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

TL;DR

This paper proves that for any linearly independent set in an infinite field, one can find a vector whose cyclic shifts, combined with this set, form a basis of the vector space, using properties of circulant matrices and Gröbner bases.

Contribution

It establishes the existence of a vector whose shifts complete a linearly independent set to a basis, employing Gröbner basis techniques on circulant matrices.

Findings

Existence of such a vector for any linearly independent set.

Circulant matrix minors form a Gröbner basis under grevlex order.

The result holds over infinite fields.

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).

Equations

$$\mathcal{F}=\{{\mathbf{f}}_1,{\mathbf{f}}_2,\dots ,{\mathbf{f}}_{d-1}\}\subseteq {\mathbb{F}}^n$$

$$\mathbf{x}=(\underset{\text{d entries}}{\underbrace{x_1,x_2,\dots ,x_d}},\underset{\text{n−d zeros}}{\underbrace{0,0,\dots ,0}})\in {\mathbb{F}}^n.$$

$${\mathbf{R}}^k\mathbf{x}=(\underset{\text{k zeros}}{\underbrace{0,0,\dots ,0}},\underset{\text{d entries}}{\underbrace{x_1,\dots ,x_d}},\underset{\text{n−d−k zeros}}{\underbrace{0,0,\dots ,0}}).$$

$$\mathcal{F}={\left\{{\mathbf{f}}_j\right\}}_{j=1}^{d-1}\subseteq {\mathbb{F}}^n$$

$$\left\{\mathbf{x},\mathbf{Rx},\dots ,{\mathbf{R}}^{n-d}\mathbf{x}\right\}$$

$$\mathbf{M}=\left[\begin{array}{cccccccc}{x}_1& x_2& \ddots & x_d& 0& 0& \ddots & 0\\ 0& x_1& x_2& \ddots & x_d& 0& \ddots & 0\\ \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots \\ 0& 0& 0& \ddots & \ddots & \ddots & x_d& 0\\ 0& 0& 0& \ddots & \ddots & \ddots & x_{d-1}& x_d\\ f_{1,1}& f_{1,2}& \dots & \dots & \dots & \dots & f_{1,n-1}& f_{1,n}\\ f_{2,1}& f_{2,2}& \dots & \dots & \dots & \dots & f_{2,n-1}& f_{2,n}\\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\ f_{d-1,1}& f_{d-1,2}& \dots & \dots & \dots & \dots & f_{d-1,n-1}& f_{d-1,n}\end{array}\right]=\left[\begin{array}{c}\mathbf{X}\\ \mathbf{F}\end{array}\right]$$

$$\mathbf{M}=\left[\begin{array}{cccccc}{x}_1& x_2& x_3& x_4& 0& 0\\ 0& x_1& x_2& x_3& x_4& 0\\ 0& 0& x_1& x_2& x_3& x_4\\ f_{1,1}& f_{1,2}& f_{1,3}& f_{1,4}& f_{1,5}& f_{1,6}\\ f_{2,1}& f_{2,2}& f_{2,3}& f_{2,4}& f_{2,5}& f_{2,6}\\ f_{3,1}& f_{3,2}& f_{3,3}& f_{3,4}& f_{3,5}& f_{3,6}\end{array}\right].$$

$$\mathbf{X}={\left[\begin{array}{c}\mathbf{x},\mathbf{Rx},\dots ,{\mathbf{R}}^{n-d}\mathbf{x}\end{array}\right]}^{⊺}$$

$$\mathbf{X}=\left[\begin{array}{cccccccc}{x}_1& x_2& \ddots & x_d& 0& 0& \ddots & 0\\ 0& x_1& x_2& \ddots & x_d& 0& \ddots & 0\\ \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots \\ 0& 0& 0& \ddots & \ddots & \ddots & x_d& 0\\ 0& 0& 0& \ddots & \ddots & \ddots & x_{d-1}& x_d\end{array}\right]$$

$$\det \mathbf{M}=\sum _{H\in {\mathcal{P}}_{n-d+1}^n}{\epsilon }^H\det {\mathbf{X}}_H\det {\mathbf{F}}_{H^{\prime }}.$$

