I’m trying to prove the following statement for any field $F$ (with no zero-divisors). I don’t know for sure that it’s true, but I’m pretty sure, and I would appreciate some advice about how to come up with an elegant proof that doesn’t involve refer to other advanced theorems.
Suppose that $\alpha_i,u_i\in F$, and that $$\alpha_1 u_1+\alpha_2 u_2+...+\alpha_n u_n=0$$ which also equals $0$ if we cyclically permute the $u_i$. It is also known that $\sum \alpha_i=0$ but $\sum u_i \ne 0$. How can we prove that either the sequence $u_1,u_2,...u_n$ is “cyclically periodic” or that all $\alpha_i=0$?
Whenever I try to prove this, I just keep going in circles. Can someone give me a hint?
NOTE: By “cyclically periodic” I mean that $u_{i+k}=u_i$ for all $i$ for some $0<k<n$, if we define $u_m=u_{m\bmod n}$ for $m > n$.
From the finite sequence $u_1,\ldots,u_n$, you can create an infinite sequence $(v_i)$ by setting $v_i := u_{i\pmod{n}}$. The linear relation you gave becomes $$\alpha_1 v_i+\cdots + \alpha_n v_{n+i}=0$$ which holds for all $i$. This shows that $(v_i)$ is a linear recurrence sequence. But on the other hand, $(v_i)$ takes values in the finite set $\{u_1,\ldots,u_n\}$.
It's not hard to show that if you have a linear recurrence with values in a finite set, then it must be periodic.