$ H $ hilbert space: Hamel dimension of $ H $ = Hilbert dimension of $ H $ $ \Leftrightarrow$ dim $ H $ is finite Clearly $\Leftarrow $ is a trivial trivial application of G-Schmidt algorithm.  I've experienced some trouble in proving the other direction. I focused my self on the fact that span($ A $)=$ H $ (it can be not closed) where A is an Hamel base, but i i didn't found nothing. I can't even use G-S because the cardinality of the hamel base can be uncountable.  Clearly the interesting case (where to find an absurd) is when H has an uncountable hilbert dimension.
Someone has some advice? Thanks in advance
 A: There are infinite-dimensional Hilbert spaces whose Hamel-dimension equals the Hilbert-dimension.
Remember that every Hilbert space is (isometrically isomorphic to) $\ell^2(\kappa)$ for some cardinal $\kappa$.
Let $\kappa = 2^{\aleph_\alpha}$ for some ordinal $\alpha$. Then $\kappa^{\aleph_0} = 2^{\max\{\aleph_\alpha,\aleph_0\}} = 2^{\aleph_\alpha} = \kappa$. Letting $\iota_S \colon \ell^2(S) \to \ell^2(\kappa)$ the canonical inclusion for a subset $S\subset \kappa$, we have
$$\ell^2(\kappa) = \bigcup_{\substack{S\subset\kappa\\ \lvert S\rvert \leqslant \aleph_0}} \iota_S(\ell^2(S)).$$
Since every $\ell^2(S)$ has cardinality $\mathfrak{c} = 2^{\aleph_0}$ (or $1$, if $S$ is empty), and there are no more than $\kappa$ countable subsets of $\kappa$, we have
$$\operatorname{card} \ell^2(\kappa) \leqslant \kappa \otimes \mathfrak{c} = \max \{\kappa,\mathfrak{c}\} = \kappa.$$
On the other hand, the cardinality is trivially at least $\kappa$, so we have equality.
The same argument for finite subsets of a basis shows that $\dim_{\text{Hamel}} V = \operatorname{card} V$ for a (real or complex) vector space $V$ if the dimension is at least $\mathfrak{c}$.
So we have $\dim_{\text{Hamel}} \ell^2(2^{\aleph_\alpha}) = \operatorname{card} \ell^2(2^{\aleph_\alpha}) = \dim_{\text{Hilbert}} \ell^2(2^{\aleph_\alpha})$ for all ordinals $\alpha$.
