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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

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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?

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  • $\begingroup$ So I just need to prove that $B=(e_1,ie_1,...e_n,ie_n)$ is a real basis right? B generates the vector space because the complex basis does: $v\in V, v=e_1a_1+...+e_na_n$ then $v$ can be written $v=e_1a_1+ie_1a_1+...+e_na_n+ie_na_n$ To prove that all vectors in B are linearly independent $e_1a_1+ie_1a_1+...+e_na_n+ie_na_n=0=(1+i)(e_1a_1+...+e_n+a_n)$ because $1+i\not= 0 \rightarrow (e_1a_1+...+e_n+a_n)=0$ and using that the complex basis is linearly indepent I get that B is a basis. Is this OK? $\endgroup$ – Shomar Feb 24 '14 at 23:06
  • $\begingroup$ Yeah, excellent! $\endgroup$ – Léo Feb 24 '14 at 23:10

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