An extremal problem for integer sparse recovery

TL;DR

This paper investigates the maximum size of integer matrices with bounded entries such that all their square submatrices are invertible, providing new bounds and solving a related geometric covering problem.

Contribution

It introduces new bounds on the size of such matrices and addresses a specific case of a grid covering problem in higher dimensions.

Findings

Established new upper bounds on matrix size d(m,k)

Derived lower bounds for the extremal problem

Solved a special case of the grid covering problem

Abstract

Motivated by the problem of integer sparse recovery we study the following question. Let $A$ be an $m \times d$ integer matrix whose entries are in absolute value at most $k$. How large can be $d=d(m,k)$ if all $m \times m$ submatrices of $A$ are non-degenerate? We obtain new upper and lower bounds on $d$ and answer a special case of the problem by Brass, Moser and Pach on covering $m$-dimensional $k \times \cdots\times k$ grid by linear subspaces.

Equations

$$m<d\le \max (k+1,k^{m/(m-1)}/2).$$

$$k^{m/(m-1)-o(1)}\le M\le C_m{k}^{m/(m-1)}.$$

$$M\ge k^{m/(m-1)}/(2m-2).$$

$$h_j$$

$$\frac{{\sum }_{j=1}^d{h}_j}{10^{2t}}\ge \frac{d}{100k\sqrt{t}}$$

$$d\le 100km\sqrt{t}=100km\sqrt{\log k}$$

$$\frac{(\Lambda +1{)}^m}{8\Lambda k\sqrt{m}}\ge \frac{25^m{k}^{m/(m-1)}}{200k^{m/(m-1)}\sqrt{m}}>2,$$

$$h_j$$

$$\frac{{\sum }_{j=1}^d{h}_j}{(\Lambda +1{)}^{2m}}\ge \frac{d}{16\Lambda k\sqrt{m}}$$

$$d\le 16\Lambda km\sqrt{m}\le 400k^{m/(m-1)}{m}^{3/2}$$

Full text

1 Introduction

Compressed sensing is a relatively new mathematical paradigm that shows a small number of linear measurements are enough to efficiently reconstruct a large dimensional signal under the assumption that the signal is sparse (see, e.g., [4] and its references). That is, given a signal ${\bf{x}}\in{\mathbb{R}}^{d}$, the goal is to accurately reconstruct $\bf{x}$ from its noisy measurements ${\bf b}=A{\bf x}+{\bf e}$. Here, $A$ is an underdetermined matrix $A\in\mathbb{R}^{m\times d}$, where $m$ is much smaller than $d$, and ${\bf{e}}\in{\mathbb{R}}^{m}$ is a vector modeling noise in the system. Since the system is highly underdetermined, it is ill-posed until one imposes additional constraints, such as the signal $\bf{x}$ obeying a sparsity constraint. We say $\bf{x}$ is $s$-sparse when it has at most $s$ nonzero entries. Clearly, any matrix $A$ that is one-to-one on $s$-sparse signals will allow reconstruction in the noiseless case when $\bf{e}=0$. However, compressed sensing seeks the ability to reconstruct efficiently and robustly even when one allows presence of noise. Motivated by this problem Fukshansky, Nadel and Sudakov [5] considered the following extremal problem, which is of independent interest.

Problem 1.1**.**

Given integers $k,m$, what is the maximum integer $d$ such that there exists $m\times d$ matrix $A$ with integer entries satisfying $|a_{ij}|\leq k$ such that all $m\times m$ submatrices of $A$ are non-degenerate?

