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 $1\le d\le n$. Let ${\vec{f}}_1,{\vec{f}}_2,\dots ,{\vec{f}}_{d-1}\in {\mathbb{F}}^n$ be $d-1$ linearly independent vectors. Let $\vec{x}=(x_1,x_2,\dots ,x_d,0,0,\dots ,0)\in {\mathbb{F}}^n$, 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. $\vec{R}\vec{x}=(0,x_1,x_2,\dots ,x_d,0,0,\dots ,0)$. Is there a vector $\vec{x}\in {\mathbb{F}}^n$, such that the $n-d+1$ vectors $\vec{x},\vec{R}\vec{x},\dots ,{\vec{R}}^{n-d}\vec{x}$ 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 ${\vec{f}}_j$, $j=1,2,\dots ,d-1$. 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. ${\left[\begin{array}{c}\vec{x},\vec{R}\vec{x},\dots ,{\vec{R}}^{n-d}\vec{x}\end{array}\right]}^{⊺}$ form a Gröbner basis with respect to the graded reverse lexicographic order (grevlex).