## Intersection of vector subspaces

Posted: February 16, 2010 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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 $0 \neq w_k \in \bigcap_{i=1}^m W_i. \ \Box$