It certainly is easy to find as many proofs as you want for the equivalence of any two norms in a finite dimensional vector space.
I was wondering what is the relationship between homeomorphisms and the equivalence of any two norms in a finite dimensional vector space.
This question urged to me since I am doing a worksheet that asked to to show that the mapping $T:(\Bbb K^n, \| \cdot \|_\infty) \rightarrow (X,\| \cdot \|)$ is a homeomorphism (more details can be seen here) and afterwards to prove that any two norms on a finite dimensional space $X$ are equivalent, based on this result.
Formal exercise. Show that in a finite-dimensional vector space $X$, every two norms are equivalent, using this result.
My attempt of resolution. Let $X$ be a finite-dimensional vector space and let $\|.\|_\beta$ and $\| \cdot \|_\gamma$ be two norms defined on $X$. Since $X$ is finite-dimensional, let $\{e_1\dots,e_n\}$ be a basis for $X$ and define a third norm on $X$ as follows:
Given $x \in X$ we can write: $$ x = \sum_{j=1}^n \alpha_je_j,$$
for some scalars $\alpha_j \in \Bbb K, \forall j \in \{1,\dots,n\}.$ Set
$$ \| x \|_\infty = \max\{|\alpha_j|: j \in \{1,\dots,n\}\}$$
This is the exact same norm used for $\Bbb K^n$ in the previous result I linked. By transitivity, it suffices to show that both $\| \cdot \|_\beta$ and $\| \cdot \|_\gamma$ are equivalent to $\| \cdot \|_\infty$. This is easy to do (I will omit it here) using norm properties (triangle inequality and homogeneity).
Now, one could finish this proof in the usual way – i.e. – considering the unit sphere $S = \{x \in X : \| x \|_\infty = 1\}$ and applying some topological arguments.
I just don't see where I am applying the result I linked… Does it allow me to make some kind of shortcut? I am not really familiar with homeomorphisms so maybe this is way easier than what I think. So, basically, what I want to do is to use the linked result to somehow help me proving this. I believe that the infinite norm definition has to stick around, but I don't see how to relate it with the homeomorphism.
Thanks for any help in advance.
Best Answer
The result you want to use is: for any $n$-dimensional $\Bbb K$-vector space $X$ ($\Bbb K=\Bbb R$ or $\Bbb C$) any linear bijection $\Bbb K^n\to X$ is an homeomorphism $(\Bbb{K}^n, \| \cdot \|_\infty) \rightarrow (X, \| \cdot \|),$ where $\| \cdot \|$ is an arbitrary norm on $X.$
Let $T:\Bbb K^n\to X$ be a linear bijection. Then, using this result twice, $T:(\Bbb{K}^n, \| \cdot \|_\infty)\to(X, \| \cdot \|_\beta)$ and $T:(\Bbb{K}^n, \| \cdot \|_\infty)\to(X, \| \cdot \|_\gamma)$ ar both homeomorphims hence by composition, $\mathrm{id}_X:(X, \| \cdot \|_\beta)\to(X, \| \cdot \|_\gamma)$ is a homeomorphism, i.e. $ \|_\gamma$ and $\cdot \|_\beta$ are equivalent.