TITLE: If there is an into isometry from $(\mathbb{R}^m,\|\cdot\|_p)$ to $(\mathbb{R}^n, \|\cdot\|_q)$ where $m\leq n$, then $p=q$?
QUESTION [12 upvotes]: Let $p,q\in [1,\infty)$.
Note that $p,q\neq\infty$.
Let $m\geq 2$ be a natural number.
The paper Isometries of Finite-Dimensional Normed Spaces by Felix and Jesus asserts that if $(\mathbb{R}^m,\|\cdot\|_p)$ is isometric to $(\mathbb{R}^m, \|\cdot\|_q)$, then $p =q$.
I am interested in the case when they have different dimensions.
More precisely,

Let $m,n\geq 2$ be natural numbers such that $m\leq n$ and $T:(\mathbb{R}^m,\|\cdot\|_p)\to (\mathbb{R}^n, \|\cdot\|_q)$ be a linear operator (Note that the dimension of domain and codomain are different).
If $T$ is an isometry (not necessarily onto), does $p = q$?

By the paper above, if $m=n$, then we have $p=q$.
However, if $m<n$, I am not sure whether the same result holds.
If there is a reference that cites this result, it would be good if someone can provide it.

REPLY [9 votes]: In the linked paper is written that Lyubich and Vasertein [9] showed that if there exists an isometric embedding of
$(\mathbb{K}^n,\|\cdot\|_p)$ to $(\mathbb{K}^m,\|\cdot\|_q)$ where $\Bbb K=\Bbb R$ or $\Bbb K=\Bbb C$ and
$1 < p, q <\infty$ then $p =2$, $q$ is an even integer, and $m$ [I guess such minimal $m$. AR.] satisﬁes the inequality
$${n+q/2-1\choose n-1}\le m\le {n+q-1\choose n-1}.$$
For instance, it can be checked that a linear map $T:(\mathbb{R}^2,\|\cdot\|_2)\to (\mathbb{R}^3, \|\cdot\|_4)$ such that
$T(0,1)=(a,b,c)$ and $T(1,0)=(b,a,-c)$, where $b=\sqrt{\frac{2\sqrt{2}-\sqrt{6}}{6}}$, $a=(2+\sqrt{3})b$, and $c=\frac{\sqrt[4]{2}}{\sqrt{3}}$, is an isometry, because $(a,b,c)$ is a solution of a system of equations
$$\cases{
a^4+b^4+c^4=1\\ 
a^3b+b^3a-c^4=0\\
12a^2b^2+6c^4=2}.$$