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It is proved that every inner product space has a basis $W$, but I am not sure if every inner product space has an orthogonal basis? It is known that every inner space has a maximal orthogonal set $S$, so if I want to prove the conclusion, we need to show that if $w\in W$ is linearly independent to $S$, then we can construct a vector that is orthogonal to $\mbox{span}\{S\}$, which leads to contradiction, but it seems needs the condition that the space is complete.

If not, can anyone give me a counterexample?

89085731
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1 Answers1

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Note: Edits made November 2016 are correcting a wrong statement pointed out by Pp.. (then overlooked) and more recently by Nate Eldredge.

In the finite dimensional case, yes.

In the infinite dimensional case, a maximal orthogonal set is not a basis in the usual sense. You can only be sure that [added: in the complete case] the linear span of a maximal orthogonal set is dense. [Added: In the incomplete case this need not be true. See comments of Pp. from Feb 2015 and Nate Eldredge from Oct 2016.]

For example, in the inner product space $\ell^2$ of square summable sequences, the set of vectors with a $1$ in one component and $0$ elsewhere is a maximal orthogonal set, but it is not a basis, because for example, $(1,1/2,1/3,1/4,\ldots)$ is not in its span. However, in a Hilbert space like $\ell^2$, every vector has a unique series expansion in terms of a maximal othogonal set.

Here is a Wikipedia reference: http://en.wikipedia.org/wiki/Hilbert_space#Orthonormal_bases

A maximal set of pairwise orthogonal vectors with unit norm in a Hilbert space is called an orthonormal basis, even though it is not a linear basis in the infinite dimensional case, because of these useful series representations. Linear bases for infinite dimensional inner product spaces are seldom useful.

Jonas Meyer
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  • I can't see why it is not in the span? – 89085731 Jul 23 '14 at 03:19
  • @89085731: Spans allow finite sums, not infinite sums. Infinite sums are limits of things in the span. – Jonas Meyer Jul 23 '14 at 03:20
  • But I think it is a basis – 89085731 Jul 23 '14 at 03:22
  • It is not a basis, because it does not span. Every element of $\ell^2$ with infinitely many nonzero components is not in the span, because it is not a finite linear combination of the vectors in the orthogonal set. This requirement of finiteness follows from the definition of basis for a vector space. (In a general vector space there is no notion of limit.) Note that $\ell^2$ has uncountable linear dimension, so no countable set could be a basis. For example, the set of vectors ${(1,t,t^2,\ldots):-1<t<1}$ is linearly independent. – Jonas Meyer Jul 23 '14 at 03:24
  • @89085731: An orthonormal basis for a Hilbert space is an example of a Schauder basis. Hamel basis means the same thing as basis for a vector space, and "Hamel basis" or "linear basis" or "vector space basis" are phrases used to emphasize that it is basis in the usual vector space sense, not using any metric approximation stuff. In an infinite dimensional Hilbert space, an orthonormal basis is never a Hamel basis. (This comment was posted to answer a comment that was deleted while I was writing. I'm leaving it at least for a little while in case it is of use.) – Jonas Meyer Jul 23 '14 at 03:31
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    This answer is incomplete. – Pp.. Feb 07 '15 at 01:24
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    Indeed, the claim at the end of the first paragraph is false: a maximal orthogonal set in an inner product space is not necessarily dense. See also http://math.stackexchange.com/a/201149/822. (It's true in a Hilbert space, of course.) – Nate Eldredge Oct 22 '16 at 20:26
  • Thanks @NateEldredge and Pp... – Jonas Meyer Nov 03 '16 at 19:50
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    Although this answer states that you can't always construct an orthogonal basis from an orthogonal set of vectors, I would like to further ask whether every inner product space, not necessarily complete, has an orthogonal basis. – James Well Dec 20 '17 at 18:11