To see the connection of this question with integer sparse recovery let $s\leq m/2$ and consider $s$-sparse signal ${\bf{x}}\in\mathbb{Z}^{d}$. We denote by $\|\bf{b}\|$ the Euclidean norm of a vector ${\bf b}=(b_{1},\dots,b_{m})\in\mathbb{R}^{m}$ and by $\|\bf{b}\|_{\infty}$ its $l_{\infty}$-norm: $\|{\bf b}\|_{\infty}=\max_{i=1,\dots,m}|b_{i}|$. Suppose we wish to decode ${\bf{x}}$ from the noisy measurements ${\bf b}=A{\bf x}+{\bf e}$ where $\|{\bf{e}}\|_{\infty}<\frac{1}{2}$ (in particular, this holds if $\|{\bf{e}}\|<\frac{1}{2}$). Note that by definition of matrix $A$ we have for any $m$-sparse integer non-zero vector $\bf{z}$ that $A{\bf z}\not={\bf 0}$ and therefore, being an integer vector, $A{\bf z}$ has $l_{\infty}$-norm at least one. So to decode $\bf{x}$ we can select the $s$-sparse signal ${\bf y}\in\mathbb{Z}^{d}$ minimizing $\|{\bf b}-A{\bf y}\|_{\infty}$. Then, since $\bf{x}$ satisfies ${\|{\bf b}-A{\bf x}\|_{\infty}=\|{\bf e}\|_{\infty}}<\frac{1}{2}$, it must be that the decoded vector $\bf{y}$ satisfies this inequality as well. Therefore, $\|A{\bf y}-A{\bf x}\|_{\infty}\leq\|{\bf b}-A{\bf y}\|_{\infty}+\|{\bf b}-A{\bf x}\|_{\infty}<1$ is an $m$-sparse vector, which guarantees that ${\bf y}={\bf x}$ so our decoding was successful. Note that if instead of error $<1/2$ we want to allow error $<C$ we can simply multiply all entries of $A$ by a factor of $2C$.

Fukshansky, Nadel and Sudakov [5] showed that a matrix $A\in{\mathbb{R}}^{m\times d}$ with integer entries $|a_{ij}|\leq k$ and all $m\times m$ submatrices having full rank must satisfy $d=O(k^{2}m)$. They also proved that such matrices exists when $d=\Omega(\sqrt{k}m)$. Their upper bound was improved by Konyagin [6] who showed that $d$ must have order at most $O(k(\log k)\,m)$ (all logarithms here and later in the paper are in base $e$) for $m\geq\log k$ and at most $O(k^{m/(m-1)}\,m^{2})$ for $2\leq m<\log k$. Improving these results further, in this paper we obtain the following new upper bound.

Theorem 1.2**.**

Let $A$ be an $m\times d$ integer matrix such that $|a_{ij}|\leq k$, and all $m\times m$ submatrices of $A$ have full rank. If $k$ is sufficiently large, then $d\leq 100k\sqrt{\log k}\,m$ for $m\geq\log k$ and $d\leq 400k^{m/(m-1)}\,m^{3/2}$ for $2\leq m<\log k$.

The lower bound construction for Problem 1.1 uses random matrices and is based on a deep result of Bourgain, Vu and Wood [3] which estimates the probability that a random $m\times m$ matrix with integer entries from $[-k,k]$ is singular. It is expected that their result is not tight and the probability of singularity for such matrix has order $k^{-(1-o(1))m}$ as $k\to\infty$. If this is the case then $m\times d$ matrices, satisfying Problem 1.1, exist for $d$ close to $km$. This suggests that our new bound for $m\geq\log k$ is not far from being optimal.

On the other hand, we get the following result.

Theorem 1.3**.**

Let $k\in\mathbb{N}$, $m\in\mathbb{N}$, $m\geq 2$, and

[TABLE]

Then there is an $m\times d$ integer matrix $A$ such that $|a_{ij}|\leq k$ and all $m\times m$ submatrices of $A$ have full rank.

We observe that this theorem improves the lower bound from [5] for $m=o(\sqrt{k})$. Moreover, the existence of required matrices in [5] was proven by using probabilistic arguments, but the matrices in Theorem 1.3 are explicit and easily computable. Also we notice that upper and lower estimate for $d$ in Theorems 1.2 and 1.3 differ by a factor $O(m^{3/2})$ depending on $m$ only.

Our result can be also used to solve a special case of a problem by Brass, Moser and Pach. In [2] (Chapter 10.2, Problem 6) they asked: what is the minimum number $M$ of $s$-dimensional linear subspaces necessary to cover $m$-dimensional $k\times\cdots\times k$ grid $K=\{{\bf x}\in\mathbb{Z}^{m}:\|{\bf x}\|_{\infty}\leq k\}$? Balko, Cibulko and Valtr [1] studied this problem and obtain upper and lower bounds for $M$. In particular in the case when $s=m-1$ they proved that

[TABLE]

Using Theorem 1.3 we obtain a new lower bound which is tight up to a constant factor.

