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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)?

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  • $\begingroup$ section 6.3 in this pdf. I hope it is what you are looking for. $\endgroup$
    – newbie
    Commented Jan 9, 2014 at 15:58
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    $\begingroup$ @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! $\endgroup$
    – Ketty
    Commented Jan 9, 2014 at 16:10
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    $\begingroup$ @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! $\endgroup$
    – Ketty
    Commented Jan 9, 2014 at 16:20
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    $\begingroup$ @Marek I do not want make a discussion here. Now you probably understand the question. That's all! $\endgroup$
    – Ketty
    Commented Jan 9, 2014 at 16:30
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    $\begingroup$ @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. $\endgroup$
    – Ketty
    Commented Jan 9, 2014 at 17:08

1 Answer 1

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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 $E$.)

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    $\begingroup$ Perfect. That is exaclty what I want! $\endgroup$
    – Ketty
    Commented Jan 9, 2014 at 17:11

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