  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).
