI$ with $z$ is an $n$th root of unity (think geometrically!). Also, $(\det A)^n=1$.

*All eigenvalues of a matrix have a corresponding eigenvector. For if $\lambda$ satisfies $\det(\lambda I-A)=0$, then $\lambda I-A$ must be singular, hence its image is a proper subspace aka nontrivial cokernel, and by the rank-nullity theorem $\lambda I-A$ has nontrivial kernel; any element of the kernel is an eigenvector.

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    My apologies, I confused "each eigenvector has an eigenvalue" which is true (and all you need for this proof) with "there is a basis of eigenvalues". I should not look at math before having coffee, but you are certainly correct, and we do not need to prove the stronger statement (that we are diagonalizable) to prove the result.2012-07-01