Why coefficients in affine combination should add up to 1

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SUMMARY

The discussion centers on the necessity for coefficients in an affine combination to sum to 1, which ensures that different origins in an affine space yield the same resultant vector. The participants illustrate this with examples using vectors a=(1, 1) and b=(0, 1), demonstrating that regardless of the chosen origin, the linear combination of these vectors remains consistent as long as the coefficients sum to 1. This property allows for meaningful translations and comparisons in affine spaces without a defined origin.

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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).
 
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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...
 
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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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