=== Sufficient Condition === Let $\mathcal L$ be an affine transformation. Let $L$ be the tangent map. Let $r \in \mathcal E$ be any point. Then by definition we have: :$\lambda p + \left({1 - \lambda}\right) q = r + \lambda \vec{r p} + \left({1 - \lambda}\right) \vec{r q}$ Thus we find: {{begin-eqn}} {{eqn | l = \mathcal L \left({\lambda p + \left({1 - \lambda}\right) q}\right)       | r = \mathcal L \left({r}\right) + L \left({\lambda p + \left({1 - \lambda}\right) q}\right)       | c = Definition of Affine Transformation }} {{eqn | r = \mathcal L \left({r}\right) + \lambda L \left({p}\right) + \left({1 - \lambda}\right) L \left({q}\right)       | c = since $L$ is linear }} {{eqn | r = \lambda \mathcal L \left({p}\right) + \left({1 - \lambda}\right) \mathcal L \left({q}\right)       | c = Definition of Barycenter }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === Suppose that for all points $p, q \in \mathcal E$ and all $\lambda \in \R$: :$\mathcal L \left({\lambda p + \left({1 - \lambda}\right) q}\right) = \lambda \mathcal L \left({p}\right) + \left({1 - \lambda}\right) \mathcal L \left({q}\right)$ Let $E$ be the difference space of $\mathcal E$. Fix a point $p \in \mathcal E$, and define for all $u \in E$: :$L\left(u\right) = \mathcal L\left(p + u\right) - \mathcal L\left(p\right)$ Let $q = p + u$. Then: :$\mathcal L \left({q}\right) = \mathcal L \left({p}\right) + L \left({u}\right)$ So to show that $\mathcal L$ is affine, we are required to prove that $L$ is linear. That is, we want to show that for all $\lambda \in k$ and all $u, v \in E$: :$L \left({\lambda u}\right) = \lambda L \left({u}\right)$ and: :$L \left({u + v}\right) = L \left({u}\right) + L \left({v}\right)$ First of all: {{begin-eqn}} {{eqn | l = L \left({\lambda u}\right)       | r = \mathcal L \left({p + \lambda u}\right) - \mathcal L \left({p}\right)       | c = Definition of $L$ }} {{eqn | r = \mathcal L \left({\left({1 - \lambda}\right) + \lambda \left({p + u}\right)}\right)       | c = Definition of Barycenter }} {{eqn | r = \left({1 - \lambda}\right) \mathcal L \left({p}\right) + \lambda \mathcal L \left({p + u}\right) - \mathcal L \left({p}\right)       | c = By Hypothesis on $\mathcal L$ }} {{eqn | r = \lambda \left({\mathcal L \left({p + u}\right) - \mathcal L \left({p}\right)}\right)       | c =  }} {{eqn | r = \lambda L \left({u}\right)       | c = Definition of $L$ }} {{end-eqn}} Now it is to be shown that :$L \left({u + v}\right) = L \left({u}\right) + \left({v}\right)$ First: :$p + u + v = \dfrac 1 2 \left({p + 2 u}\right) + \dfrac 1 2 \left({p + 2 v}\right)$ Now: {{begin-eqn}} {{eqn | l = \mathcal L \left({p + u + v}\right)       | r = \mathcal L \left({\frac 1 2 \left({p + 2 u}\right) + \frac 1 2 \left({p + 2 v}\right)}\right)       | c =  }} {{eqn | r = \frac 1 2 \mathcal L \left({p + 2 u}\right) + \frac 1 2 \mathcal L \left({p + 2 v}\right)       | c = By Hypothesis on $\mathcal L$ }} {{eqn | r = \frac 1 2 \left({\mathcal L \left({p + 2 u}\right) - \mathcal L \left({p}\right)}\right) + \frac 1 2 \left({ \mathcal L \left({p + 2 v}\right) - \mathcal L \left({p}\right)}\right) + \mathcal L \left({p}\right)       | c =  }} {{eqn | r = \frac 1 2 L \left({2 u}\right) + \frac 1 2 L \left({2 v}\right) + \mathcal L \left({p}\right)       | c = Definition of $L$ }} {{eqn | r = L \left({u}\right) + L \left({v}\right) + \mathcal L \left({p}\right)       | c = as $L$ preserves scalar multiples }} {{end-eqn}} From the above calculation: :$L \left({u + v}\right) = \mathcal L \left({p + u + v}\right) - \mathcal L \left({p}\right) = L \left({u}\right) + L \left({v}\right)$ This shows that $L$ is linear, and therefore concludes the proof. {{Qed}} Category:Affine Geometry 9splhjisbqidaxnnz69xjkwpx8v7k1c