On completion of a linearly independent set to a basis with shifts of a fixed vector

Abstract: Let $\mathbb{F}$ be an infinite field. Let $n$ be a positive integer and let $1\leq d\leq n$. Let $\vec{f}1, \vec{f}_2, \ldots, \vec{f}{d-1} \in \mathbb{F}^{n}$ be $d-1$ linearly independent vectors. Let $\vec{x}=(x_1,x_2,\ldots,x_{d},0,0,\ldots,0)\in\mathbb{F}^{n}$, with $n-d$ zeros at the end. Let $\vec{R}: \mathbb{F}^n \to\mathbb{F}^n$ be the cyclic shift operator to the right, e.g. $\vec{R}\,\vec{x} = (0,x_1,x_2,\ldots,x_{d},0,0,\ldots,0)$. Is there a vector $\vec{x} \in \mathbb{F}^n$, such that the $n-d+1$ vectors $\vec{x},\vec{R}\vec{x}, \ldots ,\vec{R}^{n-d}\vec{x}$ complete the set ${\vec{f}j}{j=1}^{d-1}$ to a basis of $\mathbb{F}^n$? The answer is in the affirmative for every linearly independent set of $\vec{f}_j$, $j=1,2,\ldots,d-1$. In order to prove this fact, we prove that the $(n-d+1)\times(n-d+1)$ minors of the $(n-d+1)\times(n-d+1)$ circulant matrix. $\begin{bmatrix} \vec{x}, \vec{R} \vec{x}, \ldots, \vec{R}^{n-d} \vec{x} \end{bmatrix}^\intercal$ form a Gr\"obner basis with respect to the graded reverse lexicographic order (grevlex).