$${\left\{\det {\mathbf{X}}_H\right\}}_{H\in {\mathcal{P}}_{n-d+1}^n}$$

$$\mathbf{X}=\left[\begin{array}{cccccccc}{x}_1& x_2& \ddots & x_d& 0& 0& \ddots & 0\\ 0& x_1& x_2& \ddots & x_d& 0& \ddots & 0\\ \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots \\ 0& 0& 0& \ddots & \ddots & \ddots & x_d& 0\\ 0& 0& 0& \ddots & \ddots & \ddots & x_{d-1}& x_d\end{array}\right]$$

$$\mathcal{X}=\left\{\det {\mathbf{X}}_H:H\in {\mathcal{P}}_{n-d+1}^n\right\}$$

$$x_1^{k_1}{x}_2^{k_2}\cdots x_d^{k_d}$$

$$k_j=\#\{i:1\le i\le n-d+1,c_i-i=j-1\}.$$

$$0\le c_i-i\le d-1.$$

$$\sum _{j=1}^d{k}_j=n-d+1.$$

$$j_{i+1}-j_i=(c_{i+1}-(i+1)+1)-(c_i-i+1)=c_{i+1}-c_i-1\ge 0.$$

$$c_{i+1}-(i+1)+1\le (c_i-i+1)-1.$$

$$c_{\ell }^{\prime }=\{\begin{array}{ll}{c}_{\ell }& \text{if ℓ∈/{i,i+1},}\\ c_{i+1}& \text{if ℓ=i,}\\ c_i& \text{if ℓ=i+1.}\end{array}$$

$$j_{\ell }^{\prime }=\{\begin{array}{ll}{j}_{\ell }& \text{if ℓ∈/{i,i+1},}\\ j_{i+1}+1& \text{if ℓ=i,}\\ j_i-1& \text{if ℓ=i+1.}\end{array}$$

$$j_i^{\prime }=c_i^{\prime }-i+1=c_{i+1}-(i+1)+2=j_{i+1}+1$$

$$j_{i+1}^{\prime }=c_{i+1}^{\prime }-(i+1)+1=(c_i-i+1)-1=j_i-1.$$

$$J=\{j_{i+1},j_{i+1}+1,j_i-1,j_i\}.$$

$$j_{i+1}<j_{i+1}+1\le j_i-1<j_i$$

$$k_{j_{i+1}}^{\prime }$$

$$k_{j_i-1}^{\prime }$$

$$(k_{j_{i+1}+1}^{\prime }$$

$$k_{j_i}^{\prime }$$

$$k_{j_{i+1}}^{\prime }$$

$$k_{j_i-1}^{\prime }$$

$$k_{j_{i+1}+1}^{\prime }$$

$$k_{j_i}^{\prime }$$

Full text

1. Background

Let $\mathbb{F}$ be an infinite field. Therefore, every polynomial $f\in\mathbb{F}[x_{1},x_{2},\ldots,x_{d}]$ is either [math] or is not identically [math] as a function.

Let $n$ be a positive integer and let $1\leq d\leq n$. Let

[TABLE]

be a subset of $d-1$ linearly independent vectors. In many problems we need to complete this set with $n-d+1$ linearly independent vectors to a basis of $\mathbb{F}^{n}$. Furthermore, we may need the completing vectors to satisfy some constraints. For instance, we can always pick the completing vectors from the standard basis of $\mathbb{F}^{n}$.

In the current paper we constrain the completing set to shifts of a fixed vector with a contiguous block of $n-d$ zeros. More precisely, let

[TABLE]

Let $\mathbf{R}:\mathbb{F}^{n}\to\mathbb{F}^{n}$ be the cyclic shift operator to the right. Thus for $k=0,1,\ldots,n-d$:

[TABLE]

The main objective of the current paper is in the theorem below:

Theorem 1**.**

For every integer $n$ and $d$, $1\leq d\leq n$ and a linearly independent set

[TABLE]

there is a vector $\mathbf{x}\in\mathbb{F}^{n}$ of the form (1) such that the set of $n-d+1$ vectors

[TABLE]

completes the set $\mathcal{F}$ to a basis of $\mathbb{F}^{n}$.

Proof.

Split into a number of lemmas that follow. ∎

The pursuit of the proof led us to considerations in ideal theory in the ring of multivariate polynomials. In particular, we find a Gröbner basis for an ideal generated by the minors of a circulant matrix. We proceed to briefly outline the transition to ideal theory which allows us to prove Theorem 1. We will only need a minimum background from Gröbner basis theory which can be found in any general reference on the subject (e.g. [2, 1]).

