Why coefficients in affine combination should add up to 1

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Discussion Overview

The discussion revolves around the concept of affine combinations in affine spaces, specifically addressing why the coefficients in such combinations must sum to 1. Participants explore the implications of this property on translations and the representation of points in affine spaces.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the necessity of coefficients summing to 1 in affine combinations and questions how this relates to the concept of an affine space lacking an origin.
  • Another participant suggests experimenting with specific examples to illustrate that regardless of the chosen origin, a linear combination of vectors will yield the same result if the coefficients sum to 1.
  • A later reply reinforces that using the same linear combination for different origins is crucial for maintaining consistency in results, while also noting that other combinations (e.g., 3/4 and 1/4) could also be valid but would not yield the same outcome.
  • One participant clarifies that the key point is that the same combination of vectors with coefficients summing to 1 allows for discussions of distance and direction without referencing an origin.

Areas of Agreement / Disagreement

Participants generally agree on the importance of coefficients summing to 1 for maintaining consistency in affine combinations, but there is some confusion regarding the implications of different combinations and their results.

Contextual Notes

Some participants highlight the need for clarity regarding the relationship between different origins and the resulting vectors, as well as the conditions under which the property of coefficients summing to 1 holds true.

martijnh
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Hello there,

I have trouble understanding why the coefficients in an affine combination should add up to 1; From the wikipedia article (http://en.wikipedia.org/wiki/Affine_space#Informal_descriptions) it's mentioned that an affine space does not have an origin, so for an translation different origins can be chosen, which will result in different translations. They then mention that because the coefficients add up to 1, different solutions to point/vector translations will result the same result?

I can see how the restriction enables the coefficients to be rewritten as translations using vectors: P = a1 * p1 + a2 * p2 => a1 = 1 - a2 => P = p1 + a2 * (p2 - p1) => P = p1 + a2 * v

Though I can follow these steps, I don't understand why expressing it using vectors would be beneficial... More specifically I do not see why this property will cause all possible solutions in affine space to describe one and the same affine structure.

I can picture visually that when I choose a different origin in my affine space, I will get different vectors when for example I add them. I also see that using scalars (coefficients) of existing vectors in affine space, I can define a result vector.

Could anyone help?

Thanks!

Martijn
 
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Have a look at the informal description on wikipedia, and then try out a simple example to convince yourself that whichever point is chosen as the origin, a linear combination of vectors will give the same result if the sum of the coefficients is 1.

eg. let a=(1 1) and b=(0 1). Consider the linear combination:1/2*a + 1/2*b. If the origin is chosen to be (0 0) then this will give a result of (1/2 1).

Now let the origin be (1 1). Then 1/2*a + 1/2*b = (1 1) + 1/2*[a-(1 1)] + 1/2*[b-(1 1)] = (1 1) + (0 0) + (-1/2 0) = (1/2 1).
 
Last edited by a moderator:
mrbohn1 said:
Have a look at the http://en.wikipedia.org/wiki/Affine_space" , and then try out a simple example to convince yourself that whichever point is chosen as the origin, a linear combination of vectors will give the same result if the sum of the coefficients is 1.

eg. let a=(1 1) and b=(0 1). Consider the linear combination:1/2*a + 1/2*b. If the origin is chosen to be (0 0) then this will give a result of (1/2 1).

Now let the origin be (1 1). Then 1/2*a + 1/2*b = (1 1) + 1/2*[a-(1 1)] + 1/2*[b-(1 1)] = (1 1) + (0 0) + (0 1/2) = (1/2 1).

Thanks for clearing that up! So this only holds when you use the same lineair combination for both origins? I got confused because fx both 1/2, 1/2 and 3/4, 1/4 would be valid affine combinations...
 
Last edited by a moderator:
Yes - only the same combination of vectors (with coefficients summing to 1) will give the same result. That is the point: we can talk about distance and direction in an affine space without needing to refer to an origin.
 

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