Consider a Hilbert space which is infinite dimensional. If it is separable, it is well known that an orthonormal basis will be countable, while a hamel basis will be uncountable (since it is a complete space), therefore the two bases cannot coincide. But how can we prove that those two bases cannot be equal if our space is not separable?

  • $\begingroup$ Nothing changes: if $\{ e_{\alpha}\}$ is an ONB, then the expansion coefficients of an $x\in H$ are the scalar products, so an infinite linear combination is not at the same time a finite linear combination. $\endgroup$ – user138530 Dec 7 '14 at 1:31
  • $\begingroup$ Basically we get that any x is perpendicular to all but finitely many e_i's. However, Tomasz's answer made it clear. It was a very simple question but I didn't try to think of such a simple approach. Thank you for your answers! $\endgroup$ – Nick Kolliopoulos Dec 7 '14 at 2:04

The argument can be made rather simpler: if you have a given ON basis $e_i$, $i\in I$, the vector $v=\sum_{n\in {\bf N}} 2^{-n}e_{i_n}$, where $i_n$ are distinct, is clearly not a finite linear combination of the $e_i$ (because if we subtract from $v$ some finite linear combination of $e_i$, we can still find some $i_n$ such that the result is not orthogonal to $e_{i_n}$).

You don't need $I$ to be countable.

On the other hand, if you have a Hilbert space of Hilbert dimension at least $\mathfrak c$, then the Hilbert dimension and Hamel dimension do coincide (but with different bases). They also coincide with the cardinality of the space, in this case.

  • $\begingroup$ Why statement in last paragraph true i.e. if you have a Hilbert space of Hilbert dimension at least c, then the Hilbert dimension and Hamel dimension do coincide $\endgroup$ – Sushil Jun 20 '15 at 4:13
  • $\begingroup$ @Sushil: I see now that I have been a little rash. You can express every element of the space, uniquely, as an infinite linear combination of a countable set of the basis vectors. This implies that if the Hilbert dimension is $\kappa$ where $\kappa^\omega=\kappa\neq 0$ (which is true for $\kappa=\mathfrak c$, or any $\kappa=\lambda^\omega$), then the cardinality of the space is equal to $\kappa$, and therefore so is the Hamel dimension. On the other hand, the cardinality of the Hilbert space of Hilbert dimension $\aleph_\omega$ is $\aleph_\omega^\omega>\aleph_\omega$. $\endgroup$ – tomasz Jun 20 '15 at 19:23
  • $\begingroup$ @Sushil: Well, that is what I think I had in mind, anyway, considering I wrote it half a year ago. $\endgroup$ – tomasz Jun 20 '15 at 19:24
  • $\begingroup$ @Sushil: So the last statement is false: it is easy to see that the cardinality of an infinite dimensional vector space is the dimension times the cardinality of a field. So the Hilbert space of (Hilbert) dimension $\kappa$, where $\kappa>\mathfrak c$ has cofinality $\omega$ will have Hamel dimension $\kappa^\omega>\kappa$. Still, this answer is not very popular, so I'd rather not needlessly bump it just to fix this mistake. $\endgroup$ – tomasz Jun 20 '15 at 20:04
  • 1
    $\begingroup$ @Sushil: More or less, yes. $\endgroup$ – tomasz Jun 25 '15 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.