The main result can be formulated in matrix form by considering the square $n\times n$ matrix of this special form:

[TABLE]

where $\mathbf{X}$ is a $(n-d+1)\times n$ matrix and $\mathbf{F}$ is a $(d-1)\times n$ matrix. If for every $\mathbf{F}$ of rank $d-1$ we can find $\mathbf{x}$ such that $\mathbf{M}$ is non-singular then the proof of Theorem 1 follows.

We make an observation that $\mathbf{X}$ is a special Toeplitz, circulant matrix [5].

Example 1**.**

We explicitly write down the matrix $\mathbf{M}$ for $n=6$ and $d=4$:

[TABLE]

We consider the matrix

[TABLE]

obtained by putting $\mathbf{x}$ and its shifts in the rows of the matrix $\mathbf{X}$, i.e. the $(n-d+1)\times n$ circulant matrix

[TABLE]

The symbols $x_{1},x_{2},\ldots,x_{d}$ are variables with range $\mathbb{F}$. Thus, the minors of the matrix $\mathbf{X}$ are polynomials in the ring of polynomials $\mathbb{F}[x_{1},x_{2},\ldots,x_{d}]$.

Lemma 1**.**

Let $\mathcal{P}_{q}^{n}$ denote the family of all subsets $H\subseteq\{1,2,\ldots,n\}$ with $q$ elements. The determinant of the special matrix $\mathbf{M}$ is a homogeneous polynomial in variables $x_{1},x_{2},\ldots,x_{d}$ of degree $n-d$ and

[TABLE]

where $\varepsilon^{H}=\pm 1$ and

1. (1)

$\mathbf{X}_{H}$* is the submatrix of $\mathbf{X}$ obtained by taking a subset of $n-d+1$ columns of $\mathbf{X}$ with indices in the subset $H$;* 2. (2)

$\mathbf{F}_{H^{\prime}}$* is the submatrix of $\mathbf{F}$ obtained from $\mathbf{F}$ by taking $d-1$ columns with indices in the complement $H^{\prime}$ of $H$.*

Proof.

We use Laplace’s expansion of the determinant by complementary minors [3]. We use the expansion along the first $n-d+1$ rows. ∎

Lemma 2**.**

Let $\mathbf{f}_{j}\in\mathbb{F}^{n}$, $j=1,2,\ldots,d-1$ be an arbitrary linearly independent set. Let the set

[TABLE]

be a linearly independent set in $\mathbb{F}_{n-d+1}[x_{1},x_{2},\ldots,x_{d}]$, the vector space of homogenous polynomials of degree $n-d+1$. Then there exists a vector $\mathbf{x}\in\mathbb{F}^{n}$ with $n-d$ zeros at the end such that $\det\mathbf{M}\neq 0$.

Proof.

By way of contradition, we assume that $\det\mathbf{M}=0$ for all $\mathbf{x}\in\mathbb{F}^{n}$ whose last $n-d$ coordinates are [math]. This implies that $\det\mathbf{M}\in\mathbb{F}[x_{1},x_{2},\ldots,x_{d}]$ is a polynomial identically equal to [math] as a function. As $\mathbb{F}$ is an infinite field, this polynomial is a zero polynomial. Therefore, linear independence of the set (3) and equation (2) imply that $\det\mathbf{F}_{H^{\prime}}=0$ for all $H\in\mathcal{P}_{n-d+1}^{n}$. Since $H^{\prime}$ traverses all sets of cardinality $d-1$, this in turn implies that the rank of the matrix $\mathbf{F}$ is strictly less than $d-1$, i.e. vectors $\mathbf{f}_{j}$, $j=1,2,\ldots,d-1$, are linearly dependent. This contradicts the assumptions. ∎

Corollary 1**.**

Theorem 1 will follow once we prove the independence of the set (3).

2. Ideal theory considerations

The objective of this section is to prove

Theorem 2**.**

The $(n-d+1)\times(n-d+1)$ minors of the circulant matrix $\mathbf{X}$ form a Gröbner basis with respect to the graded reverse lexicographic order (grevlex).

Let $\mathbf{X}$ be the $(n-d+1)\times(n+1)$ matrix

[TABLE]

Let us recall that $\mathcal{P}_{n-d+1}^{n}$ is the family of all subsets $H\subset\{1,2,\ldots,n\}$ of cardinality $n-d+1$. Also, $\mathbf{X}_{H}$ is the submatrix of $\mathbf{X}$ with columns whose indices are in $H$. The set of minors

[TABLE]

is a family of homogeneous polynomials of degree $n-d+1$ in variables $x_{1},x_{2},\ldots,x_{d}$.

