Is it necessary for coefficients of an affine linear combination to be positive? Is factorization allowed?

Question

Recently I was having some reading on geometry. I encountered a form called affine linear combination. $$c_1p_1+c_2p_2+\cdots+c_np_n$$ It looks similar to linear combination in vector space but it must satisfy $\sum c_i=1$. My question is: is it necessary for all of the $c_i$'s to be positive? $$$$ Meanwhile, somebody told me scalar multiplication is not defined on affine space, does it mean that for $\alpha+\beta+\gamma=1$, I can't make the operation like: $$\alpha x+\beta y+\gamma z=\alpha x+(\beta+\gamma)\bigg(\frac{\beta}{\beta+\gamma}y+\frac{\gamma}{\beta+\gamma}z\bigg)=\alpha x+(\beta+\gamma)u$$ for a newly defined point $u$ from affine linear combination?

Answer 1

It's not necessary for all of the $c_i$'s to be positive. If $A$ is an affine space over the real numbers and $p_1, \dotsc, p_n \in A$, then the set $$\{ c_1 p_1 + \dotsb + c_n p_n \in A \mid c_1, \dotsc, c_n \ge 0,\, c_1 + \dotsb + c_n = 1 \}$$ is the convex hull of the set $\{ p_1, \dotsc, p_n \}$. But in general, the point $$c_1 p_1 + \dotsb + c_n p_n$$ is well defined even if some of the $c_i$'s are negative, or if the field of scalars is not ordered (for example if $A$ is an affine space over the complex numbers).

The proof goes as follows. Let $q \in A$ and define $$c_1 p_1 + \dotsb + c_n p_n := q + c_1 (p_1 - q) + \dotsb + c_n (p_n - q)$$ for $c_1, \dotsc, c_n$ scalars such that $c_1 + \dotsb + c_n = 1$. This is a point of $A$, because $q \in A$ and $p_i - q$ is a vector for each $i = 1, \dotsc, n$. What we need to check is that the definition doesn't depend on the choice of $q$. So, suppose $q' \in A$. Then $$\begin{array}{l} q' + c_1 (p_1 - q') + \dotsb + c_n (p_n - q') \\ = q' + c_1 (p_1 - q + q - q') + \dotsb + c_n (p_n - q + q - q') \\ = q' + c_1 (p_1 - q) + \dotsb + c_n (p_n - q) + (c_1 + \dotsb + c_n) (q - q')\\ = q' + c_1 (p_1 - q) + \dotsb + c_n (p_n - q) + q - q' \\ = q + c_1 (p_1 - q) + \dotsb + c_n (p_n - q). \\ \end{array}$$ Thus $c_1 p_1 + \dotsb + c_n p_n$ is a well-defined point of $A$.