Let $X\subset \mathbb{R}^m$ not empty and $f: X\to\mathbb{R}^n$ an isometric immersion. Prove that there exists an isometric immersion $\varphi: \mathbb{R}^m \to\mathbb{R}^n$ such that $\varphi|_X=f$. When $X$ generates (in the sense of linear algebra) $\mathbb{R}^m$, $\varphi$ is unique. Metrics are euclidean metrics (in particular, $\mathbb{R}^m$ and $\mathbb{R}^n$ metrics comes from usual inner product).
Part of proof:
Without loss of generality we can say that $f(0)=0$ (up translations). Then $f$ satisfies $f(x+y)=f(x)+f(y)$ and $f(\lambda x)=\lambda f(x) \ \forall \lambda\in\mathbb{R}$ because $f$ preserve distances and the fact that $\mathbb{R}^n$ euclidean metric comes from an inner product, is not so hard to prove.
I need to prove the existence of $T$ linear transformation such that $T|_X=f$ (assumming $f(0)=0$). I've tried extending $f$ using values of a basis of $X$, but is not factible to prove existence of $T$.
Also, is question possible if $m>n$? or exercise is just incomplete?.
I think that if $X$ generates $\mathbb{R}^m$ the uniqueness is obvious if $T$ is linear extension.
Added: I was opened a bounty for +100 because I need to prove existence of $T$. This is the only fact that I can't be formal. If you want to explain what happend if $m>n$ better. But the priority is a formal proof of the fact of existence of $T$.