What Are Hamel & Schauder Bases and How Do They Relate to Vector Spaces?

[Rasalhague]

I'm trying to get a handle on these concepts, which are new to me.

http://planetmath.org/encyclopedia/HamelBasis.html :

A (Hamel) basis of a vector space is a linearly independent spanning set. It can be proved that any two bases of the same vector space must have the same cardinality. This introduces the notion of dimension of a vector space, which is precisely the cardinality of the basis, and is denoted by dim(V) , where V is the vector space. The fact that every vector space has a Hamel basis is an important consequence of the axiom of choice (in fact, that proposition is equivalent to the axiom of choice.)

Kreyszig offers the following example of a Schauder basis for $l^2$: $e_i=(\delta_{ij})$ (Introductory Functional Analysis with Applications, p. 43-44). Since any vector can be written as a linear combination of vectors from this set, it spans $l^2$ http://planetmath.org/encyclopedia/SpanningSet.html . Every finite subset of $\left \{ e_i \in l^2 : e_i = (\delta_{ij}) \right \}$ is linearly independent, so, by definition (Kreyszig 2.1-6, p. 53), the set itself is linearly independent. Thus, as a linearly independent spanning set, it constitutes a Hamel basis, if I've undersood the PlanetMath definition correctly. The set is countably infinite, since its elements can be indexed by the natural numbers. Therefore $\dim(l^2) = \beth_0$.

Now, according to Kreyszig, $l^2$ is a Banach space. And it's infinite dimensional. Yet Morrison writes, "Any infinite dimensional Banach space must have an uncountable Hamel basis" (Functional Analysis: An Introduction to Banach Space Theory). If my conclusion was right, I think this would contradict the PlanetMath article, quoted above, which says that "any two (Hamel) bases of the same vector space must have the same cardinality".

So presumably my reasoning was mistaken. I suspect that, at some stage, I've assumed something that's true only of finite dimensional linear algebra without realize that it doesn't generalise to the case of infinite dimensions. Any advice welcome...

P.S. Is the essential difference between a Schauder basis (countable basis) and a Hamel basis that a Schauder basis needn't be linearly independent, and in this sense is, in PlanetMath's words "not usually a basis"? I'm not sure how this works with the uniqueness of representation of a particular vector, but the only other possibility seems to be that a Schauder basis doesn't necessarily span the space; but the definition of Schauder basis requires that it does.

 

[George Jones]

Since any vector can be written as a linear combination of vectors from this set, it spans $l^2$

What do you mean by "linear combination"? See

http://en.wikipedia.org/wiki/Schauder_basis.

 

[micromass]

The term "span" means something different in Schauder and Hamel-context.

In Hamel context, it means that any vector can be written as a FINITE linear combination.
In Schauder context, it means that any vector can be written as a linear combination that is not necessairly finite. I.e. expressions as $$v=\sum_{i=1}^\infty{v_i}$$ is allowed in Schauder context, but not in Hamel context.

The set of $$e_i=(\delta_{ij})$$ is a Schauder basis. Every vector can be written as combination of the $$e_i$$, but you'll need infinite sums for that.
And although it is linearly independent, it doesn't span the set in Hamel-context. For example, the vector (1,1,1,1,...) can not be written as a FINITE sum of the $$e_i$$.

 

[Rasalhague]

Aha, thanks, folks! That's clearer. When I read PlanetMath's "More generally, the span of a set S (not necessarily finite) of vectors is the collection of all (finite) linear combinations of elements of S.", I noted the not necessarily finite but must not have properly taken in the later parenthetical finite. D'oh. I know Kreyszig has a similar parenthetical "(finitely many!)", but from his comment, "Hence if B is a [Hamel] basis for X, then every nonzero x in X has a unique representation as a linear combination of (finitely many!) B with nonzero coefficients.", I wasn't sure whether finite was part of the definition of span, or of linear combination, or of Hamel basis, or just a happy corollery.

The PlanetMath article http://planetmath.org/encyclopedia/LinearCombination.html suggests that linear combination can mean

$$\sum_{i=1}^n a_i \textbf{v}_i \enspace\enspace \text{or} \enspace\enspace \sum_{i=1}^\infty a_i \textbf{v}_i,$$

as in your post, micromass, so that the expression "linear combination" itself doesn't specify whether there's a finite or infinite quantity of terms, but span might, unless span is taken to have two senses, depending on context, in which case finiteness or otherwise would be included in the definitions of the two kinds of basis.

