I have one definition and one proposition, from which I formulated a characterization of Euclidean motion. These are:
Definition: A map $T: \mathbb{E}^n \to \mathbb{E}^n$ is an affine transformation if it is given in a coordinate system by $T(\mathbf{x})=A\mathbf{x}+\mathbf{b}$, where $A=(a_{ij})$ is an $n \times n$ matrix with nonzero determinant and $\mathbf{b}=(b_i)$ a vector; in more detail, $$ \mathbf{x} = (x_i) \mapsto \mathbf{y} = \left( \sum_{j=1}^n a_{ij}x_j+b_i \right) \text{,} \quad \text{ or } \quad \left( \begin{matrix} x_1 \\ \vdots \\ x_n \end{matrix} \right) \mapsto A \left( \begin{matrix} x_1 \\ \vdots \\ x_n \end{matrix} \right) + \left( \begin{matrix} b_1 \\ \vdots \\ b_n \end{matrix} \right) \text{.} \tag{9} $$ (original picture)
Proposition : Let $f$ be an affine transformation from $R^n$ to $R^n$ given by $f(x)=Ax+b$. Then $f$ is a Euclidean motion if and only if $A^TA=I$.
We also know that every Euclidean motion is an affine transformation. Does it follow that every Euclidean motion $f$ can be associated with the equation $f(x)=Ax+b$ for some $A$ and $b$, where $A$ admits $A^TA=I$?
Assuming that my intuition is correct, I used this fact(?) and the following Lemma in proving the following proposition (which states that Euclidean motions send Euclidean frames to Euclidean frames, whose definition is included below), is it correct?
Definition : An $\mathbf{Euclidean \; frame}$ of $R^n$ is a set of $n+1$ points $Q_0,Q_1,...,Q_n$ such that
- $d(Q_0,Q_i)=1 \; \forall i=1,...,n$
- $(Q_i-Q_0)\cdot(Q_j-Q_0)=0 \; \forall i\not=j$, where "$\cdot$" denotes the dot product on $R^n$.
Lemma : Let $T:R^n \to {R^n}$ be a linear operator on $R^n$ and assume that $A$ represents $T$ in the standard basis for $R^n$. Then the followings are equivalent:
- $T$ preserves the dot product on $R^n$.
- $A^TA=I$.
Proposition : Let $f:R^n \to R^n$ be a Euclidean motion. Let {$Q_0,Q_1,...,Q_n$} be a Euclidean frame of $R^n$. Then {$f(Q_0),f(Q_1),...,f(Q_n)$} is a Euclidean frame.
proof : Since f is a motion, f preserves distances. Thus, $\forall i=1,...,n$ we have $d(f(Q_0),f(Q_i))=d(Q_0,Q_i)=1$.
Since f is a motion, it is also an affine transformation, so $\forall x\in{R^n} f(x)=Ax+b$, where $A\in{M_{nxn}(R)}$ is such that $A^TA=I$, and $b\in{M_{nx1}(R)}$.
Define $T:R^n \to {R^n}$ by $T(x)=Ax$, i.e, T is represented by $A$ in the standard basis for $R^n$. Then for $i\not=j$, we have
$(f(Q_i)-f(Q_0))\cdot(f(Q_j)-f(Q_0))=((AQ_i+b)-(AQ_0+b))\cdot((AQ_j+b)-(AQ_0+b))=(A(Q_i-Q_0))\cdot(A(Q_j-Q_0))=(T(Q_i-Q_0))\cdot(T(Q_j-Q_0))$.
Since $A$ is the representing matrix for $T$ in the standard basis and $A^TA=I$, by the preceding Lemma, $T$ preserves the dot product, so finally we have
$(T(Q_i-Q_0))\cdot(T(Q_j-Q_0))=(Q_i-Q_0)\cdot(Q_j-Q_0)=0$.