are algebraic, then the eigenvalues of $U(t)$ must be algebraic. If we further suppose that $\theta_r / \theta_s$ is irrational, then Theorem \[gelfond\_schneider\] implies that one of $$\{ \theta_r / \theta_s, e^{it \theta_s}, e^{i t \theta_s (\theta_r/\theta_s)} \}$$ must be transcendental, which is a contradiction. Therefore $\theta_r / \theta_s$ must be rational for every pair of distinct eigenvalues $\theta_r$ and $\theta_s$ of $X$.[  ]{}

\[no\_algebraic\_reg\] If $X$ is regular and at time $t$ all entries of the transition matrix $U(t)$ are algebraic, then $X$ must have integral eigenvalues.

[*Proof.* ]{}First recall that all of the eigenvalues of $X$ are roots of a monic polynomial with integral coefficients. Therefore the eigenvalues of $X$ are algebraic integers. If we assume that $X$ is regular, then the largest eigenvalue $\theta_0$ is equal to the valency (see [@godsil01]). Using Theorem \[no\_algebraic\], we see that the nontrivial eigenvalues of $X$ must be rational. The only rational algebraic integers are the integers, and so we see that the eigenvalues of $X$ must be integral.[  ]{}

The following observation is useful in our following work.

If $U(t)$ is the complex scalar multiple of a matrix with all entries algebraic, then all entries of $U(t)$ are algebraic.

[*Proof.* 