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It is a well known fact that a symmetric bilinear form $g$ on a finite-dimensional vector space $V$ over a field $F$ of characteristic $0$ admits a orthogonal basis $\{e_i\}$, i.e. $\{e_i\}$ is a basis of $V$ and $g(e_i,e_j)=0$ for i≠j.

Does the same hold over an infinite dimensional vector space? I can't come up with a counterexample.

Here, $g$ is not necessarily non-degenerate.

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  • $\begingroup$ What is the question here, regarding the axiom of choice? $\endgroup$ – Asaf Karagila Sep 10 '14 at 21:02
  • $\begingroup$ In the theory of Hilbert spaces, an orthonormal basis of a Hilbert space is not a basis in the linear algebra sense. $\endgroup$ – Henry M- Sep 11 '14 at 12:54
  • $\begingroup$ Note: $g$ could be degenerate, that is $g(x,x)$ can be negative sometimes. Also, he says "basis", and we don't know what he means by that in the infinite-dimensional case. So, Mike is correct in some cases: in case of a positive-definite $g$, and a separable space: take a countable dense set and apply the Gram-Schmidt process to get an orthonormal set whose span is dense. $\endgroup$ – GEdgar Sep 11 '14 at 14:12
  • $\begingroup$ @GEdgar I've deleted my comment, as I hadn't read the question clearly enough (to a) notice $g$ wasn't necessarily an inner product b) notice the axiom of choice tag, making it possible that a Hilbert space might not have an orthonormal Schauder basis.) Sorry! $\endgroup$ – user98602 Sep 11 '14 at 15:00
  • $\begingroup$ @Mike: The axiom of choice tag is meant for questions which are about the axiom of choice. Not questions whose answer uses the axiom of choice. If that was the case, it would have had a whole lot more than ~650 questions. And set theory would have had way larger tags as well. $\endgroup$ – Asaf Karagila Sep 14 '14 at 3:43
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How about a non-separable inner product space, where every orthogonal set is countable (and thus not a basis for $V$).

LINK

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  • $\begingroup$ Thanks GEdar!!! can you say more about your answer, please? $\endgroup$ – Henry M- Sep 11 '14 at 13:07
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(A bit long for a comment.) From Henry's comment, I guess by "basis" he means Hamel basis.

Consider space $l_2$. Let $V$ be the subspace spanned by the canonical unit vectors $e_n$, together with one additional vector (such as $u=(1, 1/2, 1/3, 1/4, \dots)$. Then can we show that there is no orthonormal set which is a Hamel basis for it? Although $\{e_1, e_2, \dots\}$ does not span $V$, and there is no way to enlarge it to an orthonormal set, that is not the end of the story. Take the sequence in the order $u, e_1, e_2, \dots$ and apply the Gram-Schmidt process. You then get an orthonormal Hamel basis for $V$.

This should work in case of positive-definite $g$ and Hamel dimension $\aleph_0$. So there are two more interesting things to consider. (1) uncountable Hamel dimension, but countable orthogonal dimension. (2) indefinte $g$.

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    $\begingroup$ You should have used the "edit" button, not posted a new answer. $\endgroup$ – Asaf Karagila Sep 14 '14 at 3:41

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