We will use some basic techniques from computational algebraic geometry. One is that of a monomial order or term order. The only monomial order relevant to this paper is the graded reverse lexicographic order also abbreviated as grevlex. We assume that every polynomial has its terms ordered in descending (grevlex) order. The largest term (monomial) is called the leading term (monomial) of a polynomial, and will be denoted $LM(f)$ for a polynomial $f$. The family $LM(\mathcal{X})$ is the set of all leading monomials $LM(f)$ of all elements $f\in\mathcal{X}$.

Lemma 3**.**

The set $LM(\mathcal{X})$ is the set of all monomials of degree $n-d+1$ in variables $x_{1},x_{1},x_{2},\ldots,x_{d}$. In particular, both $LM(\mathcal{X})$ and $\mathcal{X}$ itself are both bases of the linear vector space $\mathbb{F}_{n-d+1}[x_{1},x_{2},\ldots,x_{d}]$ of homogeneous polynomials of degree $n-d+1$.

Proof.

Let $H=\{c_{1},c_{2},\ldots,c_{n-d+1}\}$ be the set of distinct $n-d+1$ column indices ($1\leq c_{j}\leq n$), where the sequence $c_{j}$ is non necessarily increasing. Using Leibniz formula for determinants [4], we know that the minor is a sum of products, up to the sign:

[TABLE]

where for $j=1,2,\ldots,d$:

[TABLE]

We require that for $i=1,2,\ldots,n-d+1$.

[TABLE]

Essentially, $k_{j}$ represent the frequency table of the sequence $c_{i}-i+1$, $i=1,2\ldots,n-d+1$. Besides $k_{j}\geq 0$, we also must have

[TABLE]

Moreover, $j=c_{i}-i+1$ represents the horizontal distance from the diagonal, of the entry at the intersection of row $i$ with column $c_{i}$. In addition, if we reorder the sequence $c_{i}$ to a sequence $c_{i}^{\prime}=c_{\rho(i)}$, where $\rho:\{1,2,\ldots,n-d+1\}\to\{1,2,\ldots,n-d+1\}$ is a permutation, while preserving condition (4), then the new sequence $k_{j}^{\prime}$ represents another monomial in the same minor $\det\mathbf{X}_{H}$. The question arises: which $\rho$ yields the monomial maximal in the grevlex order? There are two objectives to achieve by reordering the sequence $c_{j}$:

1. (1)

Let $k_{j}>0$ be the last $j$ with this property, i.e. $k_{j+1},k_{j+2},\ldots,k_{d}$ are all zero; we make $j$ the lowest possible. We recall that grevlex considers variables to be ordered from the highest, i.e. $x_{d}>x_{d-1}>\ldots>x_{1}$; therefore, a monomial which has variable $x_{j}$ but no variables $x_{j+1},x_{j+2},\ldots,x_{d}$, is greater than all polynomials which have one of those variables. 2. (2)

Once we achieved the first goal, we minimize $k_{j}$ i.e. the power of $x_{j}$; this is also the count of $i$ such that $c_{i}-i=j-1$. The grevlex order considers the term with the lower power $k_{j}$ greater.

We claim that the optimal ordering is achieved when the sequence $c_{i}$ is strictly increasing. The remainder of the current proof is devoted to the proof of this claim.

We note that $c_{i}$ is strictly increasing iff the sequence $j_{i}=c_{i}-i+1$ is non-decreasing. Indeed

[TABLE]

The numbers $k_{j}$ are simply the counts of the number of times $i$ such that $j_{i}=j$, i.e. level counts of the non-decreasing sequence $j_{i}$. Obviously, knowing $k_{j}$ for $j=1,2,\ldots,d$ such that $\sum_{j=1}^{d}k_{j}=n-d+1$ allows us to reconstruct the non-decreasing sequence $j_{i}$, $i=1,2,\ldots,n-d+1$ uniquely. Then the equation $j_{i}=c_{i}-i+1$ allows us to reconstruct $c_{i}$. Moreover, $c_{i+1}-c_{i}=j_{i+1}-j_{i}+1$, so $c_{i}$ is strictly increasing if $j_{i}$ is non-decreasing. Thus, we established a 1:1 correspondence between sequences $(k_{j})_{j=1}^{d}$, $(j_{i})_{i=1}^{n-d+1}$ and $(c_{i})_{i=1}^{n-d+1}$ with the specified properties.

