Let $f_1, \cdots , f_r \colon X \to Y$ $(r \ge 2)$ be affine transformations and $\alpha_1, \cdots, \alpha_r \in K$, $\sum_{i=1}^r \alpha_i = 1$. Prove that $f = \sum_{i=1}^r \alpha_i f_i : X \to Y, f(M) = \sum_{i=1}^r \alpha_i f_i(M)$ is an affine transformation.
This is quite interesting, I tried to apply different results that involve that sum of scalars but I ended nowhere and it`s a little frustrating that I study translations and projections but in the course I am following there is no passing from theoretical results towards problems. Any help would be extraordinary. Thank you.
Let $\mathbb A$ be an affine space over a $K$-vector space.
If $P_1,P_2,\ldots,P_r \in \mathbb A$, and $\alpha_1,\alpha_2,\ldots,\alpha_r$ are scalars (elements of the field $K$) such that
$$ \sum_{i=1}^r \alpha_i = 1, $$
we put
$$ \sum_{i=1}^r \alpha_iP_i := O + \sum_{i=1}^r \alpha_i(P_i-O) $$
where $O$ is any point of $\mathbb A$. We thus obtain a point which is independent of $O$.
Indeed, if $O'$ is any other point of $\mathbb A$, we have:
\begin{align} O' + \sum_{i=1}^r \alpha_i(P_i-O') &= O+(O'-O)+\sum_{i=1}^r \alpha_i\Big((P_i-O)+(O-O')\Big)=\\[1ex] &= O+(O'-O)+\sum_{i=1}^r \alpha_i(P_i-O) + \sum_{i=1}^r \alpha_i(O-O')=\\[1ex] &= O+\sum_{i=1}^r \alpha_i(P_i-O) + (O'-O) + \Big(\sum_{i=1}^r \alpha_i\Big)(O-O')=\\[1ex] &= O+\sum_{i=1}^r \alpha_i(P_i-O)+(O'-O)+(O-O') =\\[1ex] &= O+\sum_{i=1}^r \alpha_i(P_i-O),\qquad \text{q.e.d.} \end{align}
Let's go back to the main problem. The function $f\,$ is defined, as we have just seen, by
$$ f(M) = \sum_{i=1}^r \alpha_if_i(M) = O+\sum_{i=1}^r \alpha_i(f_i(M)-O), $$
where $O$ is a fixed point of $\,Y$. Consequently, if $P,Q\in X$, we have:
\begin{align} f(P)-f(Q) &= \Big(O+\sum_{i=1}^r \alpha_i(f_i(P)-O)\Big) - \Big(O+\sum_{i=1}^r \alpha_i(f_i(Q)-O)\Big) = \\[1ex] &= \sum_{i=1}^r \alpha_i\Big((f_i(P)-O)-(f_i(Q)-O)\Big) = \\[1ex] &= \sum_{i=1}^r \alpha_i(f_i(P)-f_i(Q)) = \\[1ex] &= \sum_{i=1}^r \alpha_i\,df_i(P-Q), \end{align}
where $\,df_i : \vec X\to \vec Y\,$ indicates the linear part of $f_i$ (also indicated with $\vec f_i\,$, or $f'_i, \ldots$), and $\vec X, \vec Y$ indicates the vector spaces associated to $X, Y$, respectively.
The last relation means that $\displaystyle f=\sum_{i=1}^r \alpha_i f_i\,$ is an affine function, with
$$ df = d\big(\sum_{i=1}^r \alpha_i\,f_i\big) = \sum_{i=1}^r \alpha_i\,df_i. $$