0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Likely May 15 '18 at 23:53
  • $\begingroup$ 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... $\endgroup$ – Likely May 15 '18 at 23:55
  • $\begingroup$ @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? $\endgroup$ – User May 15 '18 at 23:57
  • $\begingroup$ @Likely and this shows you (at least intuitively) the difference between these two vector spaces. $\endgroup$ – User May 16 '18 at 0:00
  • $\begingroup$ I'm not sure I see why...thinking about a complex plane with 4D!! couldn't do it. Any hints? $\endgroup$ – Likely May 16 '18 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.