It remains to be shown that the monomial obtained from an unsorted sequence $c_{i}$ is strictly smaller in grevlex order than that obtained from the sorted sequence. Thus, let us assume that the sequence $j_{i}=c_{i}-i+1$ is not a non-decreasing sequence; hence, for some $i$, $1\leq i<n-d+1$ $j_{i+1}<j_{i}$ or $j_{i+1}\leq j_{i}-1$. Equivalently

[TABLE]

Hence $c_{i+1}\leq c_{i}$. Thus $c_{i+1}<c_{i}$ and $j_{i+1}\leq j_{i}-2$. Let us consider the sequence $c_{\ell}^{\prime}$, $\ell=1,2,\ldots,n-d+1$, obtained by swapping $c_{i}$ and $c_{i+1}$. We will prove that by doing so we increased the monomial $x_{1}^{k_{1}}\,x_{2}^{k_{2}}\cdots x_{d}^{k_{d}}$. Thus

[TABLE]

Therefore $j_{\ell}^{\prime}=c_{\ell}^{\prime}-\ell+1$ is given by:

[TABLE]

Indeed,

[TABLE]

and

[TABLE]

As $j_{i+1}+2\leq j_{i}$ and $j_{i}-1\geq j_{i+1}+1$, all values remain in the range from $1$ to $d$, and the sequence $c_{\ell}^{\prime}$ is valid (satisfies $1\leq c_{\ell}^{\prime}-\ell+1\leq d$). The value of $k_{j}$ does not change unless $j\in J$, where

[TABLE]

As $j_{i+1}+1\leq j_{i}-1$, we have

[TABLE]

the set $J$ consists of either $3$ or $4$ values. It is easy to see how values $k_{j}$ change for $j\in J$. There are 2 cases.

Case $j_{i}-1=j_{i+1}+1$: when $J$ consists of $3$ elements. Then

[TABLE]

The parenthesized third equation is a copy of the second equation.

Case $j_{i+1}+1<j_{i}-1$: when $J$ consists of $4$ elements. Then

[TABLE]

Since the power of the highest-indexed impacted variable $x_{j_{i}}$ drops, the new sequence $k_{\ell}^{\prime}$, $\ell=1,2,\ldots,n-d+1$ corresponds to a monomial which is strictly greater in grevlex order. This proves that the maximum monomial is obtained when $c_{\ell}$ is sorted in ascending order.

This completes the proof of the main claim, and the entire lemma. ∎

Remark 1**.**

It follows that $LM\left(\det\mathbf{X}_{H}\right)$ with respect to the grevlex order is the product of the diagonal entries of $\mathbf{X}_{H}$.

Remark 2**.**

The set $\mathcal{X}$ is a Gröbner basis of the ideal generated by $\mathcal{X}$. This follows from Lemma 3 and the Buchberger criterion. This ideal is also $\mathfrak{m}^{n-d+1}$ where $\mathfrak{m}=\langle x_{1},x_{2},\ldots,x_{d}\rangle$ is the maximal ideal generated by $x_{1},x_{2},\ldots,x_{d}$ and is the ideal of homogeneous polynomials vanishing at [math].

Example 2**.**

We give a counterexample to the conjecture that the grevlex order could be substituted with the (graded) lexicographic order (lex or grlex). Let the matrix be:

[TABLE]

Thus $d=3$ and $n=4$. We find:

[TABLE]

We also find that with grlex or lex used instead of grevlex:

[TABLE]

Notably, $x_{2}^{2}$ is missing. We note that $x_{2}^{2}<x_{1}\,x_{3}$ with respect to the (graded) lexicographic order, but $x_{2}^{2}>x_{1}\,x_{3}$ with respect to grevlex. Therefore,

[TABLE]

Hence, consistent with the statement of Lemma 3,

[TABLE]

3. Generalizations

The assumption that $\mathbb{F}$ is an infinite field can be relaxed. If we assume that $\mathbb{F}$ is an infinite integral domain then the only identically vanishing polynomial is the zero polynomial. This is sufficient to prove Theorem 1.

Theorem 2 and Lemma 3 are valid over any ring with unity.

4. Acknowledgments

I thank Johnatan Ashbrock for the question that led to Theorem 1.