Let $\{v\_{1,j},\ldots, v\_{n,j}\}$ be a basis of the $n$-dimensional vector space $V$ for $j=1\ldots k$ (and assume $2k\leq n$). Let $V\_i$ be the subspace spanned by $v\_{i,1},\ldots, v\_{i,k}$ for $i=1\ldots n$. (So the dimension of each $V\_i$ can be anything between $1$ and $k$.) Is it true that $V\_1\cap V\_2\cap \ldots \cap V\_{2k}=0$ (or likewise any intersection of $2k$ of the $V\_i$ is trivial)?


(The answer to the question is needed to proof a quite technical lemma, the details of which I prefer not to spread in full length. In any case I want to do something like span "as much of $V$ as possible, using as few $V\_i$ as possible".)


I seem not be able to construct a counterexample, neither to do a proof by induction, though I suspect the latter should be possible. Any pointer to a reference or hint to a simple argument is appreciated!



Intersection of vector spaces
I believe the answer is no if $n\ge4$. Here is a counterexample for $n=4$, $k=2$:
Let $v$ be the sum $v\_{1,1}+v\_{2,1}+v\_{3,1}+v\_{4,1}$ of the vectors of the first basis.
Define the second basis by $v\_{i,2}=v-v\_{i,1}$. Then $v\in V\_i$ for $i=1,2,3,4$, a contradiction as $v\ne 0$.


