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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 $E=k^n$?.

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

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