Can we choose norms such that $\mathbb R^m\otimes\mathbb R^n\cong\mathbb R^{mn}$ is an isometry?

Question

$\mathbb R^m\otimes\mathbb R^n$ and $\mathbb R^{mn}$ are (algebraically) isomorphic. Is there a norm on $\mathbb R^{mn}$ such that this isomorphism is an isometry when $\mathbb R^m\otimes\mathbb R^n$ is equipped with the injective or projective tensor norm?

Answer 1

Yes, because these are finite-dimensional Hilbert spaces, the (Hilbert-Schmidt) norm on the tensor product is the standard one on $\mathbb R^{mn}$. Checking on a basis, or checking more abstractly, both succeed.

EDIT (in response to query from the original poster): what I would call the Hilbert-Schmidt norm on the algebraic tensor product of two Hilbert spaces $V,W$ is completely specified by $$ \langle v\otimes w,\;v'\otimes w'\rangle \;=\; \langle v,v'\rangle_V\cdot \langle w,w'\rangle_W $$ (This extends bilinearly to completely define the thing...)

Answer 2

Hint: if $A$ is isomorphic to $B$ as a vector space via some linear transformation $f: A \to B$ and if $B$ is equipped with a norm, there is a simple, obvious and unique way of equipping $A$ with a norm that makes $f$ an isometry.