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Unspannable vector space? 
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#1
Feb1711, 03:54 AM

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If you have a vector space you can find a set of elements and consider their span, and then look for elements that cannot be spanned by them and so add them to the set, if you can't add anymore then you have a basis.
My question is what happens if this process continues forever, do you automatically call it infinite dimensional or is there such a thing as an unspannable space. Also what happens if there is a set but it cannot be labeled nicely such as {sin(nx)}.. Thanks!! 


#2
Feb1711, 05:56 AM

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Such spaces are said to be infinitedimensional. Every vector space has a basis, so there's no need for terms like unspannable. You can use a notation like [itex]\{\mathbb R\ni x\mapsto \sin nx\in\mathbb Rn\in\mathbb Z^+\}[/itex], or be less formal and shorten it to [itex]\{\sin nxn\in\mathbb Z^+\}[/itex]. You can also e.g. define, for each n=1,2,..., [itex]u_n:\mathbb R\rightarrow\mathbb R[/itex], by [itex]u_n(x)=\sin nx[/itex] for all [itex]x\in \mathbb R[/itex], and write the set as [itex]\{u_nn\in\mathbb Z^+\}[/itex].



#3
Feb2511, 01:12 AM

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there's a theorem out there that says every finite dimensional vector space must have a basis ( and it uses Zorns lemma ) , so that should clear things up for oyu



#4
Feb2511, 05:44 AM

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Unspannable vector space?
In the finitedimensional case, the existence of a basis follows immediately from the definition of "basis" and "finite dimensional". So you only need Zorn when the vector space is infinite dimensional.



#5
Feb2511, 07:30 AM

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Like Frederik has already said: using Zorn's lemma, we can show that every possible vector space has a basis. But this uses the axiom of choice and is highly unconstructive. If we do not assume the axiom of choice, then there may be some vector spaces without a basis (for example: consider [tex]\mathbb{R}[/tex] as [tex]\mathbb{Q}[/tex]vector space).
So while we can show (using choice) that every vector space has a basis, there is something undesirable about this. Namely the fact that we can never write down the basis in any way. But the entire point of having a basis is so that we can use it to know more about the vector space. Thus having a basis of an infinite dimensional vector space seems to be a little useless. That's why some people proposed things which weren't a basis, but which did have some desirable properties. For example, a Schauderbasis is a set of elements such that every element can be written as an infinite linear combination of basis elements. In infinitedimensional (separable) spaces, the concept of Schauder basis is a good replacement for the concept of basis... 


#6
Feb2511, 07:38 AM

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If there is a metric then one can talk about infinite linear combinations of basis vectors that converge to an arbitrary vector in the space. The classic examples are L^2 normed function spaces. This is a different idea of basis. 


#7
Feb2511, 07:46 AM

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*: For the OP, the wellordering theorem is equivalent to the axiom of choice and to Zorn's lemma 


#8
Feb2511, 09:20 AM

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alemsalem, the definition of "finite dimensional" is that the space can be spanned by some finite set of vectors. So any "unspannable space" would be infinite dimensional. Yes, there do exist spaces with uncountable bases. "What happens"? You avoid them like the plague! 


#9
Feb2511, 09:25 AM

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every vector space spans itself, so strictly speaking "unspannable spaces" (with no mention of independence) do not exist.



#10
Feb2611, 07:00 AM

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