Let $V$ be a finite-dimensional (with dimension $n$) normed vector space. I want to show that $V$ is a smooth manifold. Firstly is following correct?
Firstly, $V$ is a topological space where the topology is naturally induced by the norm on $V$, in fact it doesn't matter what norm we choose, the topology will be the same regardless. Let $V$ have a basis ${v_1,...,v_n}$ and let $\varphi :V \rightarrow \mathbb{R}^{n} $ be the canonical linear isomorphism which maps each vector $v \in V$ to its basis representation. It is a homeomorphism because it is continuous and bijective. So not only are $V$ and $\mathbb{R}^{n} $ isomorphic as vector spaces, they are also homeomorphic as topological spaces. This shows that $V$ is locally Euclidean, second-countable, and Hausdorff, because all three are topological properties.
The hard part is showing its smooth. How is this done?
Hint: Consider any two linear isomorphisms $\varphi,\varphi'\colon V\to\mathbb R^n$. What can you say about the transition map $\varphi'\circ\varphi^{-1}$?