Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
@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? –  user21436 Mar 23 '12 at 5:09

3 Answers 3

up vote 5 down vote accepted

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

share|improve this answer
    
Makes good sense, just didn't realize if that was considered the "standard". –  Casey Patton Mar 22 '12 at 22:07
2  
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

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.)

share|improve this answer

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.

share|improve this answer
    
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
5  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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