# Every metric on a finite-dimensional vector space is equivalent

I am trying to prove the theorem that states that on every finite-dimensional vector space ($$E$$ with dimension $$n$$), every norm is equivalent. Going through my lecture notes the proof given, starts saying that we can suppose without losing generality that $$E =k^n$$.

Then it uses the fact that on $$k^n$$ unit balls are compact and you can easily get the bounds using $$||\cdot||_\infty$$ and other generic norm.

My doubt is: why can we asumme $$E=k^n$$?.

• For an isomorphism $f:E \to k^n$, consider the norm $\|x\|_E = \|f(x)\|_{k^n}$. – anomaly Oct 21 '18 at 19:20
• Oh you're right. It was really simple. Thanks – Johanna Oct 21 '18 at 23:05

Let $$v_1,\dots,v_n$$ be a basis of $$V$$. Then any $$v\in V$$ is uniquely represented by $$v=\alpha_1v_1+\dots+\alpha_nv_n$$. Then we identify $$v\in V$$ with $$(\alpha_1,\dots,\alpha_n)\in k^n$$. This is also topological homeomorphism.