Complexifying a vector space and its practical meaning My question is very straightforward. Let $V$ be a vector space over $\mathbb{R}$. Is there any practical difference between the complexification $V_{\mathbb{C}}$ and implicitly treating $V$ as a complex vector space rather than a real one? In applications, it might be instructive to make such distinctions but if I'm treating $V$ as an abstract vector space over $\mathbb{R}$, isn't $V_{\mathbb{C}}$ equivalent to treat $V$ as a $\mathbb{C}$-vector space?
 A: I think some confusion is arising from the fact that most "natural" real vector spaces are the real points of some other "natural" complex vector space.  For example, if one considers the canonical $n$-dimensional real vector space $\mathbb{R}^n$ consisting of $n\times 1$ column vectors, then this obviously sits inside of $\mathbb{C}^n$.
Other examples include subsets of function spaces, since $\{ f:X\rightarrow \mathbb{R}\}\subseteq \{f:X\rightarrow \mathbb{C}\}$, e.g., the collection of real polynomials of degree less than or equal to $n$ is a subset of the corresponding collection of complex polynomials.
So, perhaps it is instructive to see an example where the naive extension to $\mathbb{C}$ does not work.
Let $V = (0,\infty)$ and define $x+_V y:= xy$ and, for $r\in \mathbb{R}$, $r\cdot_V x := x^r$.  Then it's not too hard (though it is somewhat tedious to do from first principles) to verify that $V$ is real vector space with $0_V = 1$ and where the additive inverse of $x$ is $\frac{1}{x}$.
For example, $(rs)\cdot_V x = x^{rs} = (x^r)^s = s\cdot_V(r\cdot_V x)$, verifying that axiom.
(All the vector space axioms are trivially verified by simply noting that the bijection $e^x:\mathbb{R}\rightarrow (0,\infty)$ can be used to transport the usual structure on $\mathbb{R}$ to $V$).
For the naive extension to a $\mathbb{C}$ vector space, we should, for example, define $i\cdot x = x^i$.  But $x^i$ has infinitely many complex values, even if $x$ is real.  Of course, you can say "for each $x$, pick one value of $x^i$", but then this choice isn't continuous, and it will make it quite hard to verify that all the necessary axioms are satisfied.
On the other hand, $V_\mathbb{C}:=V\otimes_\mathbb{R} \mathbb{C}$ makes sense for any real vector space $V$, and gives a uniform way of complexifying all real vector spaces simultaneously.  It even has the advantage that when you apply it to "natural" real vector spaces, it gives you (something isomorphic to) the "natural" corresponding complex vector space.
A: We probably can't treat $V$ as a complex vector space. For instance, $R^n$ isn't closed under multiplication by complex numbers as you get products with nonzero imaginary components. On the other hand, the complexification $V \bigotimes C$ is guaranteed to be a vector space. Though the tensor product may be unintuitive, the resulting vector space is usually exactly what you'd expect. For instance, the complexification of $R^n$ is $C^n$.
