Binary vectors defined by remainders modulo prime numbers. What is the dimension of their span?

Question

Let numbers $n$ and $k$ such that $k \leq n$ be given.

Let $S$ be the set of prime numbers less than or equal to $k$.

We define a binary vector $v_{p, r}$ of length $n$ for each $p \in S$ and $r \in [p - 1] \cup \{0\}$ as follows.

For each $i \in [n-1] \cup \{0\}$:

• $(v_{p, r})_i = 1$ if $i \equiv r \; mod \; p$
• $(v_{p, r})_i = 0$ otherwise

Consider the set of all such binary vectors $X := \{ \; v_{p, r} \; | \; p \in S \text{ and } r \in [p - 1] \cup \{0\} \; \}$.

Question 1: Is $X$ linearly independent over $\mathbb{R}^{n}$? (No. See answer by @ChrisCulter)

Further, consider the vector space $V := span(X)$ such that $V$ is viewed as a subspace of $\mathbb{R}^{n}$.

I am trying to find bounds on $dim(V)$ in terms of $n$ and $k$. Any bounds would be greatly appreciated, but I am specifically trying to answer the following.

Question 2: What is the smallest $k$ (in terms of $n$) such that $dim(V) = n$? Can we always pick $k$ large enough to guarantee that $dim(V) = n$?

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Update

Based on @GerryMyerson's suggestion, I coded up some examples in Octave.

When $n = 400$ and $k = 61$, we have $dim(V) = n = 400$.

This suggests that we might be able to find an upper bound on the smallest $k$ (relative to $n$) such that $dim(V) = n$.

Here is my Octave code in case anyone wants to give it a try:

% Parameters
n = 400
k = 61

% Prime numbers
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113];
primes = primes(find(primes <= k));

% Compute number of rows & cols
count = 0;
for p = primes
  count += p;
end
rows = count
cols = n

% Construct matrix
M = zeros(rows, cols);
count = 1;
for p = primes
  for r = 0:(p-1)
    for c = 1:cols
      if r == mod(c, p)
        M(count, c) = 1;
      end
    end
    count += 1;
  end
end

% Print rank
rankOfM = rank(M)

Answer 1

Suppose $\dim(V) < n$. Then there is some nonzero vector $x$ orthogonal to all the $v_{r,p}$'s, which it's easy to see implies $\sum_{j=1}^n x_j z^j$ has roots at $z = e^{2\pi i \frac{m}{p}}$ for each $p \le k$ and $0 \le m \le p-1$, so since $x$ is nonzero and $\sum_{j=1}^n x_jz^j$ is a degree $n$ polynomial, we must have $1+\sum_{p \le k} (p-1) \le n$.

Answer 2

No, if $k>3$ then $V$ is not linearly independent (over any field). We can write the all-$1$s vector as two different linear combinations: $$\sum_r v_{2,r} = \sum_r v_{3,r}.$$ The same occurs if $X$ is any set containing at least two numbers.

Answer 3

For $n \ge 1$, let $k(n)$ denote the minimum value of $k$ for which the associated $V$ has $\dim(V) = n$.

Claim: For each $\epsilon > 0$, $k(n) \ge (1-\epsilon)\sqrt{n\log n}$ for all large $n$.

Proof: Less than $n$ vectors can't span an $n$-dimensional space, so we must have $n \le \sum_{p \le k} p \sim \frac{1}{2}\frac{k^2}{\log k}$. Clearly $k \le \sqrt{n}$ implies $\frac{1}{2}\frac{k^2}{\log k} \le n$. And if $k \in [\sqrt{n},(1-\epsilon)\sqrt{n\log n}]$, then $\frac{1}{2}\frac{k^2}{\log k} \le \frac{1}{2}\frac{(1-\epsilon)n\log n}{\log\sqrt{n}} = (1-\epsilon)n$. $\square$

.

Claim: For each $\epsilon > 0$, $k(n) \le (\frac{3}{4}+\epsilon)n$ for all large $n$. In particular, the answer to the second half of question 2 is "yes".

Proof: Out of pure laziness, I'll just prove $k(n) \le \frac{9n}{10}$ for all large $n$. The amount of effort to adapt the following to $(\frac{3}{4}+\epsilon)n$ is less than that to write this current sentence. By the prime number theorem, we may take some prime $p \in (\frac{9n}{10},n)$. Then, for $\frac{n}{10} \le r \le p-1$, $v_{p,r} = e_r$ (where $e_j = (0,\dots,0,1,0,\dots,0)$, the $1$ in index $j$). Take some $q \in (\frac{n}{2},\frac{8n}{10})$. Then, for $0 \le r \le \frac{n}{10}$, $v_{q,r} = e_r+e_{q+r}$, so $v_{q,r}+v_{p,q+r} = e_r$ (note $q+r < p$, so $v_{p,q+r}$ is valid). Finally, for $p \le r \le n-1$, $v_{q,r-q} = e_{r-q}+e_r$, so $v_{q,r-q}+v_{p,r-q} = e_r$. $\square$

.

Now some minor results and speculations.

Let $v_1 = (1,1,1,\dots,1)$ and $v_2,v_3,v_4,\dots = v_{2,1},v_{3,1},v_{3,2},v_{5,1},v_{5,2},v_{5,3},v_{5,4},v_{7,1}\dots,v_{7,6},v_{11,1}\dots$.

Conjecture: For $n \ge 4$, $v_1,\dots,v_n$ are linearly independent in $\mathbb{R}^n$.

This is the natural conjecture to make in light of the obvious dependence pointed out by Chris Cutler ($n=2,3$ are too small for silly reasons).

Unfortunately, this conjecture is false, and $n=11$ is the smallest counter-example. However, it appears from code to be nearly true. If the conjecture were true, then the minimum value of $k$ for which $\dim(V) = n$ is the minimum value of $k$ for which $1+ \sum_{p \le k} (p-1) \ge n$. By partial summation and the prime number theorem, one can see that $\sum_{p \le k} p \sim \frac{1}{2}\frac{k^2}{\log k}$, so the answer would be basically $k = \sqrt{n\log n}$. Since the conjecture seems to be nearly true, I would guess $k = \Theta(\sqrt{n\log n})$ is the answer to question 2.

Answer 4

We can use the same approach as was presented by @Ilya Bogdanov here: https://mathoverflow.net/questions/343355/do-the-following-binary-vectors-span-mathbbrn

Following the same format, we get a polynomial $$P_k(x) = \prod_{\substack{ p \, \in \, \mathtt{PRIMES \hspace{0.08em} \cup \hspace{0.08em} \{1\}}\\ p \leq k }} \Phi_p(x).$$

Therefore, the degree of $P_k$ is equal to $$1 + \sum_{\substack{ p \, \in \, \mathtt{PRIMES}\\ p \leq k }} (p - 1).$$

Further, we get that the degree of $P_k \sim \frac{k^2}{2 \log k}$ by applying results from: What is the sum of the prime numbers up to a prime number $n$?

We have that $dim(V) = n$ precisely when degree of $P_k \geq n$.

Therefore, $k(n) = \Theta(\sqrt{n \log n})$ as conjectured by @mathworker21.

Thank you very much @Ilya Bogdanov and @mathworker21!