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Let $A,A'$ two affine subspaces of a finite Euclidean Vectorspace $V$. Let $p,p'$ two points, such that $d(A,p)=d(A',p')$. $\dim(A)=\dim(A')$

I would like to show that there exists a movement $\alpha:V->V$ such that $\alpha(A)=A'$ and $\alpha(p)=p'$

I only know, if $\alpha$ is a movement, then the following properties hold:

(1) $d(\alpha(v),\alpha(w))=d(v,w)$

(2) $\alpha$ is affine

(3) $\alpha$ affine and for a basis $a_1,..,a_n$ it holds that $d(\alpha(a_i),\alpha(a_j))=d(a_i,a_j)$

How can I use this to prove the statement above ?

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What is $d$? Euclidean distance? – user1551 Dec 2 '12 at 21:33
yes, d is the euclidean – Voyage Dec 2 '12 at 21:33
up vote 0 down vote accepted

Affine functions are just compostions of translations and linear transformations. So you may just adhere some translations to an orthogonal transformation to achieve your goal.

Specifically, take an orthonormal basis $\{u_1,\ldots,u_m\}\ (m=\dim A)$ of $A$ and an orthonormal basis $\{v_1,\ldots,v_m\}$ of $A'$. Complete them to two orthonormal bases $\{u_1,\ldots,u_n\}\ (n=\dim V)$ and $\{v_1,\ldots,v_n\}$ of $V$. For any $x\in V$, define $\alpha(x) = p' + \sum_i \lambda_iv_i$ whenever $x-p=\sum_i \lambda_iu_i$. Now you are done.

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