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(I'm assuming full axiom of choice in this post)

Theorem: Let $V$ be a vector space and $\beta$ be a basis for $V$ and $S$ be a linearly independent subset of $V$. Then, there exists $H\subset \beta$ such that $H\cup S$ is linearly independent.

Is there a direct way to prove that "If a vector space has bases $\beta_1$&$\beta_2$, then $|\beta_1|=|\beta_2|$" as a corollary of the given theorem above?

I'm asking this because if $V$ is finite-dimensional, then this corollary can be directly proved from the "replacement theorem"(Dimension theorem for finite-dimensional vector space).

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1 Answer 1

1) if $\,S\,$ is a maximal lin. ind. set (i.e., a basis) then $\,H=\emptyset\,$ makes the trick (in fact, it makes it in any case, so I'd assume we don't want trivial cases), otherwise

2) It can't be that $\,S\cup\{b\}\,$ is lin. dependent for all $\,b\in\beta\,$ , since we know that $\,S\cup\{b\}\,$ lin. dependent iff $\,b\in\operatorname{Span}{S}\,$ , but then we'd get that $\,\beta\subset\operatorname{Span }S\,$ and thus $\,S\,$ would be a generating set and thus maximal lin. ind.-- against the assumption--, so there must be some element $\,b\in\beta\;\;s.t.\;\;S\cup\{b\}\,$ lin. ind.

And yes: from this we get that any two basis of a vector space have the same cardinality, otherwise we'd get, using the above argument, a contradiction to the fact (either theorem or equivalent definition) that a basis is a maximal linear independent set, which in turn is equivalent to "a basis is a minimal generating set...)

Added on request: WLOG, assume $\,w_1\ge w_2\,$ and infinite cardinalities (as finite ones is rapidly dealed with, generally in basic linear algebra courses), and look at the basis $\,\beta_1\,$ and at the lin. ind. set $\,\beta_2\,$ .

According to the proven lemma, there exists $\,H\subset \beta_1\,\,\,s.t.\,\,\,\beta_2\cup H\,$ is lin. ind. Since $\,\beta_2\,$ is a maximal lin. ind. set this implies $\,H=\emptyset\,$ , but by the theorem mentioned in (2) in my answer (after "since we know") , this means

$$b_1\in\operatorname{Span}{\beta_2}\,\,,\,\forall\,b_1\in\beta_1\Longrightarrow\operatorname{Span}{\beta_1}\subset\operatorname{Span}{\beta_2}$$

and we're done since $\,w_i=\left|\operatorname{Span}{\beta_i}\right|\,$ for infinite cardinalities , and thus $\,w_1\le w_2\,$

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I don’t think that Katlus is worried about proving the theorem at the top of the question; the question is whether that result can be used to give a direct proof that any two bases of $V$ have the same cardinality. Your third paragraph addresses the real question, but I think that you need to add a lot of detail to make it convincing. How exactly are you going to use the quoted theorem to show, for instance, that a vector space $V$ over $\Bbb R$ can’t have disjoint bases $\beta_1$ and $\beta_2$ of cardinalities $\omega_1$ and $\omega_2$, respectively? –  Brian M. Scott Feb 20 '13 at 8:02
    
I'm adding something to my answer which, I believe, could have been done by the OP. –  DonAntonio Feb 20 '13 at 14:20
    
Don't you mean $\beta \subset \operatorname{span}S$ rather than $\beta \subset S$? –  dfeuer Jan 5 at 6:44
    
Yup, thanks @dfeuer –  DonAntonio Jan 5 at 15:35

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