I want to show that all $f \in Isom(\mathbb{R^n})$ can be written as $f(x) = Q(x) +v$ with $Q \in o(n) $ and $v \in \mathbb{R^n}$
This is how the proof goes: Let us set G(x) = f(x) -v
We want to show 2 things:
G is linear
G is orthogonal
I get that it is necessary to show that G is orthogonal but why to we have to show G's linearity?
The proof follows quickly from a "crucial" lemma, namely the following:
Lemma let $G$ be an isometry fixing the origin, i.e., with $G(0)=0$. Then $G$ is a linear, orthogonal map.
Proof: Using the so-called polarization formula one obtains \begin{equation} \begin{aligned} 2\langle x,y\rangle & =||x||^2+||y||^2-||x-y||^2 \\ &= ||G(x)||^2+||G(y)||^2-||G(x)-G(y)||^2 \\ & = 2\langle G(x),G(y)\rangle . \end{aligned} \end{equation} To show that $G$ is a linear map, let $(e_1,\ldots ,e_n)$ be an orthonormal basis. Then also $(G(e_1),\ldots ,G(e_n)$ is an orhonormal basis. For each $x$ we have \begin{equation} \begin{aligned} x & =\sum_{k=1}^n \langle x,e_k\rangle e_k, \\ f(x) & = \sum_{k=1}^n\langle G(x),G(e_k)\langle G(e_k) \\ & = \sum_{k=1}^n \langle x,e_k\rangle G(e_k). \end{aligned} \end{equation} This implies that $$ G\left(\sum_{k=1}^nx_ke_k\right)=\sum_{k=1}^n x_kG(x_k), $$ which says that $G$ is linear. Together we have that $G$ is a linear map which keeps invariant the scalar product, i.e., which is an element of the linear orthogonal group $O_n(\mathbb{R})$.