dimension of $\mathbb{R}^n$ over $\mathbb{C}$

What can i say about dimension of $\mathbb{R}^n$ over $\mathbb{C}$ ??

I know dimension of $\mathbb{C}$ over $\mathbb{R}$ is 2 where basis is $\{1, i\}$ but what about $dim_{\mathbb{C}}{\mathbb{R}}$

• When you say that $\mathbb{C}$ has dimension 2 over $\mathbb{R}$, what you are really saying is that the vector space $(\mathbb{C}, +, \cdot,\mathbb{R})$ has dimension two. You are asking for the dimension of something like $(\mathbb{R}, +', \cdot',\mathbb{C})$, so who are $+'$ and $\cdot'$? – Smurf Aug 13 '17 at 22:36

$\mathbb{R}^n$ is not a $\mathbb{C}$-vector space in itself. $i(1,0,0)\notin\mathbb{R}^3$, so it fails to be closed. But, if you have an even-dimensional vector space $\mathbb{R}^{2n}$, you can manufacture an isomorphism (of real vector spaces) to $\mathbb{C}^n$. Can you construct it?
• how i construct if i have even dimensional vector space $\mathbb{R}^{2n}$ , elaborate plz.. – RAM_3R Aug 13 '17 at 22:35
• You can identify $\mathbb{C}$ with $\mathbb{R}^2$ as real vector spaces via $a+ bi \mapsto (a,b)$. Hence you can identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ via $$(a_1 + b_1 i, ... , a_n + b_ni) = (a_1, b_1, ... , a_n, b_n)$$ – D_S Aug 14 '17 at 0:34
• so dimension of $\mathbb{R}^{2n}$ ober $\mathbb{C}$ is $n$. – RAM_3R Aug 14 '17 at 19:42
It makes no sense. If $V$ is a vector space over $\Bbb C$ then, $zv$ must belong to $V$ for every $z\in \Bbb C$ and $v\in V$, but for $x\in \Bbb R^n\setminus\{0\}$, $ix\notin \Bbb R^n$.
• Or in other words, $\Bbb R^n$ is not a $\Bbb C$-space. So its dimension is not defined. – Lubin Aug 13 '17 at 22:31
• $\mathbb{R}$ can be a $\mathbb{C}$-space using appropriate definitions for the operations. I know this seems very pedantic but I think is important to point out that his question makes no sense until he defines properly the vector space he wants to study. – Smurf Aug 13 '17 at 22:39