Let $u_0,\dots,u_n$ be vectors in $\mathbb{R}^n$ such that $u_1-u_0,\dots,u_n-u_0$ are linearly independent and similarly let $v_0,\dots,v_n$ be vectors in $\mathbb{R}^n$ such that $v_1-v_0,\dots,v_n-v_0$ are linearly independent. Further suppose that $$ |u_i-u_j| = |v_i-v_j| \: \text{for all} \: i,j. $$

I want to show that there is a unique Euclidean isometry $\phi$ which takes $u_1,\dots,u_n$ to $v_1,\dots,v_n$ respectively.

The only idea I've had is to set $\phi(u_i-u_0) = v_i-v_0$ for $i=1,\dots,n$ and then extend linearly to get a linear isomorphism. But this needn't be distance preserving - if there is to be a unique isometry, I should need to map $u_i-u_0$ to a unique ordering of the $v_i-v_0$ (geometrically, e.g. in $\mathbb{R}^2$, fixing particular vertices $u_0$ and $v_0$ of each triangle, there should be only one way of matching up the pairs of edges coming out from those vertices). But I cannot see any convenient method of finding that ordering, or a method of defining the isometry which is order independent.

EDIT: The stricken through text is incorrect, for each choice of ordering we get an isometry, it is unique once we fix an ordering and a fixed one is given by the question statement.

Thank you in advance.

  • $\begingroup$ Consider $u_0 = v_0 = \mathbf{0}$, and let $u_i = v_i = \hat{e}_i$, that is let the sets of linearly independent vectors be the standard basis. Then there are $n!$ distinct Euclidean isometries among the vectors. (Half that many if orientation preservation was also intended.) So uniqueness is going to be very hard to guarantee in general. $\endgroup$ Jul 10, 2014 at 5:32
  • $\begingroup$ @EricTowers: However, the map $\phi$ above is unique, and in your example is just the identity map. $\endgroup$
    – copper.hat
    Jul 10, 2014 at 5:38
  • $\begingroup$ Your condition $|u_i-u_j| = |v_i-v_j|, \forall i,j$ implies that you already know that $u_i \leftrightarrow v_i$ is the correct matching -- that they're already sorted in the right order. Perhaps you mean that $\{|u_i-u_j| \mid i,j \in [1,n]\} = \{|v_i-v_j| \mid i,j \in [1,n]\}$ as multisets? $\endgroup$ Jul 10, 2014 at 5:38
  • $\begingroup$ @copper.hat: Yes. The apparent uncertainty in how to proceed implies some degree of freedom, but the written constraints imply none. Consequently, performing the experiment of putting this dissonance in sharp relief is useful. $\endgroup$ Jul 10, 2014 at 5:43
  • $\begingroup$ @EricTowers I've just realised your point, the correct way to match up the $u_i$'s and $v_i$'s is given by the specified ordering and distance constraint. Thanks. $\endgroup$
    – user163602
    Jul 10, 2014 at 5:47

1 Answer 1


First deal with existence:

Let $x_i = u_i -u_0, y_i = v_i-v_0$, and define the linear $Q x_i = y_i$. We have $\|x_i\| = \|Q x_i \|$, and $\|x_i-x_j\| = \|Q (x_i-x_j) \|$ for $i,j =1,...,n$.

Since $\|x_i-x_j\|^2=\|x_i\|^2+\|x_i\|^2-2 \langle x_i, x_j \rangle $, we see that $\langle x_i, x_j \rangle = { 1\over 2} \left( \|x_i\|^2+\|x_i\|^2- \|x_i-x_j\|^2\right)$, from which it follows that $\langle x_i, x_j \rangle = \langle Qx_i, Qx_j \rangle $ for $i,j =1,...,n$, and hence $Q^TQ = I$, so $Q$ is orthogonal.

Now define $\phi(u) = v_0 + Q(u-u_0)$, which is easily seen to be affine and distance preserving.

Now uniqueness:

Suppose $\eta$ is an isometry and $\eta(u_i) = v_i$ for $i,j =0,...,n$. We have $\|\eta(u_i)-\eta(u_j)\|=\|u_i-u_j\|$. Define $A(x)= \eta(x+u_0)-\eta(u_0)$. A little work shows that $A$ preserves the inner product and that $A$ is linear. Consequently, we have $\eta(u) = v_0+A(u-u_0)$, and we see that $\eta$ is affine. Since $\eta$ is affine and $u_0,...,u_n$ are affinely independent, it follows that $\eta = \phi$, hence it is unique.

  • $\begingroup$ Thanks a lot :) Just what I needed, just one point: I said in the question that we need a unique isometry taking $u_1,\dots,u_n$ to $v_1,\dots,v_n$ excluding the $u_0$ and $v_0$ because the place I saw this question at excluded them but in fact it is necessary to include them (e.g. otherwise for two equilateral triangles adjoined along a common edge we could take the identity map or the flip across the common edge, both of which preserve the vertices of the common edge) $\endgroup$
    – user163602
    Jul 10, 2014 at 10:13
  • $\begingroup$ Sorry that I can't upvote your answer, I don't have enough reputation it seems. $\endgroup$
    – user163602
    Jul 10, 2014 at 10:14
  • $\begingroup$ No problem at all, glad to be of help. $\endgroup$
    – copper.hat
    Jul 10, 2014 at 14:38

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