Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $V$ is a vector space of dimension $n$ and $S$ is a subset of $V$ such that $\operatorname{span}(S)=V$. Prove there exists a basis for $V$ in $S$ without assuming $S$ is a finite set.

I'm not sure what direction to take when $S$ is infinite. I know a bunch of facts. I ultimately want to show that it's possible to pick a finite subset of $S$ that generates $V$ and is linearly independent.

share|improve this question
2  
The key property to use is that only finitely many vectors are involved in a given linear combination. –  Qiaochu Yuan Sep 7 '12 at 3:46

2 Answers 2

Inductively: choose $\,0\neq s\in S\,$ (why is there such a non-zero element?), and use the following (after, I presume, you properly prove it):

Theorem: Let $\,W\,$ be any vector space, $\,A\subset W\,$ a linearly independent set. Then, for any

$$w\in W\,\,\,,\,\,w\in Span A\Longleftrightarrow A\cup \{w\}\,\,\text{ is linearly *dependent*}$$

So, if there's some $\,s'\in S\,\,s.t.\,\,s'\notin Span\{s\}\,$ , then by the above $\,\{s,s'\}\,$ is linearly independent.

Continue in this fashion and, after at most $\,n\,$ steps, you'll get a basis (why?)

share|improve this answer

Let $B$ be a basis of $V$. Each element of $B$ is generated by $S$. That means there exists finite subset $T$ of $S$ such that $B \subseteq span(T)$. Make $T$ linearly independent by removing some elements. Then the final $T$ is your answer.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.