A conjecture about vector space 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 all these vectors are linearly-independent to each other $\{u,v\}$ is a linearly-independent set for any $u,v\in X$ such that $u\neq v$. Moreover, any (2s+2−p) vectors in X are not in some (s+1)-dimensional subspaces 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.
P.S. 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.  
 A: First, I'll point out a simpler way to pose the conjecture, for me at least. By "rank" of a set of vectors, I mean the rank of the matrix composed of the vectors, or equivalently the dimension of the linear span of the vectors.

Let $n \geq 3$ and $0 \leq k \leq n - 3$. Suppose $X$ is a set of $n + k + 1$ vectors in an $n$-dimensional vector space such that, for each
  $s = 0, \dotsc, k$, any $n - k + 2s$ of the vectors have rank at least
  $n - k + s$. Then (we conjecture) $X$ can be partitioned into two
  disjoint linearly independent subsets, of size $n$ and $k + 1$.

In particular, the condition says that any $n - k$ of the vectors are linearly independent, and any $n + k$ vectors span the space. (In terms of the original formulation, $n = r + 1$ and $k = r - p - 1$.)
As a comment over at MathOverflow noted, this is really a problem about matroids. Matroids are an abstraction of vector configurations, like $X$, that forget the coordinates and only pay attention to which subsets are linearly independent. In this language, the conjecture is:

If $X$ is a rank-$n$ matroid with $n + k + 1$ elements such that any
  subset of $n - k + 2s$ elements has rank at least $n - k + s$, then
  $X$ can be partitioned into two independent subsets, of size $n$ and
  $k + 1$.

This leads us to the article Minimum Partition of a Matroid into Independent Subsets by Jack Edmonds (1964). The principal result is that a matroid can be partitioned into $k$ independent sets if and only if every subset $A$ has cardinality at most $k \operatorname{rank}(A)$.
In our case, for any subset $A$, let $s$ be the maximum such that $n - k + 2s \leq |A|$;
then $\operatorname{rank}(A) \geq n - k + s$,
and $2\operatorname{rank}(A) \geq 2(n-k) + 2s \geq n - k + 2s + 1 \geq |A|$.
Therefore, by Edmonds' theorem, $X$ can be partitioned into two independent subsets, say $X_1$ and $X_2$.
However, these subsets need not be of size $n$ and size $k + 1$; of course, $n$ is the maximum possible size, but they could be of size $n - 2$ and $k + 3$, for example.
But we know that $X$ contains a basis, and so either set (say $X_1$) can be extended to a basis by adding vectors from $X$. Then we've extended $X_1$ to a basis of $n$ elements,
and the remaining elements of $X_2$, being a subset of an independent set, are still independent.
So the conjecture is true. In fact, we see that it remains true even if we allow $k$ to range up to $n - 1$, or in the original form, if we allow $-1 \leq p \leq r - 1$.
