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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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    You need to show that there are actually eigenvectors corresponding to the eigenvalues for this proof to work. One way is that $A^n-1=0$, and since $x^n-1$ has no repeated roots, neither does the minimal polynomial. A second way is that, since there are only two eigenvalues and their product is one, $|\lambda_2|=1/|\lambda_1|=1$.2012-07-01
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    @Aaron: Correct me if I'm wrong, but don't all eigenvalues have a corresponding eigenvector (though not necessarily with the same multiplicity, unless we move to generalized eigens)? If $\lambda$ is an eigenvalue (a root of the characteristic polynomial), then $(A-\lambda I)$ is not invertible (otherwise its determinant would be nonzero), so it must have nontrivial cokernel and hence nontrivial kernel and hence $\lambda$ has an eigenvector, right?2012-07-01
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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