Let \(S\subset\mathbb R^n\) be finite with \(0\in S\). Suppose \(\varphi:S\to S\) satisfies \[ \varphi(0)=0 \] and \[ d(\varphi(s),\varphi(t))=d(s,t) \] for all \(s,t\in S\), where \(d\) is the Euclidean metric. Prove that there is a linear map \[ F:\mathbb R^n\to\mathbb R^n \] whose restriction to \(S\) is \(\varphi\).
Solution
Write \(\langle\cdot,\cdot\rangle\) for the Euclidean inner product.
1. For all \(s,t\in S\),\[ \langle\varphi(s),\varphi(t)\rangle = \langle s,t\rangle. \]
Because \(\varphi(0)=0\) and distances are preserved, \[ \norm{\varphi(s)} =d(\varphi(s),0) =d(s,0) =\norm{s}. \] Using \[ 2\langle u,v\rangle =\norm u^2+\norm v^2-\norm{u-v}^2 \] and the preservation of the distance between \(s\) and \(t\) gives the claim.
2. Every linear relation among elements of \(S\) is carried to thesame linear relation among their images.
Suppose \[ \sum_{j=1}^r a_js_j=0, \qquad s_j\in S. \] By step 1, \[ \begin{aligned} \norm{\sum_{j=1}^r a_j\varphi(s_j)}^2 &= \sum_{i,j=1}^r a_i a_j \langle\varphi(s_i),\varphi(s_j)\rangle\\ &= \sum_{i,j=1}^r a_i a_j \langle s_i,s_j\rangle\\ &= \norm{\sum_{j=1}^r a_js_j}^2 =0. \end{aligned} \] Hence \[ \sum_{j=1}^r a_j\varphi(s_j)=0. \]
3. There is a well-defined linear map\[ T:\operatorname{span}(S)\longrightarrow\RR^n \]such that \(T(s)=\varphi(s)\) for every \(s\in S\).
For \[ v=\sum_{j=1}^r a_js_j \] define \[ T(v)\coloneqq\sum_{j=1}^r a_j\varphi(s_j). \] Step 2 shows that two representations of \(v\) give the same value, so \(T\) is well-defined. Its linearity is immediate from the definition.
4. The map \(T\) extends to a linear map\[ F:\RR^n\longrightarrow\RR^n. \]
Choose a complementary subspace \[ \RR^n=\operatorname{span}(S)\oplus U. \] Define \[ F(v+u)\coloneqq T(v) \qquad (v\in\operatorname{span}(S),\ u\in U). \] The direct-sum decomposition makes this well-defined and linear. By step 3, for every \(s\in S\), \[ F(s)=T(s)=\varphi(s). \]
5. Q.E.D.
Step 4 provides the required linear extension.