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affine combination
Definition
Let $V$ be a vector space over a division ring $D$. An affine combination of a finite set of vectors $v_{1},\ldots,v_{n}\in V$ is a linear combination of the vectors
$k_{1}v_{1}+\cdots+k_{n}v_{n}$ 
such that $k_{i}\in D$ subject to the condition $k_{1}+\cdots+k_{n}=1$. In effect, an affine combination is a weighted average of the vectors in question.
For example, $v=\frac{1}{2}v_{1}+\frac{1}{2}v_{2}$ is an affine combination of $v_{1}$ and $v_{2}$ provided that the characteristic of $D$ is not $2$. $v$ is known as the midpoint of $v_{1}$ and $v_{2}$. More generally, if $\operatorname{char}(D)$ does not divide $m$, then
$v=\frac{1}{m}(v_{1}+\cdots+v_{m})$ 
is an affine combination of the $v_{i}$’s. $v$ is the barycenter of $v_{1},\ldots,v_{n}$.
Relations with Affine Subspaces
Assume now $\operatorname{char}(D)=0$. Given $v_{1},\ldots,v_{n}\in V$, we can form the set $A$ of all affine combinations of the $v_{i}$’s. We have the following
$A$ is a finite dimensional affine subspace. Conversely, a finite dimensional affine subspace $A$ is the set of all affine combinations of a finite set of vectors in $A$.
Proof.
Suppose $A$ is the set of affine combinations of $v_{1},\ldots,v_{n}$. If $n=1$, then $A$ is a singleton $\{v\}$, so $A=0+v$, where 0 is the null subspace of $V$. If $n>1$, we may pick a nonzero vector $v\in A$. Define $S=\{av\mid a\in A\}$. Then for any $s\in S$ and $d\in D$, $ds=d(av)=da+(1d)vv$. Since $da+(1d)v\in A$, $ds\in S$. If $s_{1},s_{2}\in S$, then $\frac{1}{2}(s_{1}+s_{2})=\frac{1}{2}((a_{1}v)+(a_{2}v))=\frac{1}{2}(a_{1}+a_% {2})v\in S$, since $\frac{1}{2}(a_{1}+a_{2})\in A$. So $\frac{1}{2}(s_{1}+s_{2})\in S$. Therefore, $s_{1}+s_{2}=2(\frac{1}{2}(s_{1}+s_{2}))\in S$. This shows that $S$ is a vector subspace of $V$ and that $A=S+v$ is an affine subspace.
Conversely, let $A$ be a finite dimensional affine subspace. Write $A=S+v$, where $S$ is a subspace of $V$. Since $\operatorname{dim}(S)=\operatorname{dim}(A)=n$, $S$ has a basis $\{s_{1},\ldots,s_{n}\}$. For each $i=1,\ldots,n$, define $v_{i}=ns_{i}+v$. Given $a\in A$, we have
$\displaystyle a$  $\displaystyle=$  $\displaystyle s+v=k_{1}s_{1}+\cdots+k_{n}s_{n}+v$  
$\displaystyle=$  $\displaystyle\frac{k_{1}}{n}(v_{1}v)+\cdots+\frac{k_{n}}{n}(v_{n}v)+v$  
$\displaystyle=$  $\displaystyle\frac{k_{1}}{n}v_{1}+\cdots+\frac{k_{n}}{n}v_{n}+(1\frac{k_{1}}{% n}\cdots\frac{k_{n}}{n})v.$ 
From this calculation, it is evident that $a$ is an affine combination of $v_{1},\ldots,v_{n}$, and $v$. ∎
When $A$ is the set of affine combinations of two distinct vectors $v,w$, we see that $A$ is a line, in the sense that $A=S+v$, a translate of a onedimensional subspace $S$ (a line through 0). Every element in $A$ has the form $dv+(1d)w$, $d\in D$. Inspecting the first part of the proof in the previous proposition, we see that the argument involves no more than two vectors at a time, so the following useful corollary is apparant:
$A$ is an affine subspace iff for every pair of vectors in $A$, the line formed by the pair is also in $A$.
