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Let $E$ be an $n$-dimensional vector space and $F_1,\ldots, F_k$ be linear subspaces of $E$ such that $$\sum_{i=1}^k \dim F_i > n(k-1)$$

Prove $\displaystyle \cap_{i=1}^k F_i \neq \{0\}$.

I find this problem quite puzzling. The condition $\sum_{i=1}^k \dim F_i > n(k-1)$ says that $\sum_{i=1}^k \dim F_i$ is close to its best upper bound so the $F_i$ have relatively big dimensions.

I haven't made any progress on this one. I feel there's some trick involved.

2 Answers 2

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Consider the obvious map

$$T \colon E \to \prod_{i = 1}^k E/F_i.$$

The rank formula gives

$$n = \dim E = \dim (\operatorname{im} T) + \dim (\ker T).$$

We have

$$\dim (\operatorname{im} T) \leqslant \sum_{i = 1}^k \dim (E/F_i) = \sum_{i = 1}^k (n - \dim F_i) = nk - \sum_{i = 1}^k \dim F_i < nk - n(k-1) = n,$$

hence $\dim (\ker T) > 0$.

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    I'm not familiar with quotient spaces. I'm going to read up on these and hope the map $T$ becomes intuitive for me. – 2017-01-11
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    Same as quotient groups, only we also have a scalar multiplication here, not only addition. As long as no topology is involved, one can also think of complementary subspaces. If $G_i$ is a complementary subspace for $F_i$, then we have an isomorphism $G_i \cong E/F_i$. (When we consider topological vector spaces, not every subspace has a topological complement, so we can't necessarily identify $E/F$ with a subspace of $E$ as a topological vector space.) – 2017-01-11
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It is probably easier to work with codimensions and not that the codimension of the intersection is at most the sum of codimensions.

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    ... because $n-\dim U\cap V\leq(n-\dim U)+(n-\dim V)$ is equivalent to $\dim(U+V)\leq n$. +1 – 2017-01-11