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It is quite simple to show that all finite-dimensional vector spaces with inner product have an orthogonal basis, with the standard definition of a basis from linear algebra. However, I am in trouble to find any reference about infinite-dimensional case.

Do all infinite-dimensional vector space with inner product have an orthogonal basis (again with the standard definition of a basis from linear algebra)?

Ketty
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  • section 6.3 in this pdf. I hope it is what you are looking for. – newbie Jan 09 '14 at 15:58
  • @ketty That takes care of positive definite complex Hermitian inner products, anyhow. You'll have to be more specific if you're looking for more general answers. – rschwieb Jan 09 '14 at 16:01
  • I already have seen this file. It is not talking about general vector spaces, but Hilbert spaces. – Ketty Jan 09 '14 at 16:02
  • @rschwieb: I would like to know in general. – Ketty Jan 09 '14 at 16:02
  • The standard definition of basis from linear algebra isn't very useful in infinite-dimensional vector spaces. There is a notion of a Hamel basis but such a basis must be necessarily uncountable. It is much better to weaken the notion of linear independence by using infinite summation, which leads to the Schauder basis. But to be able to do summation you need to have norm on your vector space and preferrably a complete one, so this leads directly to Banach spaces. If you are interested in orthogonality as well then you need inner product and such a space is actually a Hilbert space. – Marek Jan 09 '14 at 16:07
  • TL;DR: if you want an inner product space that is useful, then we're talking about Hilbert spaces. If you explicitly want to learn about incomplete inner product spaces, please state that in your question. – Marek Jan 09 '14 at 16:10
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    @Marek I understand all you are saying. But the question is still fair in the context of the standard definition of basis from linear algebra (e.e only finite sums). We are not talking about which basis is better! – Ketty Jan 09 '14 at 16:10
  • @Ketty: we are talking about which basis is better since the answer depends on which kind of basis (Hamel, Schauder or perhaps something else) you want. Also, you want to talk about the inner product. Every vector space with compatible inner product is a pre-Hilbert space and its completion is a Hilbert space. So if you know the answer for Hilbert spaces, that's all there is to know. – Marek Jan 09 '14 at 16:18
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    @Marek: You are talking about which basis is better because you want. Moreover, you are also judging if the vector space is useful or not because you want. It is not necessary! Whenever we talk about a vector space, with or without an inner product, we are assuming the word BASIS in the standard way of linear algebra. I totally disagree that I have to state that I am interested on incomplete vector spaces! – Ketty Jan 09 '14 at 16:20
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    @Marek I do not want make a discussion here. Now you probably understand the question. That's all! – Ketty Jan 09 '14 at 16:30
  • @Ketty: the standard definition of basis for inner product spaces is not the same as the one for linear algebra, (see e.g. here http://en.wikipedia.org/wiki/Inner_product_space#Orthonormal_sequences). I'm voting to close because of your uncooperativeness and refusal to make yourself and your question clear even after much discussion with your peers. – Marek Jan 09 '14 at 16:47
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    @Marek Well. You are understanding what you want. I totally disagree in a Math site that one person can say "the only usefull spaces are Hilbert". Moreover, wikipedia reference for "inner product space and orthonormal sequence" is not a Linear algebra reference. Moreover, you are quite quick in judging of people and question. – Ketty Jan 09 '14 at 17:08
  • @Ketty I would like to know in general General what? Inner products that aren't positive definite? Over general fields? botH? – rschwieb Jan 09 '14 at 17:10
  • @rschwieb Once said "inner product", it is an inner product and not anything else. There is a very good answer below! – Ketty Jan 10 '14 at 11:30

1 Answers1

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From your comments (correct me if I am wrong), it appears that you are using "basis" in the following sense:

a linearly independent subset $B$ of a vector space $V$ such that every vector in $V$ may be written as a finite linear combination of vectors from $B$.

In the context of infinite dimensional vector spaces, people usually call such a set a Hamel basis.

In that case, the answer to your question is:

Some infinite-dimensional inner product spaces admit an orthogonal Hamel basis; others do not.

For one that does, consider the vector space of polynomials, equipped with the inner product $\langle p,q\rangle = \int_{-1}^1 p(x) q(x)\,dx$ (i.e. the $L^2([-1,1])$ inner product. Then the Legendre polynomials form an orthogonal Hamel basis.

For one that does not, consider any infinite-dimensional separable Hilbert space $H$ (such as $\ell^2$). It's a consequence of the Baire category theorem that any Hamel basis for $H$ is necessarily uncountable. On the other hand, because of separability, any orthogonal set is necessarily at most countable. (By rescaling we can assume all vectors in the set are unit vectors; then they are all at distance $1/\sqrt{2}$ from one another. If we put a ball of radius $1/2\sqrt{2}$ around each one, those balls are disjoint. But being separable, our space has a countable dense subset $E$, and each of these balls must contain an element of $E$.)

Nate Eldredge
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