# Proving that every finite-dimensional normed space has the same norm as the euclidean n-dimensional space

I have a problem where I have to show that every $$n$$-dimensional normed space $$E$$ has the same norm as the euclidean space $$E_n$$.

Here's what I've got:

Since $$E$$ is $$n$$-dimensional then for the basis $$(e_1, ..., e_n)$$ of $$E$$ there's a unique representation for every $$x \in E$$ i.e. $$x=\sum_{i=1}^{n}{\lambda_i e_i}$$.

First I show for every $$x \in E$$ the function $$g(x)=(\lambda_1, ... \lambda_n) \in E_n$$ is an isomorphism.

Then I show that $$\lVert x\rVert_{E} \le C\lVert v\rVert_{E_n}$$ and thus the image is continuous.

After that I show that$$\lVert x\rVert_{E_n} \le C_2\lVert v\rVert_{E}$$ and thus the image is homeomorphic.

I also conclude that $$m\lVert v\rVert_{E_n} \le \lVert x\rVert_{E} \le M\lVert v\rVert_{E_n}$$ for every $$x$$.

I'm not clear on how to conclude that the norms are equivalent. How can I use the last inequality to show that?

What you want to prove is not true. For instance take $$E_2$$, and consider the norms $$\|(a,b)\|_1=|a|+|b|,\ \ \|(a,b)\|_\infty=\max\{|a|,|b|\}.$$ Then $$\|(1,0)\|_1\ne\|(1,1)\|_1,$$ while $$\|(1,0)\|_\infty=\|(1,1)\|_\infty.$$