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Let $n$ be a positive integer. Denote by $B_n$ the set of $n\times(n+1)$-matrices of rank $n$ and with coefficients in $\{0,1\}$. I would like to measure how "complex" the coefficients of a linear combination of the columns of a matrix of $B_n$ can be. More precisely, I'd like to compute (or estimate the asymptotic behaviour of) $$ P_n := \max \left\{ \prod_{i=1}^{n+1}|\lambda_i|,\; M\in B_n,\; \sum_{i=1}^{n+1}\lambda_i m_{\star,i} = 0,\; \lambda_1,\dots,\lambda_{n+1}\in \mathbb{Z},\; \gcd(\lambda_1,\dots,\lambda_{n+1})=1 \right\} $$

where $m_{\star,i}$ stands for the $i$-th column vector of the matrix $M$.

$P_2=1$ and $P_3 = 2$, and for every $1\le k\le n-2$ with $\gcd(k,n-1)=1$, $P_n$ is bounded below by $(n-1)(n-(k+1))^{k}k^{n-k}$ (indeed, setting $v$ the vector whose $k$ first coefficients are 1 and $n-k$ last coefficients are 0 and $\hat e_i$ the vector whose only 0 coefficient is at line i and whose other coefficients are 1, we have the following combination: $ (n-1) v + (n-(k+1))(\hat e_1 +\dots + \hat e_k) = k(\hat e_{k+1} +\dots + \hat e_n).) $

AlexandreV
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1 Answers1

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Let $n$ be any natural number and $M\in B_n$ be any matrix. If for some natural $i\le n$ the $i$-th column vector of the matrix $M$ is zero, then we can pick $\lambda_i=0$, so we suppose that $M$ has no zero column vectors. Since $B$ has rank $n$, then the sequence $(\lambda_1,\dots,\lambda_{n+1})$ is determined up to the factor $(-1)$. Then Lemma 2 from this answer implies that $|\lambda_i|\le n^{n/2}$ for each $i$, so $P_n\le n^{n(n+1)/2}$.

Alex Ravsky
  • 106,166