Proof that if the dimension of the complex space is n, then dimension of the real one is 2n If a a set of vectors can be algebraized as an $\mathbb R$-vector space or a $\mathbb C$-vector, prove that if the dimension of the complex space is $n$, then dimension of the real one is $2n$.
My idea was to try and prove it with the property $\dim(S_1+S_2)=\dim(S_1)+\dim(S_2)-\dim(S_1\cap S_2)$ because a complex number can be written as $(a,b)$ with $a,b \in \Re$ 
So if $S_1$ is the subset generated by $(1,0)$ and $S_2$ by $(0,1)$ then I get $\dim(S_1+S_2)=1+1-0=2$
While if I try to algebraize a given complex number as a C-vector field I only need 1 vector
But this is just proving the theorem for the vector space $(\mathbb C,+,.)$ and $(\mathbb R,+,.)$ how do I extend it for any vector space?
Thanks

Hint:
Let $V$ be your vector space, with $(e_1,...,e_n)$ a complex basis.
What do you think about the family $(e_1,ie_1,...e_n,ie_n)$ on the real numbers?
