Apparently the orthonormal basis $(e_n)_{n\in \mathbb{N}}$ of the Hilbert space $H$ (in special case, infinitly dimensional) is not a basis of $H$ as a vectorspace. Is there a way to prove this?
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You mean Hamel basis? – Jan 21 '19 at 12:40
3 Answers
Every vector space basis of an infinite dimensional Banach or Hilbert space is necessarily uncountable. This is actually a non-trivial result and follows from the Baire category theorem, see e.g. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.

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First show or recall that every $x\in H$ has a unique representation $$x=\sum a_ne_n,$$with $\sum|a_n|^2<\infty$. So if $$x=\sum_{n=1}^\infty e_n/n$$then the uniqueness shows that $x$ is not a (finite) linear combination of the $e_n$.

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A basis means that for everu $u\in H$ you have $u=a_{i_1}e_{i_1}+...+a_{i_n}e_{i_n}$ for $n$ finite, and orthogonal basis means that $u=\sum_{n\geq 0}a_ne_n$ where the sequence $\sum_{i=0}^{i=n}a_ie_i$ converges, so for orthogonal basis, all the coefficients may not be zero.

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1@Judith think about $v=(1, 1/2^2, \ldots, 1/n^2, \ldots)$. Then, $v$ can't be written as a finite linear sum of $ (e_{n})$ (why?). But $\sum_{i=1}^{n} (1/i^2) e_{i}$ converges to $v$.Also, $(e_{n})$ are the Schauder's basis not the Hamel basis for the given Hilbert space. – Shubham Namdeo Jan 21 '19 at 12:57