Intersection of vector subspaces

Posted: February 16, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: ,

Problem. Let V be a vector space over some field F and \dim_F V=n < \infty. Let \mathcal{A} be a finite collection of the subspaces of V and suppose that for any \emptyset \neq \mathcal{B} \subseteq \mathcal{A} with |\mathcal{B}| \leq n we have \bigcap_{W \in \mathcal{B}} W \neq \{0 \}. Prove that \bigcap_{W \in \mathcal{A}} W \neq \{0 \}.

Solution. The proof is by induction over |\mathcal{A}|. There is nothing to prove if |\mathcal{A}| \leq n. Suppose \mathcal{A}=\{W_1, \cdots , W_m \}, \ m > n, and put W'_j=\bigcap_{i \neq j} W_i. By the induction hypothesis W'_j \neq \{0\}, for all 1 \leq j \leq m. So for any 1 \leq j \leq m, there exists some 0 \neq w_j \in W'_j. Since m > n, the set \{w_1, \cdots , w_m \} is linearly dependent. Thus there exists 1 \leq k \leq m such that w_k=\sum_{i \neq k} c_iw_i, for some c_i \in F. But if i \neq k, then w_i \in W_k and so w_k \in W_k. We also have that w_k \in W'_k \subseteq W_i, for all i \neq k. Therefore 0 \neq w_k \in \bigcap_{i=1}^m W_i. \ \Box

  1. Abdulrasool Azizi says:

    It seems that this problem is also true for every finitely generated torsion-free module over an integral domain. We should replace “dimension” with “rank” of the module, and use localization.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s