Are the normed spaces $ \mathbb{R}^{n^2} $ and $ M_n(\mathbb{R}) $ isometric?

Question

Consider the spaces $ \mathbb{R}^{n^2} $ with euclidean norm and $ M_n(\mathbb{R}) $ of $n\times n$ matrices with the norm defined by $ \Vert A\Vert = \sup\limits_{\Vert x\Vert \le 1}\Vert Ax\Vert$. They are isomorphic as linear spaces, but is there also a linear isometry between them with the given norms?

Answer 1

Assuming that the operator norm is taken with respect to the Euclidean norm on $\mathbb{R}^n$, we have

$$\left\lVert \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}\right\rVert = 1 = \left\lVert \begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}\right\rVert,$$

hence

$$2\left\lVert\begin{pmatrix}1 & 0\\0&0 \end{pmatrix}\right\rVert^2 + 2 \left\lVert\begin{pmatrix} 0&0\\0&1\end{pmatrix}\right\rVert^2 = 4 \neq 2 = \left\lVert\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}\right\rVert^2 + \left\lVert \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}\right\rVert^2,$$

and we see that the parallelogram identity doesn't hold, so the operator norm is not induced by an inner product, and the two spaces are not isometrically isomorphic.

If the operator norm is taken with respect to other norms on $\mathbb{R}^n$, I am almost sure it is still true in all cases, but may need less obvious examples to establish the fact.