Prove the following lemma: Lemma 0.3 Let $(\mathbb{R}^n,d)$ be a metric space. If $\{a_0,…,a_n\}$ and $\{b_0,…,b_n\}$ are affinely independent in $\mathbb{R}^n$ with $d(a_i,a_j)=d(b_i,b_j)$ for $0\leq i, j \leq n$, then there exists the unique isometry $f: \mathbb{R}^n \to \mathbb{R}^n $ s.t. $f(a_i)=b_i$ for all $i=0,…,n$

Prove the following lemma: Lemma 0.3 Let $(\mathbb{R}^n,d)$ be a metric space. If $\{a_0,…,a_n\}$ and $\{b_0,…,b_n\}$ are affinely independent in $\mathbb{R}^n$ with $d(a_i,a_j)=d(b_i,b_j)$ for $0\leq i, j \leq n$, then there exists the unique isometry $f: \mathbb{R}^n \to \mathbb{R}^n $ s.t. $f(a_i)=b_i$ for all $i=0,…,n$

sol) Since $\{a_i-a_0 : i=1,2,…,n\}$ and $\{b_i-b_0 : i =1,2,…n\}$ are basis’, there exists matrix $A$ such that $A(a_i-a_0)=b_i-b_0$.
Let $f(x) = A(x-a_0)+b_0$, then $f(a_i)=b_i$ for all $i=0,1,2,…,n$. By assumptions $\lVert a_i-a_j \rVert^2=\lVert b_i-b_j\rVert^2 = \lVert b_i-b_0 +b_0-b_i\rVert^2 = \lVert b_i-b_0\rVert^2 +\lVert b_j-b_0\rVert^2 -2 (b_i-b_0)^T(b_j-b_0) =\lVert a_i-a_0\rVert^2 +\lVert a_j-a_0\rVert^2 -2 (b_i-b_0)^T(b_j-b_0)$, therefore, $(b_i-b_0)^T(b_j-b_0) = (a_i-a_0)^T(a_j-a_0)$ for all $0\leq i, j \leq n$. Let $x,y \in \mathbb{R}^n$, then $x=\sum_{i=1}^n c_i (a_i-a_0)$ and $y=\sum_{j=1}^n d_j (a_j-a_0)$ for some $c_i, d_j$. $d(x,y)^2 = \sum_{i=1}^n\sum_{j=1}^n c_i d_j (a_i-a_0)^T(a_j-a_0)$. $d(f(x),f(y))^2=\sum_{i=1}^n \sum_{j=1}^n c_i d_j (a_i-a_0)^TA^TA(a_j-a_0)=\sum_{i=1}^n \sum_{j=1}^n c_i d_j (b_i-b_0)^T(b_j-a_0)=\sum_{i=1}^n \sum_{j=1}^n c_i d_j (a_i-a_0)(a_j-a_0)=d(x,y)^2$.
Therefore, $f(x)$ is an isometry. Since every isometry has form $Mx+m$ for $M \in O(n)$ and $m \in

 

출처- 기하학1 (GIST-강현석교수님 강의) Homework 1 문제 중 일부



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