# How to understand the decomplexification of a vector space

Specifically, I'm trying to solve the following problem:

Let $$V$$ be a finite-dimensional complex vector space and $$e_1, \cdots,e_n\in V$$ be a basis. Show the following:

(i) addition $$V\times V\rightarrow V$$ and restriction of scaling $$\mathbb{C}\times V$$ to $$\mathbb{R}\times V$$ make $$V$$ a real vector space;

(ii) the vectors $$e_1,\cdots,e_n,ie_1,\cdots,ie_n$$ are a basis for $$V$$ as a real vector space.

Generally speaking, I intuitively understand this as the decomplexification of a vector space where we "ignore" multiplication by complex values. I'm struggling to formally solve this. Here's my attempt:

(i): The restriction of scaling implies that $$V_{\mathbb{C}}\ni u+iv\mapsto u+v$$. The desired result follows.

(ii): Suppose $$\lambda_1e_1+\cdots+\lambda_ne_n+\lambda_1ie_1+\cdots+\lambda_nie_n=0$$ for some $$\lambda_1,\cdots,\lambda_n\in\mathbb{R}$$. Then $$\lambda_1e_1+\cdots+\lambda_ne_n=-i(\lambda_1e_1+\cdots+\lambda_ne_n)$$, which implies that $$\lambda_1=\cdots=\lambda_n=0$$. Hence, $$e_1,\cdots,e_n,ie_1,\cdots,ie_n$$ are linearly independent in $$V_{\mathbb{R}}$$.

The questions I have are as follows:

1. What does the addition $$V\times V\rightarrow V$$ contribute to part (i) aside from the fact that for $$V$$ to be a real vector space, addition must be defined?
2. What is a better way to express what the restriction of scaling is doing? I used the \mapsto symbol, but there isn't really a mapping going on.
3. Aside from questions 1 and 2, am I missing anything to finish solving part (i)?
4. Is my justification that $$e_1,\cdots,e_n,ie_1,\cdots,ie_n$$ are linearly independent valid? It feels sketchy to repeat scalars.
5. How can I show that $$e_1,\cdots,e_n,ie_1,\cdots,ie_n$$ span $$V_{\mathbb{R}}$$? It seems to follow directly from the fact that $$e_1,\cdots,e_n$$ are a basis for $$V_{\mathbb{C}}$$ and the "decomposition" of complex vectors into real parts by restriction of scaling.

1. The point is $$+$$ must be defined on any vector space, and in this case we can trivially define the $$\Bbb R$$-space $$+$$ in terms of the $$\Bbb C$$-space $$+$$.
2. We're restricting which field of scalars the space is on. We no longer consider non-real complex scalars "legal". For example, we no longer consider $$ie_n$$ to be a multiple of $$e_n$$.
4. Expanding on my point in $$2$$, linear independence means $$\sum_{j=1}^na_je_j+\sum_jb_jie_j=0$$ with $$a_j,\,b_j\in\Bbb R$$ implies $$a_j=b_j=0$$. This is trivial because we already know $$\sum_j(a_j+ib_j)e_j=0\implies a_j+ib_j=0$$.