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After reading the following post (What is the "standard basis" for fields of complex numbers?), I tried to confirm that the suggested standard basis for $\mathbb C^2$ over $\mathbb C$ spans all of $\mathbb C^2$. The standard basis suggested by several answers is the set: $\{ (1,0), (0,1) \}$.

I played around with this for a while...and just wanted to make sure that my understanding is correct.

Before moving on, consider the following color coding scheme...as things will get messy. Let $\color{blue}{blue}$ represent an object that is an element of $\mathbb C$ and let $\color{red}{red}$ represent an object that is an element belonging to $\mathbb R$. Continuing...

When people say the set $\{ (1,0),(0,1)\}$ forms the basis for $\mathbb C^2$ over $\mathbb C$, they are, in fact, saying that $(1,0)$ and $(0,1)$ are both $ \in \mathbb C^2$. Using our color notation, we would say $(\color{blue}{1},\color{blue}{0}) \in \mathbb C^2$ and $(\color{blue}{0},\color{blue}{1}) \in \mathbb C^2$. Given that each one of those blue objects belongs to $\mathbb C$, and recalling that a complex number is an ordered pair of real numbers, we could then rewrite the basis accordingly:

$(\color{blue}{1},\color{blue}{0}) = \big((\color{red}{1},\color{red}{0}),(\color{red}{0},\color{red}{0})\big)$ and $(\color{blue}{0},\color{blue}{1}) = \big((\color{red}{0},\color{red}{0}),(\color{red}{1},\color{red}{0})\big)$.

For further clarity to the readers:

$\color{blue}{1} \mapsto (\color{red}{1},\color{red}{0})$ and $\color{blue}{0} \mapsto (\color{red}{0},\color{red}{0})$

Is this the correct understanding of what is meant when people say $\{ (1,0), (0,1) \}$ is a basis for $\mathbb C^2$ over $\mathbb C$?

Thanks!

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  • $\begingroup$ What it means is that every element of $\mathbb{C}^2$ is of the form $c_1(1,0)+c_2(0,1)$ where $c_{1,2}\in\mathbb{C}$. In other words it is just the orthonormal basis vectors of $\mathbb{C}^2$ as a $\mathbb{C}$ vector space. $\endgroup$
    – Μάρκος Καραμέρης
    Sep 8, 2020 at 21:21
  • $\begingroup$ I believe what you are trying to say is that $\mathbb{C}^2$ is a vector space of dimension 4 over $\mathbb{R}$ $\endgroup$
    – Μάρκος Καραμέρης
    Sep 8, 2020 at 21:24
  • $\begingroup$ Yes, this is the correct understanding. $\endgroup$
    – Teebro Prokash
    Sep 8, 2020 at 21:24
  • $\begingroup$ $(1,0)$ is the complex number $1+0(i)$ and $(0,1)$ is $0+1(i)$. I think you are somehow trying to express $1$ in complex numbers as a pair of ordered pairs $((1,0),(0,0))$ which is not the best way to write.. Think of $a+bi=a(1,0)+bi(0,1)$, so the scalars used $a$ and $bi$ are from complex field. $\endgroup$
    – Anurag A
    Sep 8, 2020 at 21:25

3 Answers 3

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Yes, that is entirely correct, from a logical point of view (assuming that you defined the set $\Bbb C$ as $\Bbb R^2$). However, it is certainly not the best way of dealing with it. The set $\{(1,0),(0,1)\}$ is a basis of $\Bbb C^2$ simply because each $v\in\Bbb C^2$ can be written in one and only one way as a linear combination of $(1,0)$ and $(0,1)$. In fact, if $v=(z_1,z_2)$, then the only way of writing $v$ as a linear combination of $(1,0)$ and $(0,1)$ is $v=z_1(1,0)+z_2(0,1)$. And this works also over $\Bbb R$, over $\Bbb Q$ and in fact over any field whatsoever.

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I think that explanation is correct.

But it's very formalistic, in particular there's too many parentheses and commas and its not close to how people actually think and write. If you would like to get closer to the way that people actually think and write when they work over complex numbers, you might write $$\color{blue}{1} = \color{red}{1} + \color{red}{0} \color{green}i $$ and $$\color{blue}{0} = \color{red}{0} + \color{red}{0} \color{green}i $$

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You're right. I would add, regarding this part:

recalling that a complex number is an ordered pair of real numbers

All the difficulty here stems from thinking that a complex number is an ordered pair, i.e. that $\mathbb C = \mathbb R^2$. It's probably healthier to treat the set-theoretic implementation of $\mathbb C$ as opaque. There is a standard isomorphism $\mathbb C\cong\mathbb R^2$ but they don't have to be literally the same set. Maybe $\mathbb C$ is secretly $\mathbb R^2$, or maybe it's the quotient ring $\mathbb R[x]/(x^2+1)$; it shouldn't matter for most purposes.

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