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My question:

why don't we get a Hamel basis (a maximal linearly independent set) instead of a maximal orthonormal set for a Hilbert space. In what dimension can we use a Hamel basis and in which we can't?

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Hamel bases exist in any dimensions, assuming the Axiom of Choice. The problem is that Hamel need not be orthonormal in the infinite dimensional case, which is why we rather work with Hilbert bases, which always are. The tradeoff is that we must work with approximations instead of actual equalities, but the orthonormality of the Hilbert bases makes the tradeoff worthwhile most of the time. – Arturo Magidin Sep 20 '11 at 16:10
In fact, Hamel bases can't be orthonormal in the infinite-dimensional case. – Robert Israel Sep 20 '11 at 16:30
@Robert: I believe you...but maybe the proof of your above comment would make for a nice answer? – Pete L. Clark Sep 20 '11 at 16:32
A Hamel basis of an infinite-dimensional Hilbert space cannot be orthonormal, but some Hamel bases of some infinite-dimensional inner product spaces (that are not complete) are orthonormal. The space of trigonometric polynomials is an example. – Michael Hardy Sep 20 '11 at 16:54
@PeteL.Clark: If a Hilbert space $\cal H$ has an orthonormal Hamel basis $B$, then any member of $\cal H$, being a linear combination of finitely many members of $B$, is orthogonal to all the other members of $B$. So take the sum of a convergent series involving a countable subset of $B$, say $\sum_{j=1}^\infty \frac{b_j}{j^2}$, and you have a contradiction. – Robert Israel Sep 20 '11 at 19:52

We know the existence of a dense countable orthonormal set i.e. $\fbox{1}\langle x,e_i\rangle \,$ for $\forall i \in \mathbb{N}$ $\, \Rightarrow x=0.\,$
Since a Hamel basis on a Hilbert space is uncountable we can extend the linearly independent set $ \{ e_i\ | \forall i \in \mathbb{N} \}\,$ to a Hamel basis in a non trivial way. So $\fbox{1} \,$ yields a contradiction.

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