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$V$ is a finite euclidean vectorspace and $\sigma:V->V$ is a motion, this means that $d(\sigma(a_i),\sigma(a_j))=d(a_i,a_j)$ for an affine coordinate system $a_0,...,a_n$

I know the following two things: $\sigma^2=id_V$ and $A:=\{v\in V:\sigma(v)=v\}$

I already proved that A is an affine subspace, but I would like to know why A is non-empty.

My second question is, why $\sigma$ is identical to a reflection $\sigma_A(v):=\sigma_{V_A}(v-a)+a$ where $\sigma_{V_A}$ is an orthogonal reflection on $V_A$ and $a\in A$ is arbitrary.

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up vote 2 down vote accepted

I assume that $\sigma$ is an affine map, and $V$ is endowed by a scalar product.

In this case, $\sigma(x)=Mx+b$ for some $M:V\to V$ linear map and $b=\sigma(0)\in V$. Since $\sigma(b)=\sigma(\sigma(0))=0$, we have $Mb+b=0$, so the midpoint of $0$ and $b$, $b/2$ will be in $A$. (Check it!)

(Actually, over a field of characteristic $2$, we can have that $A$ is empty, for example by any translation..)

For the second question, reduce it to the case when $0\in A$, ie. $\sigma$ is actually linear. Then, using that it keeps the distances of a basis, you could prove that it keeps all distances (keeps norm), using the following identity: $$||x-y||^2=||x||^2+||y||^2-2\langle x,y\rangle .$$ (By hypothesis, and assuming $a_0=0$ we have that $||\phi(a_i)||=||a_i||$ and $||\phi(a_i-a_j)||=||a_i-a_j||$ for all $i,j$.)

So we also have that $\sigma$ keeps the scalar product: $\sigma\in O(V)$, that is, fixing an orthonormal basis, its matrix will be an orthonormal matrix, satisfying $\sigma^2=id$, hence having $\pm 1$ as eigenvalues, ...

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Thank you very much for your answer, I already thought that affine geometry is not so interesting for most people (Recently I put on an other question relating to affine geomtry but no answer yet: Anyway, I am not sure if I understand the last part. Instead of $d(\sigma(a_i),\sigma(a_j))=d(a_i,a_j)$ you say that $d(\sigma(v_i),\sigma(v_j))=d(v_i,v_j)$ is also true, so $||a_j-a_i||=||v_j-v_i||$ ? Where can I use the parallelogramm identity $||x+y||^2+||x-y||^2=2(||x||^2+||y||^2)$ – Montaigne Nov 26 '12 at 23:07
Sorry, it is not the parallelogramma identity I thought of, but similar, I corrected – Berci Nov 28 '12 at 10:52

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