Isomorphism of a vector space with $R^n$ and the role of inner product

Question

It is often stated that a choice of a basis for a vector space over $R$ gives an isomorphism between $V$ and $R^n$. However, I am having a hard time showing this and confused over some issues. For simplicity, let's stick to a two-dimensional space $V$. Let's choose a basis $\{e_1, e_2\}$ for the vector space. Then we have the linear map $\phi: R^2 \to V; \phi(v_1,v_2)=v_1 e_1+v_2 e_2$. To prove the isomorphism between $R^2$ and $V$, I need to construct a map $\phi^{-1}: V\to R^2: \phi^{-1} \phi=Identity$. I don't see any obvious way to construct such a map. However, if there is an inner product (denoted here by ".") on $V$, we can construct such a map in the following way. Suppose $e_1,e_2$ is an orthonormal basis. We then set $\phi^{-1}: V\to R^2; \phi^{-1}(v)=(e_1.v, e_2.v)$. It is obvious that $\phi^{-1}\phi(v_1,v_2)=(v_1,v_2)$. So my question is, is it true that we need an inner product and the existence of an orthonormal basis to prove the isomorphism between $V$ and $R^2$? Or I have misunderstood something here?

Edit: I also would like to mention that this question originated from noticing the fact that although a given vector can be written as $v=v_1e_1+v_2e_2$, there is no natural way to find the coefficients $v_1,v_2$. The inner product seems to determine the values of $v_1,v_2$.

Answer 1

The inner product is not relevant to the question.

In order that $\phi\colon\mathbb{R}^2\to V$ is an isomorphism it is necessary and sufficient that it is injective and surjective (besides being linear, of course).

Since $\{e_1,e_2\}$ is a spanning set, the map is surjective.

Since $\{e_1,e_2\}$ is linearly independent, the map is injective.

Answer 2

You don't need an inner product. For $\{e_1,e_2,...,e_n\}$ to be a basis of the finite-dimensional vectorspace $V$ (by definition of a basis or as a consequence of it being a maximal linearly independent set) there exists for every vector $v\in V$ a $\textbf{unique}$ $n$-tuple $$\begin{pmatrix}v_1\\v_2\\ \cdot\\\cdot\\\cdot\\v_n\end{pmatrix}\in\mathbb{R}^n$$ such that $v=\sum_{j=1}^nv_je_j$ and you define $\phi^{-1}(v)=\begin{pmatrix}v_1\\v_2\\ \cdot\\\cdot\\\cdot\\v_n\end{pmatrix}$. Now it is easy to check that $\phi\circ\phi^{-1}=id_{V}$ and $\phi^{-1}\circ\phi=id_{\mathbb{R}^n}$.