Corollary 1.4**.**

For $k\geq m\geq 2$ we have

[TABLE]

Indeed, suppose that we cover $K$ by $M$ hypersubspaces $P_{1},\dots,P_{M}$. We consider the columns of the matrix $A$ constructed in Theorem 1.3. Since any $m$ of them are linearly independent, every subspace $P_{i}$ contains at most $m-1$ of these columns. Thus, $d\leq M(m-1)$, and the corollary follows.

For matrices $A$ constructed in Theorem 1.3, Ryutin [7] suggested an effective algorithm for recovery of a sparse vector $\bf x$ via $A\bf x$. See also [8].

2 Proof of Theorem 1.2

Let $t=[\log k]$ for $m\geq\log k$ and $t=m$ for $2\leq m<\log k$. In the first case we suppose that $d>100k\sqrt{t}m$ and in the second case we suppose that $d>400k^{m/(m-1)}\,\sqrt{t}m$. Let $v_{1},\dots,v_{t}$ be the first $t$ rows of the matrix $A$. Take $\Lambda=9$ if $m\geq\log k$ and $\Lambda=[25k^{1/(m-1)}]$ otherwise. Given a vector of integer coefficients ${\bf{\lambda}}=(\lambda_{1},\ldots,\lambda_{t})$ such that $0\leq\lambda_{i}\leq\Lambda$ denote by $v_{\bf\lambda}$ a linear combination $\sum_{i}\lambda_{i}v_{i}$. Our goal is to find two combinations ${\bf\lambda}\not={\bf\lambda}^{\prime}$ such that corresponding vectors $v_{\bf\lambda}$ and $v_{\bf\lambda}^{\prime}$ agree on at least $m$ coordinates. This will show that a linear combination of first $t$ rows of matrix $A$ with coefficients ${\bf\lambda}-{\bf\lambda}^{\prime}\not=\bf{0}$ has at least $m$ zeros and therefore the $m\times m$ submatrix of $A$ whose columns correspond to these zeros is degenerate, since its first $t$ rows are linearly dependent.

Consider ${\bf\lambda}$ chosen uniformly at random out of $(\Lambda+1)^{t}$ possible vectors and look at a value of a fixed coordinate $j$ of the vector $v_{\bf\lambda}$. This value is a random variable $X$ which is a sum of the $t$ independent random variables $X_{i}$, where $X_{i}$ is a value of the $j$-th coordinate of $\lambda_{i}v_{i}$. Since $|a_{ij}|\leq k$, we have that $|X_{i}|\leq\Lambda k$ and therefore its variance $\operatorname{Var}(X_{i})\leq\mathbb{E}(X_{i}^{2})\leq\Lambda^{2}k^{2}$. This implies that $\operatorname{Var}(X)=\sum_{i}\operatorname{Var}(X_{i})\leq\Lambda^{2}k^{2}t$. Thus, by Chebyshev’s inequality, with probability at least $3/4$, the value of $X$ belongs to an interval $I$ of length $4\sqrt{\operatorname{Var}(X)}\leq 4\Lambda k\sqrt{t}$. Hence there are at least $0.75\cdot(\Lambda+1)^{t}$ values of $\bf\lambda$ giving a linear combination $v_{\bf\lambda}$ whose $j$-th coordinate belongs to $I$. For every integer $s$ let $h_{j}(s)$ be the number of values for $\bf\lambda$ giving linear combinations $v_{\bf\lambda}$ whose $j$-th coordinate is $s$, and let $h_{j}$ be the number of ordered pairs ${\bf\lambda}\not={\bf\lambda}^{\prime}$ such that $v_{\bf\lambda}$ and $v_{\bf\lambda}^{\prime}$ agree on $j$-th coordinate. By definition $0.75\cdot(\Lambda+1)^{t}\leq\sum_{s\in I}h_{j}(s)\leq(\Lambda+1)^{t}$ and $h_{j}=\sum_{s}h_{j}(s)(h_{j}(s)-1)$.

