This post is copied from math.SE in the following link:


I have posted the question two days ago, but receive no answer yet. Therefore I think maybe I can post it on MO and invite more attention. If this violates some policy, let me know and I'll delete the post.

I am a college lecturer in Macau, and this conjecture is based on some discussions with my colleagues. We think this problem can be set for some math competitions for college students, however we have not reached conclusion regarding this conjecture. Therefore I post it out and invite your attention. My description of the conjecture using English may not look professional, and I welcome your editing to make it more sound. Thank you very much.

Let $V$ be a $(r+1)$-dimensional vector space, and $p$ be a positive integer and $1\leq p\leq r-1$. Let $$X=\{v_1,\cdots,v_{2r+1-p}\}\subseteq V$$ be a finite set containing $(2r+1-p)$ different vectors, and $\{u,v\}$ is a linearly-independent set for any $u,v\in X$ such that $u\neq v$. Moreover, for any set of $(2s+2−p)$ vectors in $X$ there exists no $(s+1)$-dimensional subspace that contains said set, where $s=p,p+1,\cdots,r-1$.

Prove or disprove the following conjecture: $X$ can be divided into two non-intersecting non-empty subsets $$X=X_1\cup X_2$$ such that $X_1$ consists of $(r+1)$ linearly-independent vectors and $X_2$ consists of $(r-p)$ linearly-independent vectors.

  • $\begingroup$ Hi Thomas: I think this is a good question. Math.SE is a better site for it, because the question is not, strictly speaking, about research mathematics. But it's not far off. To my non-expert eye, this looks like a problem in matroids, not vector spaces. $\endgroup$ – Theo Johnson-Freyd Aug 3 '13 at 6:10
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    $\begingroup$ Is this problem actually under consideration for a contest? (I'm not sure whether "We think this problem can be set for some math competitions for college students" was hypothetical.) If so, posting it on the internet seems risky. The students may not find it in time to have any impact on the contest, but I bet they'll find it by web searches afterwards, in which case they will wonder whether anyone got an unfair advantage. $\endgroup$ – Henry Cohn Aug 3 '13 at 14:34
  • $\begingroup$ @HenryCohn: Thank you for your concern. We were indeed considering such possibilities, but now the problem seems quite difficult so we have already changed our minds. Even we do not have the answer so far! Therefore it is now purely out of curiosity. $\endgroup$ – Thomas Tam Aug 3 '13 at 19:33

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