Note, however, that the $A$ in the above corollary is not assumed to be finite dimensional.
Remarks.

Given $A=\{k_{1}v_{1}+\cdots+k_{n}v_{n}\mid v_{i}\in V,k_{i}\in D,\sum k_{i}=1\}$, the subset
$\{k_{1}v_{1}+\cdots+k_{n}v_{n}\in A\mid k_{i}=0\}$ is also an affine subspace.
Affine Independence
Since every element in a finite dimensional affine subspace $A$ is an affine combination of a finite set of vectors in $A$, we have the similar concept of a spanning set of an affine subspace. A minimal spanning set $M$ of an affine subspace is said to be affinely independent. We have the following three equivalent characterization of an affinely independent subset $M$ of a finite dimensional affine subspace:
1. $M=\{v_{1},\ldots,v_{n}\}$ is affinely independent.
2. every element in $A$ can be written as an affine combination of elements in $M$ in a unique fashion.
3. for every $v\in M$, $N=\{v_{i}v\mid v\neq v_{i}\}$ is linearly independent.
Proof.
We will proceed as follows: (1) implies (2) implies (3) implies (1).
(1) implies (2). If $a\in A$ has two distinct representations $k_{1}v_{1}+\cdots+k_{n}v_{n}=a=r_{1}v_{1}+\cdots+r_{n}v_{n}$, we may assume, say $k_{1}\neq r_{1}$. So $k_{1}r_{1}$ is invertible with inverse $t\in D$. Then
$v_{1}=t(r_{2}k_{2})v_{2}+\cdots+t(r_{n}k_{n})v_{n}.$ 
Furthermore,
$\sum_{{i=2}}^{n}t(r_{i}k_{i})=t(\sum_{{i=2}}^{n}r_{i}\sum_{{i=2}}^{n}k_{i})=% t(1r_{1}1+k_{1})=1.$ 
So for any $b\in A$, we have
$b=s_{1}v_{1}+\cdots+s_{n}v_{n}=s_{1}(t(r_{2}k_{2})v_{2}+\cdots+t(r_{n}k_{n})% v_{n})+\cdots+s_{n}v_{n}.$ 
The sum of the coefficients is easily seen to be 1, which implies that $\{v_{2},\ldots,v_{n}\}$ is a spanning set of $A$ that is smaller than $M$, a contradiction.
(2) implies (3). Pick $v=v_{1}$. Suppose $0=s_{2}(v_{2}v_{1})+\cdots+s_{n}(v_{n}v_{1})$. Expand and we have $0=(s_{2}\cdotss_{n})v_{1}+s_{2}v_{2}+\cdots+s_{n}v_{n}$. So $(1s_{2}\cdotss_{n})v_{1}+s_{2}v_{2}+\cdots+s_{n}v_{n}=v_{1}\in A$. By assumption, there is exactly one way to express $v_{1}$, so we conclude that $s_{2}=\cdots=s_{n}=0$.
(3) implies (1). If $M$ were not minimal, then some $v\in M$ could be expressed as an affine combination of the remaining vectors in $M$. So suppose $v_{1}=k_{2}v_{2}+\cdots+k_{n}v_{n}$. Since $\sum k_{i}=1$, we can rewrite this as $0=k_{2}(v_{2}v_{1})+\cdots+k_{n}(v_{n}v_{1})$. Since not all $k_{i}=0$, $N=\{v_{2}v_{1},\ldots,v_{n}v_{1}\}$ is not linearly independent. ∎
Remarks.

If $\{v_{1},\ldots,v_{n}\}$ is affinely independent set spanning $A$, then $\operatorname{dim}(A)=n1$.

