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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.2011-09-20
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    In fact, Hamel bases _can't_ be orthonormal in the infinite-dimensional case.2011-09-20
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    @Robert: I believe you...but maybe the proof of your above comment would make for a nice answer?2011-09-20
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    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.2011-09-20
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    @RobertIsrael: Thank you; of course. Somehow I was thinking "Direct sum of copies of $F$", but although that's an infinite dimensional inner product space, it's not a Hilbert space. Silly me.2011-09-20
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    @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.2011-09-20
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    A related question: [An orthonormal set cannot be a basis in an infinite dimension vector space?](http://math.stackexchange.com/questions/13641/an-orthonormal-set-cannot-be-a-basis-in-an-infinite-dimension-vector-space)2012-06-09
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    These are different concepts: Linear Basis: $v=\alpha_1b_1+\ldots\alpha_n b_n$ Approximate Linear Basis: $v\approx\lambda_1e_1+\ldots+\lambda_n e_n$2015-02-23

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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.