Why coefficients in affine combination should add up to 1

martijnh
Messages
8
Reaction score
0
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
 
Physics news on Phys.org
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.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top