More generally, a set $M$ (not necessarily finite) of vectors is said to be affinely independent if there is a vector $v\in M$, such that $N=\{wv\mid v\neq w\in M\}$ is linearly independent (every finite subset of $N$ is linearly independent). The above three characterizations are still valid in this general setting. However, one must be careful that an affine combination is a finitary operation so that when we take the sum of an infinite number of vectors, we have to realize that only a finite number of them are nonzero.

Given any set $S$ of vectors, the affine hull of $S$ is the smallest affine subspace $A$ that contains every vector of $S$, denoted by $\operatorname{Aff}(S)$. Every vector in $\operatorname{Aff}(S)$ can be written as an affine combination of vectors in $S$.
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Comments
Homework Problem Help
If anyone can help me with this proof, that would be great.
I need to show that if v_0,...,v_n are n+1 affineindependent points in R^n, and if there exits x,y in R^n such that x  v_i = y  v_i for i = 0,...,n, then x = y.
Re: Homework Problem Help
Start with the identity xv_i = (xy)+(yv_i). Square it (scalar product by itself), after simplification you get:
xy^2 = 2(xy).(yv_i) for all i, so (xy).(v_iv_j) = 0 for all i and j.
Since the n+1 vectors v_i are affine independent, v_iv_j generate the whole space R^n. That means that xy is orhogonal to every vector in R^n.
Re: Homework Problem Help
Hi there, I follow the argument until you say
> so (xy).(v_iv_j) = 0 for all i and j
how does this follow from
> xy^2 = 2(xy).(yv_i) for all i
? Thanks for all the help!
Re: Homework Problem Help
Since xy^2 = 2(xy).(yv_i) for all i, 0 = xy^2  xy^2 = 2(xy).(yv_i)  (2(xy).(yv_j)) = 2(xy)(v_iv_j).
Re: Homework Problem Help
wow ok that's rather clever...thanks so much; this was a big help!!!
Help with proof
I need to prove two things
1) If x_0,...x_n are n+1 affineindependent points in R^n and if x_i  x_j = y_i  y_j for all i,j then y_0,...,y_n are also affine independent.
2) If x_0,...x_n are n+1 affineindependent points in R^n and A:R^n >R^n is an affine tranformation such that for all i,j
A(x_i)  A(x_j) = x_i  x_j
then A is a Euclidean isometry.
Thanks so much for the help!
Re: Help with proof
I figured out (2); since A is affine, there is a linear transformation L and a vector v such that A(x) = L(x) + v for all x. It suffices to show that L is orthogonal which is to say that
L(x).L(y) = x.y for all x,y. Now notice that by hypothesis we have
(x_i  x_j).(x_i  x_j) = x_i  x_j^2
= A(x_i)  A(x_j)^2
= L(x_i)  L(x_j)^2
= L(x_i  x_j)^2
= L(x_i  x_j).L(x_i  x_j)
since x_0,...,x_n are affineind., the vectors x_0x_i form a basis for R^n so we can expand x = sum_i(a_i(x_0  x_i)) and y = sum_j(b_j(x_0  x_j)). Using the above computation we have
L(x).L(y) = L(sum_i(a_i(x_0  x_i))).L(sum_j(b_j(x_0  x_j)))
= sum_ij(a_ib_jL(x_0  x_i).L(x_0  x_j))
= sum_ij(a_ib_j(x_0  x_i).(x_0  x_j))
= (sum_i(a_i(x_0  x_i))).(sum_j(b_j(x_0  x_j)))
= x.y
Re: Help with proof
I also figured out (1) with the help of a professor. It's not hard to show that the set v_i  v_0 has the same Gramian matrix as w_i  w_0 by using the constraints v_i  v_j = w_i  w_j. It follows that since the set v_i  v_0 is linearly independent, so is the set w_i  w_0 and thus w_0,...,w_n are affine independent.
If anyone finds any errors in the proofs of either of the two parts of the original post, then I'd be grateful to know about them.