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In the study of finite dimensional vector spaces, one can prove the following theorem:

Let $\beta$ be an (ordered) basis for an $n$ dimensional vector space, $V$, defined over field $F$. Let \begin{equation*} \phi_{\beta} : V \longrightarrow F^n \quad \text{such that} \quad \phi_{\beta}(\mathbf{u}) = [\mathbf{u}]_{\beta} \quad \text{for every} \quad \mathbf{u} \in V, \end{equation*} where $[\mathbf{u}]_{\beta}$ is the coordinate vector of $\mathbf{u}$ relative to the basis $\beta$. Prove that $\phi_{\beta}$ is an isomorphism.

I'm not sure how to use this theorem to prove that $\mathbb{C}$ and $\mathbb{R}^2$ are isomorophic.

My concern: the underlying fields which define vector spaces are different. $\mathbb{C}$ can be thought of as a vector space defined over itself; whereas $\mathbb{R}^2$ can be thought of as a vector space defined over $\mathbb{R}$. How can one then show the existence of a bijective linear transformation between the two spaces? Should we keep the field to be the same, $\mathbb{R}$, and introduce the $i$ by hand?

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    You're right: $\dim_{\mathbb C}(\mathbb C)=1$, but $\dim_{\mathbb R}(\mathbb C)=2$. – 2017-02-09
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    In this context, we're meant to think of $\Bbb C$ as a vector space over $\Bbb R$. That is, we "forget" that you can multiply two elements of $\Bbb C$. – 2017-02-09
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    Isomorphism between vector spaces is defined only for vector spaces over the same field. But, just as $\mathbb C$ can be thought of as a vector space over itself, it can also be thought of as a vector space over $\mathbb R$. – 2017-02-09
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    $\Bbb R^2$ is **not** isomorphic to $\Bbb C$ as vector spaces because the fields are different. – 2017-02-09

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Consider the elementary basis of $\mathbb{R}^2$, $B=\{e_1,e_2\}$,

And the isomorphism (of vector spaces) $\phi:\mathbb{R}^2 \to \mathbb{C}$, given by $\phi((x,y))=x+iy \in \mathbb{C}$,

It sends $e_1$ to $1+i0=1$, and $e_2$ to $0+i1=i$,

Now it is easy to see that $\{1,i\}$ is indeed a basis for the complex plane (any $z=a+ib$ is a linear combination of the basis elements, using scalars from $\mathbb{R}$ - $a,b\in \mathbb{R}$).

Regarding your theorem: $\phi_\beta:\mathbb{C}\to\mathbb{R}^2$, is simply $\phi^{-1}$.

$\phi_\beta(z)=\phi_\beta(a+ib)=(a,b)\in \mathbb{R}$ where $(a,b)$ is your $[u]_\beta$

hope it helps...