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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.
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  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 $e_n$.
  3. I have nothing to add here.
  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 $\sum_j(a_j+ib_j)e_j=0\implies a_j+ib_j=0$.
  5. That's basically it.
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