# Why the sum of all coefficients is 1 in Affine Combination?

In Vector Space $$V$$, any vector $$v$$ can be written in $$\textbf{linear combination}$$ of a basis $$\{e_1, e_2, \dots e_n\}$$ such as $$v = \sum_{i=1}^{n} \alpha_i e_i$$

In Affine Space, any point $$p$$ can be written in $$\textbf{affine combination}$$ of $$\{p_1, p_1, \dots p_n\}$$ such as $$p = \sum_{i=1}^{n} \beta_i p_i \quad \mbox{ where } 1 = \sum_{i=1}^{n} \beta_i$$

My Questions: Why we need the condition in $$\textbf{Affine combination}$$ such as $$1 = \sum_{i=1}^{n} \beta_i$$ What is wrong with that if we DO NOT have the constraint

• This is just a definition. If the sum is not 1 then the sum is a linear combination and not affine. The question is illposed. Why is a banach space complete? – user515599 Feb 11 '19 at 18:58
• when people come up a definition, there is reason why they need some conditions or constraints. you can not come up some random definitions and expect some reasonable "useful" mathematic objects. The question is not illposed. – 1234 Feb 11 '19 at 19:11
• Note bthat is the first case you have a vector space of dimension $n$, whereas in the second case, you have a affine space of dimension $n-1$. In an affine space, a point is represented as a weighted mean of the points in an affine basis – Bernard Feb 11 '19 at 19:11
• @1234 You don't define things and then find stuff that fit the definition. You work with an object and come up with a word for convenience. – user515599 Feb 11 '19 at 19:15
• Non one stop you come up some non-linear combination on Vector Space, but you can do not use "matrix" over the vector space any more. It might be useful in some area of mathematic that I did not know about it. One reason we use linear combination over a Vector Space is "matrix" can be used. – 1234 Feb 11 '19 at 19:29

Think of the affine combination as a linear combination of position vectors, which we want to specify a point. Now, a position vector has two parts; a base vector to whatever point we're calling zero, and a displacement from that. Suppose you have coordinates based on the street grid numbers in a city - the vector says to go to the "zero point" downtown, and then move away from that a specified amount in each direction.

What happens when we apply an affine combination to these? Every vector in the combination has the same base vector, so we add a total of $$1$$ times that. The displacement vectors vary, and we get a new displacement vector from that. Overall, it's a position vector in the same form.

If, instead, we applied a linear combination without requiring the sum of coefficients to be $$1$$, we would multiply the base vector by that sum, whatever it is. Our new position vector would go to somewhere completely new as its base, then displace from there.

So what that sum condition means? It means that we can take an affine combination of position vectors for various points and get the position vector for a new point, in a way that doesn't depend on exactly which coordinate system we were using. Translating our coordinate system (choosing a new base point) translates that affine combination in exactly the same way, so that it still represents the same point.

• so the affine combination does not depend on what origin that you choose? – 1234 Feb 11 '19 at 21:17
• Yes, that's the point. For an example that uses affine combinations extensively, consider the geometry of a triangle. We can describe a point related to the triangle by an affine combination $xA+yB+zC$, and we want this to be independent of the "origin". After all, what would we choose as the origin? The center? Which center? – jmerry Feb 11 '19 at 22:28

Perhaps a small example is useful. Considider the two standard vectors in $$\mathbb{R}^n$$: $$\begin{pmatrix}1 \\ 0\end{pmatrix}$$ and $$\begin{pmatrix}0 \\ 1\end{pmatrix}$$. The set of linear combinations is $$\mathbb{R}^2$$ while the set of affine combinations is the line through the two vectors: $$\left\{\begin{pmatrix}x \\ 1-x\end{pmatrix} : x \in \mathbb{R}\right\}$$. The set of affine combinations is the smallest affine set that contains the vectors.

It's helpful to recognize that, since affine combinations are particular linear combinations with the property given for the coefficients, every affine hull sits inside of a linear span. Conversely, if we are handed a vector space $$V$$ we can recover any affine space sitting inside of it by 1) choosing a 1-form $$\omega : V\to k$$, and 2) selecting a scalar $$\lambda: \ast \to (k)$$. The affine space is the set of vectors $$v:V$$ satisfying $$\omega (v) = \lambda$$. If $$\lambda \neq 0$$, then this set is a hyperplane in $$V$$ of codimension 1 which is offset from the origin. This corresponds to the description of affine space as a vector space which "forgot" its origin. If $$\lambda = 0$$, then you always have a preferred affine combination whose coefficients are all zero, so the reason zero has this exceptional quality (and hence why a generic nonzero $$\lambda$$ is conventionally used) is because zero is the additive identity in $$k$$ and the "affinity" of a linear combination is characterized by an additive expression in the coefficients.

The property of having a fixed sum $$\lambda$$ of the coefficients is what allows us to take differences of affine points and obtain vectors, which is a geometric restatement of the associativity and commutativity of scalar addition in the ring $$k$$. The statement $$\lambda \neq 0$$ is what prevents us from taking sums of affine points; if we add the coordinates pointwise then the resulting coordinates will instead sum to $$2\lambda$$, which is not equal $$\lambda$$ if we chose $$\lambda \neq 0$$. The important part is just that $$\lambda$$ be nonzero, because this obstructs our attempts to add points as vectors, and through this obstruction we formalize the idea that an affine space has lost its additional structure of having a distinguished origin, or additive identity element.