1
$\begingroup$

What is the dimension of the complex linear space $\mathbb{C}^n$ over $\mathbb{R}$? I know that the dimension of $\mathbb{C}^n$ over $\mathbb{C}$ is $n$. I also know that the dimension of $\mathbb{C}$ over $\mathbb{R}$ is $2$. So, I think the dimension of $\mathbb{C}^n$ over $\mathbb{R}$ is $2n$. But I cannot make it precise. Any help would be appreciated.

$\endgroup$

2 Answers 2

3
$\begingroup$

In general, if $K$ is a field and $L$ is a field extension of $K$, then $L$ is a $K$-vector space and every $L$-vector space $V$ is also a $K$-vector space, and it holds that $$\dim_KV = \dim_L V \dim_KL.$$So: $$\dim_{\Bbb R}\Bbb C^n = \dim_{\Bbb C}\Bbb C^n \dim_{\Bbb R}\Bbb C = 2n.$$

$\endgroup$
1
  • $\begingroup$ It is nice.thank u. $\endgroup$ May 7, 2017 at 2:23
3
$\begingroup$

Let $\{e_1,\dots,e_n\}$ be the standard basis of $\Bbb C^n$ over $\Bbb C$, so $e_i=(0,\dots,1,\dots,0)$, with the $1$ in the $i$-th position. Try to show that $\{e_1,ie_1,\dots,e_n,ie_n\}$ is a basis of $\Bbb C^n$ over $\Bbb R$.

$\endgroup$
3
  • $\begingroup$ Yes,I also thought the same .Now it is clear. $\endgroup$ May 6, 2017 at 14:38
  • $\begingroup$ If you consider $C^2$ as a vector space over C then basis can be {(1, 0), (01) }. I just want to know if a basis can be {(1, i), (i, 1) }.. I am just having a doubt the scalars need to be from the field but the numbers which come inside the basis should also be from the same field or not. Like say if you would take C^2 as a field over R then a basis vectors are (1, 0), (0, i) and so on. But field is R yet an "i" is coming in (0, i)... Could you please help in these 2 particular doubts. $\endgroup$
    – Shashaank
    Sep 23, 2020 at 16:38
  • $\begingroup$ Yes, $\{(1,i),(i,1)\}$ is a basis for $\Bbb C^2$ as a $\Bbb C$-vector space. And yes $\{(1,0),(0,1),(i,0),(0,i)\}$ is a basis of $\Bbb C^2$ over $\Bbb R$, it's not an issue that the $i's$ are involved. For comparison, you may have seen the example of the $\Bbb R$-vector space given by polynomials of degree $\le n$ for some $n$, say $n=2$. Then a basis is given by $\{1,x,x^2\}$; it is not an issue that $x\notin\Bbb R$, it's just an element of the abstract vector space and so it also is with $i\in\Bbb C$. $\endgroup$ Sep 25, 2020 at 6:02

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.