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For a finite-dimensional inner product space $V$, since it has a finite basis, then we can do the Gram-Schimidt process to produce an orthonormal basis. However, the Gram-Schmidt process does not work with infinitely many vectors. So it is natural to ask, does every infinite-dimensional inner product space have an orthonormal basis? If the answer is yes, how to prove it?

PS: For "basis", I mean the Hamel basis.

Eric
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  • Do you mean a basis in the sense of a Hamel basis, or just one that spans a dense subspace? – Asaf Karagila Sep 12 '17 at 15:10
  • @AsafKaragila Well, I haven't learned Hamel basis or dense subspace. I originally refers to a inner-product space $V$ over a field $F$ in linear algebra course. So does it depends on other chacterization that I should specify more information on that space? – Eric Sep 12 '17 at 15:13
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    @Eric in infinite-dimensional settings, there are two common notions of a basis. In a sense, the question amounts to whether you count infinite sums among the "linear combinations" of your basis elements. – Ben Grossmann Sep 12 '17 at 15:14
  • There are more ways than one to think about a basis for a topological vector space (e.g., an inner product space). It just happens that for finite dimensions these notions coincide. – Asaf Karagila Sep 12 '17 at 15:15
  • Hmm.. I see. I haven't learned these course. In many linear algebra books, they defined a basis for a vector space(including infinite dimension case) to be a set that spans $V$ and linearly independent right? If I adope this definition of basis, is it enough? (or still ambiguous..? lol) – Eric Sep 12 '17 at 15:18
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    @Eric the question is what does "spans $V$" mean to you? Do we include "infinite linear combinations" or not? – Ben Grossmann Sep 12 '17 at 15:19
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    Not. I use the definition that the span of a set $S$ means the set of all linear combination of any finitely vectors in $S$. So there're no truly infinite linear combinations. – Eric Sep 12 '17 at 15:21
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    @Eric In that case, you're asking about a Hamel basis (as opposed to a "Schauder basis"). – Ben Grossmann Sep 12 '17 at 15:23
  • @Omnomnomnom Thanks for pointing out that, I edited! – Eric Sep 12 '17 at 15:26
  • The answer is yes as Chessantor's answer shows. If you want the space to be complete (Hilbert space), then no. – Jonas Meyer Sep 12 '17 at 15:30
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    You need to reword your question. Right now it is confusing. Does the "any" in your question mean "one" or "all/every"? The title seems to suggest the former, but the question body seems to suggest the latter. As Chessantor's answer and Omnomnomnom's comment beneath it show, users have already been interpreting your question differently. – user1551 Sep 12 '17 at 17:06
  • @user1551 I edited. I mean "every". Sorry about that, I just found that this is a grammar mistake... (my native language is not English.) – Eric Sep 12 '17 at 17:08

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Edit: this answer was given before an edit to the question and does not validly answer the new question.

You can define such a space from the ground up: Let $V$ be the vector space of finite (i.e. eventually always zero) linear combinations of a countable set $\{e_i | i \in \mathbb{N}\}$. Then define the inner product as $\langle \sum x_i e_i , \sum y_i e_i \rangle = \sum x_i \bar{y_i}$, where this sum always exists because eventually both the $x_i$ and the $y_i$ become $0$.

In this vector space, the basis vectors $e_i$ are orthonormal.

Chessanator
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