Bonjour/bonsoir à toutes et à tous.

This may really be a very basic question, but... Let $\mathbf{X} \equiv (X, \|\cdot\|_X)$ and $\mathbf{Y} \equiv (Y, \|\cdot\|_Y)$ be surjectively isometric (1) *normed* spaces (over the real or the complex field).

Question 1.Do $\mathbf{X}$ and $\mathbf{Y}$ need to share the same Hamel dimension?

The answer is clearly yes if the scalar field is the real numbers, since then any surjective isometry $\mathbf{X} \to \mathbf{Y}$ is, a fortiori, an affine transformation $X \to Y$ (via the Mazur-Ulam theorem). But what about the complex case?

**Added later.** As pointed out in the comments below, the answer to Question 1 is still yes if $\mathbf{X}$ and $\mathbf{Y}$ are (real or complex) Banach spaces essentially because the Hamel dimension of an infinite-dimensional (real or complex) Banach space is always at least the cardinality of the continuum (even if the CH fails) as proved in H. E. Lacey, *The Hamel Dimension of any Infinite Dimensional Separable Banach Space is c*, The American Mathematical Monthly, Vol. 80 (1973), p. 298 (click). This is why I resolved to edit the OP and drop the earlier assumption on the completeness of $\mathbf{X}$ and $\mathbf{Y}$.

Question 2.Now assuming that $\mathbf{X}$ and $\mathbf{Y}$ are Banach, do they need to share the same (extended) Schauder dimension as defined in J. W. Evans and R. A. Tapia,Hamel Versus Schauder Dimension, The American Mathematical Monthly, Vol. 77, No. 4 (Apr., 1970), pp. 385-388 (click)?

**Notes.** (1) I'm using the term *isometry* to refer, herein, to both linear and non-linear isometries.

Real isomorphic complex Banach spaces need not be complex isomorphic, in Proc. AMS, vol 96, n.2 (1986) where an example is given of two complex Banach spaces which are isometric but not linearly isomorphic (over complex numbers). I don't remember the details, so I'm not sure it has really something to do with your question, but it could be a start. – Samuele Aug 25 '11 at 15:13The Hamel Dimension of any Infinite Dimensional Separable Banach Space is $\mathfrak{c}$, The AMM, Vol. 80 (1973), p. 298. I confess, this is somehow surprising for me as I really thought the answer should not depend on the completeness of the space, and I'm editing the OP accordingly to pose the question in theright(normed) setting. – Salvo Tringali Sep 13 '11 at 13:34Note.By error, I deleted my previous comment (dated 25 Aug 2011). Now, I've tried to remember what it should be and posted it again. Sorry for the inconvenience! @Samuele. Not really what I was seeking, but still useful. Thanks! – Salvo Tringali Sep 13 '11 at 13:38