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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.

Arturo Magidin
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Casey Patton
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    @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. – Qiaochu Yuan Mar 22 '12 at 22:05
  • @QiaochuYuan, yes, sorry, that wasn't a particularly relevant response! – Sid Raval Mar 22 '12 at 22:11
  • The title still sounds vague. Will someone please edit it? –  Mar 23 '12 at 05:09
  • @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.. – Shashaank Aug 02 '20 at 19:44

3 Answers3

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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.)

Jonas Kibelbek
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  • 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.. – Shashaank Aug 02 '20 at 19:42
  • @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 $\mathbb{R}$. – Charalambos Kioulos Feb 17 '21 at 18:21
  • @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? – Shashaank Feb 17 '21 at 18:28
  • @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 $n \in \mathbb{N}$. – Charalambos Kioulos Feb 17 '21 at 18:36
  • @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 – Shashaank Feb 17 '21 at 18:46
  • @Shashaank It is still an n-tuple of complex numbers. However, the difference is that $\mathbb{C}^2 = \text{Span}_{\mathbb{R}}{ (1,0), (i, 0), (0,1), (0,i) }$ is generated as an $\mathbb{R}$-Span, that is, any element of $\mathbb{C}^2$ can be written as $$ a_1 (1,0) + a_2 (i, 0) + a_3 (0,1) + a_4 (0,i) $$ where $a_i \in \mathbb{R}$ for each $ i = 1,2,3,4$. In the other case it is a $\mathbb{C}$-Span, that is, the coefficients are allowed to be complex numbers. – Charalambos Kioulos Feb 17 '21 at 19:06
  • @CharalambosKioulos ohh yes, I see. Many thanks. And just a last thing. Do the quantities “i” or 0 in (i,0) just referred as a entry in the n-tuple or do some books call it a “component” as well? Do all the books call it an entry in the n-tuple? I have vague memory of a book calling it a component as well. – Shashaank Feb 17 '21 at 19:17
  • @Shashaank Can be called components as well. – Charalambos Kioulos Feb 17 '21 at 19:51
  • @CharalambosKioulos but then then they are components with respect to what? As far as I understand components are with respect to a basis vectors ( or component along a basis vector). What will be the “i” in (i,0) be a component with respect to? – Shashaank Feb 17 '21 at 19:54
  • @Shashaank Components of the n-fold Cartesian product $\mathbb{C}^n$ – Charalambos Kioulos Feb 17 '21 at 19:59
  • @CharalambosKioulos okay I am bit confused as to what is exactly meant by the n fold Cartesian product. I guess I will go with “elements of a tuple” – Shashaank Feb 17 '21 at 20:02
  • $\mathbb{C}^2$ over $\mathbb{Q}$ is the set of all complex numbers with rational components, correct? If so I fail to see how all examples of sets of basis vectors are not constructed from the axiom of choice (AOC). What does it mean for the existence of the basis to depend on the AOC? – Gerald Nov 28 '22 at 04:27
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The "most standard" basis is also $\left\lbrace(1,0),\, (0,1)\right\rbrace$. You just take complex combinations of these vectors. Simple :)

Juan Bermejo Vega
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  • Makes good sense, just didn't realize if that was considered the "standard". – Casey Patton Mar 22 '12 at 22:07
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    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. – Bill Cook Mar 22 '12 at 22:11
  • @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. – krismath Oct 09 '14 at 16:30
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    @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? – Juan Bermejo Vega Oct 10 '14 at 20:18
  • @JuanBermejoVega Oh I see. Thanks. – krismath Oct 11 '14 at 00:23
  • 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.. – Shashaank Aug 02 '20 at 19:42
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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.

Andrea Mori
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    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? – Casey Patton Mar 22 '12 at 22:28
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    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 $4$. – Andrea Mori Mar 22 '12 at 22:37
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    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. – Casey Patton Mar 23 '12 at 23:04
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    Doesn't $\Bbb C^2$ shows complex vector space always ? How can it shows real vector space? – Pranita Gupta Dec 19 '18 at 13:17
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    Complex vector space refers to vector space over the complex field, similarly real vector space is over the real field. See Jonas' answer. – Chris Wang Apr 08 '20 at 17:08
  • 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.. – Shashaank Aug 02 '20 at 19:42