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Un-spannable vector space?

by alemsalem
Tags: space, unspannable, vector
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alemsalem
#1
Feb17-11, 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!!
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Fredrik
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Feb17-11, 05:56 AM
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Such spaces are said to be infinite-dimensional. 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 R|n\in\mathbb Z^+\}[/itex], or be less formal and shorten it to [itex]\{\sin nx|n\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_n|n\in\mathbb Z^+\}[/itex].
dexdt
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Feb25-11, 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

Fredrik
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Feb25-11, 05:44 AM
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Un-spannable vector space?

In the finite-dimensional 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.
micromass
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Feb25-11, 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 Schauder-basis is a set of elements such that every element can be written as an infinite linear combination of basis elements. In infinite-dimensional (separable) spaces, the concept of Schauder basis is a good replacement for the concept of basis...
lavinia
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Feb25-11, 07:38 AM
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Quote Quote by alemsalem View Post
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!!
There is no general algorithm for finding the basis of an infinite dimensional vector space. In this case a basis means a set of independent vectors that span the entire space by finite linear combination. That means that an arbitrary vector in the space is a linear combination of finitely many basis vectors. Since there is no algorithm one must make a postulate that allows one to conclude that there is a basis. A typical such postulate is the Axiom of Choice.

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.
Hurkyl
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Feb25-11, 07:46 AM
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Quote Quote by micromass View Post
But this uses the axiom of choice and is highly unconstructive.
It is an eye of the beholder thing. e.g. with the well-ordering theorem* in hand, the algorithm quoted in the opening post can be continued transfinitely to produce a basis. I wouldn't call it highly unconstructive, but I know opinion differs on such points.


*: For the OP, the well-ordering theorem is equivalent to the axiom of choice and to Zorn's lemma
HallsofIvy
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Feb25-11, 09:20 AM
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Quote Quote by dexdt View Post
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
More correctly, Zorn's Lemma shows that every vector space has a basis but isn't really needed for finite dimensional vector spaces. Did you mean to say "infinite dimensional"?

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!
mathwonk
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Feb25-11, 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.
HallsofIvy
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Feb26-11, 07:00 AM
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Quote Quote by mathwonk View Post
every vector space spans itself, so strictly speaking "unspannable spaces" (with no mention of independence) do not exist.
Absolutely true. But since alemsalem said "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." I assumed he was referring to spanning by a finite set.
mathwonk
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Feb27-11, 05:45 PM
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yes indeed halls, your answer was more useful whereas mine was just picky. but thats my strong suit!


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