Are two Hilbert spaces with the same algebraic dimension (their Hamel bases have the same cardinality) isomorphic? We know that two Hilbert spaces that have orthonormal bases of the same cardinality are isomorphic (as an inner product spaces). 
My question is: what can we say when we know that their Hamel bases have the same cardinality? It clearly implies they are isomorphic as vector spaces (just send a basis to a basis and extend linearly), but are they isomorphic also as inner product spaces? (i.e. via an isomorphism that also preserves the inner-product).
 A: No, Hilbert spaces of different Hilbert dimensions can have the same algebraic dimension as vector spaces over $\mathbb R$ (or $\mathbb C$, take your pick).
For a cardinal $\kappa$, let $H_{\kappa}$ be the $\kappa$-dimensional real (or complex, whatever) Hilbert space, i.e., it has an orthonormal basis of cardinality $\kappa$. Let $\mathcal H=\{H_{\kappa}:\aleph_0\le\kappa\le2^{\aleph_0}\}$. The set $\mathcal H$ contains at least two nonisomorphic Hilbert spaces ($H_{\aleph_0}$ and $H_{2^{\aleph_0}}$) and maybe as many as $2^{\aleph_0}$ of them depending on your set theory. I claim that they are all algebraically isomorphic because they all have algebraic dimension $2^{\aleph_0}$ as vector spaces over $\mathbb R$.
First, since the number of points in the space $H_{\kappa}\in\mathcal H$ with orthonormal basis $\mathcal B$ is $|H_{\kappa}|\le|\mathcal B|^{\aleph_0}2^{\aleph_0}\le(2^{\aleph_0})^{\aleph_0}2^{\aleph_0}=2^{\aleph_0}$, the algebraic dimension of each $H\in\mathcal H$ is at most $2^{\aleph_0}$.
Next, we show that the infinite-dimensional separable Hilbert space $H_{\aleph_0}$ has algebraic dimension at least $2^{\aleph_0}$, by exhibiting a continuum of (algebraically) linearly independent elements in the space $\ell^2$ of square-summable sequences.
Let $(r_n:n\in\mathbb N)$ be an enumeration of the rational numbers. For $t\in\mathbb R$ and $n\in\mathbb N$ define $\varepsilon(t,n)$ to be $1$ if $r_n\lt t$ and $0$ otherwise. Finally, let $x_t=\left<\dfrac{\varepsilon(t,n)}n:n\in\mathbb N\right>\in\ell^2$. It is easy to see that every finite subset of $\{x_t:t\in\mathbb R\}$ is linearly independent.
