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

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 $$v_1,v_2$$.

• Send your basis to a basis. That‘s induces an isomorphism. – Qi Zhu Sep 27 '20 at 12:44
• +1 for the edit. That's true the inner product gives you a simple method to "pick" the suitable coefficients – Peter Melech Sep 27 '20 at 17:52

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.
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 $$\phi^{-1}\circ\phi=id_{\mathbb{R}^n}$$.