Is $\mathbb{C}^n = \mathbb{R}^{2n}?$

Question

I am trying to show that some $n$-dimensional vector space over $\mathbb{C}$ is complete. Which is isomorphic to $\mathbb{C}^n.$ (Is that the right word in this context? Isomorphic?) But $\mathbb{R}^2 = \mathbb{C}.$ So can we say that $\mathbb{C}^n = \mathbb{R}^{2n}$ and since $\mathbb{R}^{2n}$ is complete, my vector space is complete?

Answer 1

To be more precise you could say that the map from $\mathbb{C}^n$ to $\mathbb{R}^{2n}$ given by $(z_1, \dots, z_n) \mapsto (\Re z_1, \Im z_1, \dots, \Re z_n , \Im z_n)$ is not only an isomorphism of $\mathbb{R}$-vector spaces, but also an isometry with respect to the 'usual' norms, that is $\sqrt{\sum_{i=1}^n z_i\overline{z_i}}$ and the Euclidean norm, respectively. Thus, $\mathbb{C}^n$ is isometric to a complete space and thus itself complete.

Some would object to the statement that $\mathbb{C}^n$ and $\mathbb{R}^{2n}$ are equal. The first naturally carries the structure of a $\mathbb{C}$-vectorspace while the latter does not (of course one can still defined one on the latter too, but it is not natural).

Answer 2

A vector space is not defined only by the set of vectors ( $\mathbb{C}^n$ or $ \mathbb{R}^{2n}$ in your case) but also by the field of scalars over wich the vector space is defined.

So, We can have a vector space $\mathbb{C}^n$ ove $\mathbb{C}$ that is a vector space of dimension $n$ or a vector space $\mathbb{C}^n$ over $\mathbb{R}$ that is a vector space of dimension $2n$, and this is isomorphic (but not the ''same'') as the vector space $ \mathbb{R}^{2n}$ over $ \mathbb{R}$.

All these vector space are complete with respect to the euclidean norm.

Note that we cannot define a vector space $ \mathbb{R}^{2n}$ over $ \mathbb{C}$.