Invertible operator not preserving Hilbert dimension

It is known that for a bijective linear operator $T:X\to Y$ the algebraic dimensions of the linear spaces $X$ and $Y$ coincide.

I am asking for an example of an invertible (bounded) linear operator between (infinite dimensional) Hilbert spaces such that their Hilbert dimension is different.

-

Notice that if you have a $\kappa$-(Hilbert) dimensional space, it has a dense subset of cardinality $\kappa$ (for infinite $\kappa$) -- the finite rational combinations of orthonormal basis vectors.
On the other hand, every such space has a collection of $\kappa$ disjoint, nonempty open sets: balls of radius $\frac{1}{2}$ centered at each vector of the ON basis, so it has no dense subset of any smaller cardinality.
Therefore, it is impossible to find a homeomorphism between two Hilbert spaces of different dimensions, so it is certainly not possible to find a linear one (though, of course, there are linear isomorphisms between spaces of appropriate dimensions -- I think iff $\kappa,\lambda$ are such that $\kappa^\omega=\lambda^\omega$).
Great answer. Many thanks. The proof can be found somewhere ? Or it is just your idea ? I admit that I do not understand your last remark (in the parenthesis). However, if the Hilbert spaces $X$ and $Y$ have the same Hilbert dimension then there exists a unitary operator (linear isomorphism) between them (this is known). – aly Sep 6 '12 at 23:38
@aly: I did not take it from anywhere, but I'm pretty sure I've seen something like that done in case where $\kappa=\aleph_0$ during my functional analysis course. The last remark is supposed to say that when two spaces have Hilbert dimensions $\kappa,\lambda$ satisfying the equality I've wrtiten, then their algebraic dimensions should coincide (and be equal to $\kappa^\omega=\lambda^\omega$), so that we will have a linear isomorphism (e.g. if $\omega\leq \kappa,\lambda\leq \mathfrak c$). – tomasz Sep 7 '12 at 1:35