A problem about a matrix in $SL_2(\mathbb Q)$ 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?
 A: 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 $|\mathrm{tr}(A)|<m$ 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.
