Let $A=\begin{bmatrix}
0 & 0 & \dots & 0 & p \\
1 & 0 & \dots & 0 & 0 \\
0 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & 1 & 0
\end{bmatrix}$ be the $n\times n$ companion matrix whose characteristic function is $x^n-p$ (For convenience let's assume $p$ is prime so that the polynomial is irreducible, this might be true in more general cases). I wonder if any nonzero vector $v\in \mathbb Q^n$ is a cyclic vector. i.e. the set $\{v, Av, A^2v, …, A^{n−1}v\}$ is linearly independent.
Let $x=(x_1,\dots,x_n)$ be a nonzero vector in $\mathbb Q^n$. Then basically $A$ permutes the entries of $x$ around and multiply the first coordinate by $p$. It seems to me this will create a linear independent set $\{v, Av, A^2v, …, A^{n−1}v\}$ but I don't know how to prove it.
I think $p$ being a prime is vey crucial. In the case when $n=2$, $(x,y)$ and $(y,px)$ being linearly independent implies that $px^2=y^2$ never holds. In $n=3$ case, the nonzero determinant means $x^3-p^2z^3-py^3-3pxyz=0$ has no nontrivial rational solutions.
Best Answer
The answer is yes. Let $v\in \mathbb{Q}^n\setminus\{0\}$; the vector-space $U=span((A^iv)_i)$ is $A$-invariant.
When $gcd(m,n)=1$, the polynomial $x^n-p^m$ is irreducible over $\mathbb{Q}$. Thus, here, $\chi_A$ is irreducible over $\mathbb{Q}$ and $A$ has no proper invariant vector-space over $\mathbb{Q}$; finally $U=\mathbb{Q}^n$ and we are done.