Theorem: Any two norms on a finite-dimensional vector space $V$ are equivalent.
My book begins the proof by noting that we can identify $V$ with $\mathbb R^n$ for some $n\in\mathbb N$, and therefore we only need to prove the theorem for $\mathbb R^n$. However, I don't see why this is true. I know that there is a bijection between $V$ and $\mathbb R^n$ for some $n\in\mathbb N$, or equivalently, there is a bijection between a basis in $\mathbb R^n$ and a basis in $V$. But how to proceed from there on?
I've read somewhere that an isomorphism is basically just the relabelling of the elements... but how do I know that things like inequalities and the like stay preserved between two identified elements? Things like "the vector space structure is preserved" sounds slightly vague to me, so is there something more rigorous to say?
Let $(V, \Vert \cdot \Vert_V$) be a finite-dimensional $\mathbb{R}$-vector space. Then there exists an isomorphism of vector spaces (i.e. a bijective $\mathbb{R}$-linear map) $\phi: V \rightarrow \mathbb{R}^n$ for some $n\in \mathbb{N}$. As you wrote above, you want this map to preserve the norm. So you want to have a norm $\Vert \cdot \Vert_{\phi}$ on $\mathbb{R}^n$ such that for every $v\in V$ holds
$$ \Vert \phi(v)\Vert_{\phi} = \Vert v \Vert_{V}.$$
In fact this norm is (check that it really is a norm)
$$ \Vert x \Vert_{\phi} = \Vert \phi^{-1}(x) \Vert_V.$$
Let now $\Vert \cdot \Vert_1$, $\Vert \cdot \Vert_2$ be two norms on $V$ and $\Vert \cdot \Vert_{\phi, 1}$ and $\Vert \cdot \Vert_{\phi,2}$ be the respective norms on $\mathbb{R}^n$. If you know that all norms on $\mathbb{R}^n$ are equivalent, then you have
$$ c^{-1} \Vert x \Vert_{\phi, 1} \leq \Vert x \Vert_{\phi, 2} \leq c \Vert x \Vert_{\phi,1}.$$
For $x=\phi(v)$ you get
$$ c^{-1} \Vert v \Vert_{1} = c^{-1} \Vert \phi(v) \Vert_{\phi, 1} \leq \Vert \phi(v) \Vert_{\phi, 2} = \Vert v \Vert_2 = \Vert \phi(v) \Vert_{\phi, 2} \leq c \Vert \phi(v) \Vert_{\phi,1} = c \Vert v \Vert_1. $$