What is the "standard basis" for fields of complex numbers?

For example, what is the standard basis for $\Bbb C^2$ (two-tuples of the form: $(a + bi, c + di)$)? I know the standard for $\Bbb R^2$ is $((1, 0), (0, 1))$. Is the standard basis exactly the same for complex numbers?

P.S. - I realize this question is very simplistic, but I couldn't find an authoritative answer online.

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    $\begingroup$ @Sid: I don't see what that has to do with anything. I assume $\mathbb{C}^2$ is to be understood as a complex vector space. $\endgroup$ Mar 22 '12 at 22:05
  • $\begingroup$ @QiaochuYuan, yes, sorry, that wasn't a particularly relevant response! $\endgroup$
    – Sid Raval
    Mar 22 '12 at 22:11
  • $\begingroup$ The title still sounds vague. Will someone please edit it? $\endgroup$
    – user21436
    Mar 23 '12 at 5:09
  • $\begingroup$ @QiaochuYuan Your answer very helpful. Can $C^2$ with $C$ as the field have such a basis vector that contains $i$ as a possible component in place of the 1's, 0's Or not.. $\endgroup$
    – Shashaank
    Aug 2 '20 at 19:44

The "most standard" basis is also $\left\lbrace(1,0),\, (0,1)\right\rbrace$. You just take complex combinations of these vectors. Simple :)

  • $\begingroup$ Makes good sense, just didn't realize if that was considered the "standard". $\endgroup$ Mar 22 '12 at 22:07
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    $\begingroup$ Yes. In fact, that is "the standard basis" for $\mathbb{F}^2$ where $\mathbb{F}$ is any field: $\mathbb{F}=\mathbb{R},\mathbb{C},\mathbb{Q},\mathbb{Z}_p,$ etc. $\endgroup$
    – Bill Cook
    Mar 22 '12 at 22:11
  • $\begingroup$ @JuanBermejoVega How is the set ${(1,0), (0,1)}$ a basis for $\mathbb{C}$? It only spans the real part of each of the complex plane. $\endgroup$
    – krismath
    Oct 9 '14 at 16:30
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    $\begingroup$ @krismath Are you only taking real combinations of those vectors? In a complex vector space you should take complex combinations. Does this answer your question? $\endgroup$ Oct 10 '14 at 20:18
  • $\begingroup$ @JuanBermejoVega Oh I see. Thanks. $\endgroup$
    – krismath
    Oct 11 '14 at 0:23

Just to be clear, by definition, a vector space always comes along with a field of scalars $F$. It's common just to talk about a "vector space" and a "basis"; but if there is possible doubt about the field of scalars, it's better to talk about a "vector space over $F$" and a "basis over $F$" (or an "$F$-vector space" and an "$F$-basis").

Your example, $\mathbb{C}^2$, is a 2-dimensional vector space over $\mathbb{C}$, and the simplest choice of a $\mathbb{C}$-basis is $\{ (1,0), (0,1) \}$.

However, $\mathbb{C}^2$ is also a vector space over $\mathbb{R}$. When we view $\mathbb{C}^2$ as an $\mathbb{R}$-vector space, it has dimension 4, and the simplest choice of an $\mathbb{R}$-basis is $\{(1,0), (i,0), (0,1), (0,i)\}$.

Here's another intersting example, though I'm pretty sure it's not what you were asking about:

We can view $\mathbb{C}^2$ as a vector space over $\mathbb{Q}$. (You can work through the definition of a vector space to prove this is true.) As a $\mathbb{Q}$-vector space, $\mathbb{C}^2$ is infinite-dimensional, and you can't write down any nice basis. (The existence of the $\mathbb{Q}$-basis depends on the axiom of choice.)

  • $\begingroup$ Your answer very helpful. Can $C^2$ with $C$ as the field have such a basis vector that contains $i$ as a possible component in place of the 1's, 0's Or not.. $\endgroup$
    – Shashaank
    Aug 2 '20 at 19:42
  • $\begingroup$ @Shashaank Indeed, one can "rotate" the standard basis of $\mathbb{C}^2 $ viewed as a v.s. over $\mathbb{C}$ to obtain the basis $\{ (i,0), (0,i) \}$ just as one could rotate any basis of $\mathbb{R}^2$ over $\mathbb{R}$. $\endgroup$ Feb 17 at 18:21
  • $\begingroup$ @CharalambosKioulos Thanks for your comment. I also wanted to know that when the basis vectors are written like this -(i,0) etc, then what is exactly “i” or “0” here. Is it a component or what? Because my understanding is that in a term like 2(i,0), 2 is the component along that specific basis vector. But what exactly is “i” or “0” in that term? $\endgroup$
    – Shashaank
    Feb 17 at 18:28
  • $\begingroup$ @Shashaank as any vector in $\mathbb{C}^n$ over $\mathbb{C}$ is an n-tuple of complex numbers, it is exactly the complex number $i$, in the first entry of this n-tuple. In the case being discussed, $n = 2$, but in general $n \in \mathbb{N}$. $\endgroup$ Feb 17 at 18:36
  • $\begingroup$ @CharalambosKioulos but in the answer above $C^2$ as a V.S over $R$ has a basis vector (i,0) and (0,i). When the field is of real numbers, why does the tuple have an “i” in it ( as in (i,0) ). Your comment suggests that $C^2$ as a v.s over R should be an n tuple of just real numbers. So the “i” in (i,0) shouldn’t be there as it is a complex number and the field is of real numbers. Am I wrong? Can you tell me what am I missing $\endgroup$
    – Shashaank
    Feb 17 at 18:46

I'm not sure that this is what you want, but under the usual Argand-Gauss identification $\Bbb C=\Bbb R^2$ the standard basis of $\Bbb C$ would be $\{1,i\}$, the standard basis of $\Bbb C^2$ would be $\{(1,0),(i,0),(0,1),(0,i)\}$ and so on.

  • $\begingroup$ This is a little confusing, because the previous answer gave me a basis of dimension 2 and this answer gives me a basis of dimension 4. How can this be possible? $\endgroup$ Mar 22 '12 at 22:28
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    $\begingroup$ What is not clear (to me, at least) from your question is that you consider $\Bbb C^2$ as a real or complex vector space. As a complex vector space it has dimension $2$, as a real vector space it has dimension $4$. $\endgroup$ Mar 22 '12 at 22:37
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    $\begingroup$ Ah gotcha. Well....being a student in an introductory Linear Algebra class, I haven't actually learned what those terms mean yet! Hence the confusing question. $\endgroup$ Mar 23 '12 at 23:04
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    $\begingroup$ Doesn't $\Bbb C^2$ shows complex vector space always ? How can it shows real vector space? $\endgroup$ Dec 19 '18 at 13:17
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    $\begingroup$ Complex vector space refers to vector space over the complex field, similarly real vector space is over the real field. See Jonas' answer. $\endgroup$
    – Chris Wang
    Apr 8 '20 at 17:08

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