If I've got this right now, we can say that each nonzero vector in any vector space has a unique representation as a finite linear combination of vectors from a Hamel basis (this basis being possibly infinite); and when a set of vectors is said to span the whole space, this means that any nonzero vector can be so represented, making the set a Hamel basis, providing it's linearly independent. Probably PlanetMath take this kind of basis as the only kind meriting the unqualified name "basis" (hence "a countable [i.e. Schauder] basis is not usually a basis"), and take this sense of span as their only sense of span.

Meanwhile, in certain kinds of vector space, a Schauder basis can be constructed, in which every nonzero vector can be expressed as unique (possibly infinite) linear combination of countably many Schauder basis vectors--always linearly independent?--although they don't span the space in the usual, Hamel sense.

I wonder what would be an example of a Hamel basis for $l^2$.

 

[micromass]

Well, if you ever meet the terms "basis" and "span" then they always mean the Hamel-definitions of it. If they mean it in Schauder-sense, then it is always mentioned.

Also note that a Schauder basis is always countable, I forgot to mention this, but it is quite important.
There are other kinds of bases, e.g. orthonormal basis,... All these beses coincide in the finite-dimensional case. But infinitely dimensional all these notions diverge and they are all quite useful...


You ask a nice question there, find a basis for l². I can't give you an explicit examples, but these examples exists. If your interested then I would suggest reading about the Haar-wavelet, this provides a basis for L² (and hence also for l²). Also note that these basis are necessairily uncountable...

So one of the advantages of Hamel bases is that they always exist. Indeed, for every vector space we can find a Hamel base. But their construction is always somewhat messy.
Schauder bases don't always exist. They only exist in separable Banachspaces (and not even in all of them). Luckily, Schauder bases are always easy constructable, so that's a nice advantage...

 

[Hurkyl]

There are other kinds of bases, e.g. orthonormal basis,...

Most bases are not orthonormal -- even for those vector spaces for which the word "orthonormal" makes sense! (i.e. inner product spaces)

 

[g_edgar]

It is best to use the term "linear combination" only for the case where finitely many coefficients are nonzero.

 

[Alesak]

If anyone would be so kind and told me if I got it correctly: Hamel basis of vector space V is a set of linearly independent vectors, such that every vector of V can be written as finite linear combination. On wikipedia they say that every Banach space that doesn't have finite basis have uncountable Hamel basis. It means that in l^p space, the usual e_i=δ_ij base won't be Hamel base, but it's Hamel base will be some uncountable monstrous construct. Is that right?


Now, the Schauder basis is not defined for general vector space but only for at least complete normed vector space(=Banach space). We say such space has Schauder basis when it spans(with countable linear combinations) the whole set. I originaly wanted to say "when it spans some dense subset" based on situation in L^p, but then I've read that L^p has as vectors equivalence classes.

And finaly, in Hilbert space a basis is usualy meant Schauder basis that is orthonormal.

 

[Bacle2]

The term "span" means something different in Schauder ........ For example, the vector (1,1,1,1,...) can not be written as a FINITE sum of the $$e_i$$.

I don't follow you, (1,1,1,...) is not in l².

An important difference is that in a Schauder basis you use convergence, while in a HAmel basis you have equality in an actual sum. When you have an infinite sum you need to bring up the topology of the space.

 

[micromass]

I don't follow you, (1,1,1,...) is not in l².

Yes, of course not. I'm not sure what I was thinking...

 

[Alesak]

An important difference is that in a Schauder basis you use convergence, while in a HAmel basis you have equality in an actual sum. When you have an infinite sum you need to bring up the topology of the space.

Yes, it makes sense. How else could one value infinite sense?

Yes, of course not. I'm not sure what I was thinking...

Maybe you meant l^∞?

 

[micromass]

Maybe you meant l^∞?

Probably not, since $\ell^\infty$ does not have a Schauder basis.

 

[micromass]

And finaly, in Hilbert space a basis is usualy meant Schauder basis that is orthonormal.

Not exactly, a Schauder basis is supposed to be a sequence $(e_n)_n$. So it is implied that a Schauder basis is countable. An orthonormal basis does not need to be countable.

 

[Alesak]

Probably not, since $\ell^\infty$ does not have a Schauder basis.

I suppose that is because even if we represent some vector in $\ell^\infty$ with usual basis it doesn't converge to it in $\ell^\infty$ norm.

Not exactly, a Schauder basis is supposed to be a sequence $(e_n)_n$. So it is implied that a Schauder basis is countable. An orthonormal basis does not need to be countable.

I'll keep that in mind, thanks.