Every finite dimensional normed (real) space is isometrically isomorphic to $\mathbf R^n$

Question

Let $(E,\|•\|)$ be a n-dimensional real normed space. So there exists $T:E\longrightarrow\mathbf{R}^n$ linear isomorphism that for a well known result is also a topological homeomorphism. How could I show that there exist an isometric isomorphism $T:(E,||•||)\longrightarrow(\mathbf{R}^n,\|•\|_k)$ where $\|•\|_k$ is anyone norm?

It is simple to show that there exists a norm $\|•\|_\infty$ such that $T:(E,\|•\|_\infty)\longrightarrow({R}^n,\|•\|_\infty)$, but I want this result with the original norm, not with another one.

Answer 1

This is not true. If $X=\mathbb R^{2}$ and the two norms are the usual norm and the norm $\|(x,y)\|=\max \{|x|,|y|\}$ there can be no isometric isomorphism. This is because $X$ is a Hilbert space with the usual norm so parallelogram identity holds. If there is an isometric isomorphism then parallelogram identity would hold in the other norm also, which is not true.