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!


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 $v\ne 0$.

  • $\begingroup$ So I overlooked a simple counterexample again... Thank you very much! $\endgroup$
    – Abel Stolz
    Nov 6 '12 at 14:40

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