# Linear combinations over the complex field, why complex coefficients?

Assume we have a vector space $V$ over the field $F$. A linear combination of elements $v_1, \dots, v_k$ of $V$ is an expression of the form $$c_1 v_1 + \dots + c_k v_k, \quad \text{where } c_i \in F.$$ My question is: what's the point of $c_i$ being in $F$?

My guess is to be able to represent all elements of $V$ using linear combinations of $v_i$. However, this guess got me confused when $F$ is the complex field.

For the sake of argument, assume $V=\mathbb{R}^2$. It is clear to me that if we consider the linear combination (with linearly independent $v_1, v_2 \in V$) $$c_1 v_1+ c_2 v_2, \quad \text{with } c_i \in F = \mathbb{N} \text{ (natural numbers)}$$ then there would be elements of $V=\mathbb{R}^2$, which we cannot represent using this linear combination, since all possible combinations can't fill all of $V=\mathbb{R}^2$. Hence, we resort to $c_i$ in $F = \mathbb{R}$ that does the job.

I fail, however, to have a similar understanding when $V=\mathbb{C}^2$ for why should $F = \mathbb{C}$ and not $\mathbb{R}$. That is, I can't see which elements of $V = \mathbb{C}^2$ can't be represented using $$c_1 v_1+ c_2 v_2, \quad \text{with } c_i \in F = \mathbb{R} \text{ (real numbers)}$$ I guess my imagination fails me when it comes to complex numbers... or maybe they're not called complex for nothing after all :-)

# Edit

Thinking about a simpler case where $V = \mathbb{C}$, linear combinations (actually scaling a single element) $$c_1 v_1, \quad \text{with } c_i \in F = \mathbb{R} \text{ (real numbers)}$$ would not allow to fill up all the complex plane. So this might be a reason for taking $c_i \in F = \mathbb{C}$.

Any further comments, corrections are welcome.

Disclaimer: I wanted this to be a comment, but it is too long for a comment.

We discard $\mathbb N$ because it is not a field to begin with.

Given an abelian group $V$, there are virtually many fields over which we can consider $V$ a vector space. Formally speaking, a vector space is a 2-tuple $(V,F)$, where $V$ is an abelian group and $F$ is a field acting on $V$ in such a way that etc. So, in some sense, the field over which you are considering your object to be a vector space is not intrinsically related to your object; you choose it (out of all the possible candidates).

For the case where $V=\mathbb C^2$, since both $\mathbb R$ and $\mathbb C$ are fields that act on $V$ appropriately (i.e. according to the axioms of a vector space), they are both candidates for a field over which $V$ can be considered a vector space. However, $\mathbb C^2$ as a vector space over $\mathbb R$ is not the same thing as $\mathbb C^2$ as a vector space over $\mathbb C$.

I don’t think I understand your question, but I thought I’d point these out because I sensed that you might have a misunderstanding of some sort. I hope this helped.

• You are right, my confusion comes, in part, from my misunderstanding of what you pointed out. Could you elaborate on how is $\mathbb{C}^2$ as a vector space different depending on chosen field? – Likely May 15 '18 at 23:53
• For the example I gave when $V=\mathbb{C}^2$ and $F=\mathbb{R}$, is it possible to recover all vectors of $V$, which then would explain why I couldn't think of a counterexample... – Likely May 15 '18 at 23:55
• @Likely yes, it is possible, but you need more vectors when $F = \mathbb R$. When $F= \mathbb C$, you only need two vectors, e.g. $v_1 = (1,0)$ and $v_2= (0,1)$ to recover $\mathbb C^2$. However, when $F = \mathbb R$, you need at least four vectors, e.g. $v_1 = (1,0)$, $v_2 = (i,0)$, $v_3 = (0,1)$ and $v_4 = (0,i)$. Do you see why? – User May 15 '18 at 23:57
• @Likely and this shows you (at least intuitively) the difference between these two vector spaces. – User May 16 '18 at 0:00
• I'm not sure I see why...thinking about a complex plane with 4D!! couldn't do it. Any hints? – Likely May 16 '18 at 0:02