If $m\geq\log k$, in which case $\Lambda=9$ and $t=[\log k]$, then using the Cauchy-Schwarz inequality, together with the facts that $10^{t}>k^{2}\geq k\sqrt{\log k}=k\sqrt{t}$ for sufficiently large $k$ and the interval $I$ contains at most $4\Lambda k\sqrt{t}+1$ integer points, we have

[TABLE]

Since the number of ordered pairs ${\bf\lambda}\not={\bf\lambda}^{\prime}$ is at most $10^{2t}$ and the number of coordinates $j$ is $d$, by averaging we obtain that there is a pair ${\bf\lambda}\not={\bf\lambda}^{\prime}$ which agrees on at least

[TABLE]

coordinates. If the pair ${\bf\lambda}\not={\bf\lambda}^{\prime}$ agrees on $\frac{d}{100k\sqrt{t}}\geq m$ coordinates, then as we explain above, this implies that $A$ has an $m\times m$ degenerate submatrix. Thus, we must have

[TABLE]

when $m\geq\log k$.

Now we consider the case $2\leq m<\log k$. Then, due to the inequality

[TABLE]

we have

[TABLE]

Since the number of ordered pairs ${\bf\lambda}\not={\bf\lambda}^{\prime}$ is at most $(\Lambda+1)^{2m}$ and the number of coordinates $j$ is $d$, by averaging we obtain that there is a pair ${\bf\lambda}\not={\bf\lambda}^{\prime}$ which agrees on at least

[TABLE]

coordinates. If the pair ${\bf\lambda}\not={\bf\lambda}^{\prime}$ agrees on $\frac{d}{16\Lambda k\sqrt{m}}\geq m$ coordinates, then as we explain above, this implies that $A$ has an $m\times m$ degenerate submatrix. Thus, we must have

[TABLE]

when $2\leq m<\log k$. This completes the proof of the theorem. $\Box$

3 Proof of Theorem 1.3

We have to construct the required $m\times d$ matrices with $d\geq k+1$, provided that $k\geq m$, and with $d\geq k^{m/(m-1)}/2$, provided that $k^{m/(m-1)}/2>m$.

First we will construct an $m\times d$ matrix with $d\geq k+1$. By Bertrand’s postulate, there exists an odd prime $d$ with $k+1\leq d\leq 2k+1$. We define the matrix $A$ by taking $a_{i,j}\equiv j^{i-1}(\bmod d)$ with $|a_{i,j}|\leq(d-1)/2\leq k$. Considering the matrix $A$ modulo $d$ we find that any submatrix of $A$ of size $m\times m$ is a Vandermonde matrix modulo $d$. Hence, its determinant is not zero modulo $d$. This implies that this submatrix has full rank.

Next we will construct an $m\times d$ matrix with $d\geq k^{m/(m-1)}/2$. We can assume that $k^{m/(m-1)}/2>k+1$ and, in particular, $k\geq 3$. By Bertrand’s postulate, there exists a prime $d$ with $k^{m/(m-1)}/2\leq d<k^{m/(m-1)}$. For $u\in\mathbb{R}$ we denote by $\|u\|$ the distance from $u$ to the nearest integer. Note that if we can multiply $u$ by some integer $\ell$ such that $\ell u$ is also an integer then the distance of $\ell u$ to the closest multiple of $\ell$ is exactly $\ell\|u\|$. Therefore we can find and integer $z$ such that $z\equiv\ell u(\bmod\ell)$ and $|z|=\ell\|u\|$.

We first consider the $m\times d$ matrix $A^{\prime}$ with entries $a_{i,j}^{\prime}=j^{i-1}$. Again, the determinant of any $m\times m$ submatrix of $A^{\prime}$ is not zero modulo $d$. The idea is to multiply the columns of $A^{\prime}$ by appropriate integers not divisible by $d$ such that when we replace all entries by their residues modulo $d$, the absolute values of this residues will be bounded by $k$. Clearly, the operation of multiplying by integers not divisible by a prime $d$ and taking residues$\mod d$ preserve the property that determinants of all $m\times m$ submatrices are not divisble by $d$.

Using Dirichlet’s theorem on simultaneous approximations (see, e.g., [9], Chapter 2, Theorem 1A), we find that for every $j=1,\dots,d$ there is a positive integer $l_{j}<d$ such that $\|l_{j}j^{i-1}/d\|\leq d^{-1/m}$ for $i=1,\dots,m$. Hence, for any $i$ there is an integer $a_{i,j}$ such that $a_{i,j}\equiv l_{j}j^{i-1}(\bmod d)$ and $|a_{i,j}|\leq d^{1-1/m}\leq k$ as required. This completes the proof of Theorem 1.3. $\Box$

Acknowledgement.  We thank the anonymous referees for many useful suggestions.