affine combination
Definition
Let $V$ be a vector space^{} over a division ring $D$. An affine combination of a finite set^{} of vectors ${v}_{1},\mathrm{\dots},{v}_{n}\in V$ is a linear combination^{} of the vectors
$${k}_{1}{v}_{1}+\mathrm{\cdots}+{k}_{n}{v}_{n}$$ 
such that ${k}_{i}\in D$ subject to the condition ${k}_{1}+\mathrm{\cdots}+{k}_{n}=1$. 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 $\mathrm{char}(D)$ does not divide $m$, then
$$v=\frac{1}{m}({v}_{1}+\mathrm{\cdots}+{v}_{m})$$ 
is an affine combination of the ${v}_{i}$’s. $v$ is the barycenter of ${v}_{1},\mathrm{\dots},{v}_{n}$.
Relations with Affine Subspaces
Assume now $\mathrm{char}(D)=0$. Given ${v}_{1},\mathrm{\dots},{v}_{n}\in V$, 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 ${v}_{1},\mathrm{\dots},{v}_{n}$. 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 $\mathrm{dim}(S)=\mathrm{dim}(A)=n$, $S$ has a basis $\{{s}_{1},\mathrm{\dots},{s}_{n}\}$. For each $i=1,\mathrm{\dots},n$, define ${v}_{i}=n{s}_{i}+v$. Given $a\in A$, we have
$a$  $=$  $s+v={k}_{1}{s}_{1}+\mathrm{\cdots}+{k}_{n}{s}_{n}+v$  
$=$  $\frac{{k}_{1}}{n}}({v}_{1}v)+\mathrm{\cdots}+{\displaystyle \frac{{k}_{n}}{n}}({v}_{n}v)+v$  
$=$  $\frac{{k}_{1}}{n}}{v}_{1}+\mathrm{\cdots}+{\displaystyle \frac{{k}_{n}}{n}}{v}_{n}+(1{\displaystyle \frac{{k}_{1}}{n}}\mathrm{\cdots}{\displaystyle \frac{{k}_{n}}{n}})v.$ 
From this calculation, it is evident that $a$ is an affine combination of ${v}_{1},\mathrm{\dots},{v}_{n}$, 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.

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If one of ${v}_{1},\mathrm{\dots},{v}_{n}$ is the zero vector, then $A$ coincides with $S$. In other words, an affine subspace is a vector subspace if it contains the zero vector.

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Given $A=\{{k}_{1}{v}_{1}+\mathrm{\cdots}+{k}_{n}{v}_{n}\mid {v}_{i}\in V,{k}_{i}\in D,\sum {k}_{i}=1\}$, the subset
$$\{{k}_{1}{v}_{1}+\mathrm{\cdots}+{k}_{n}{v}_{n}\in A\mid {k}_{i}=0\}$$ 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.
$M=\{{v}_{1},\mathrm{\dots},{v}_{n}\}$ 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 $v\in M$, $N=\{{v}_{i}v\mid v\ne {v}_{i}\}$ 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 ${k}_{1}{v}_{1}+\mathrm{\cdots}+{k}_{n}{v}_{n}=a={r}_{1}{v}_{1}+\mathrm{\cdots}+{r}_{n}{v}_{n}$, we may assume, say ${k}_{1}\ne {r}_{1}$. So ${k}_{1}{r}_{1}$ is invertible^{} with inverse^{} $t\in D$. Then
$${v}_{1}=t({r}_{2}{k}_{2}){v}_{2}+\mathrm{\cdots}+t({r}_{n}{k}_{n}){v}_{n}.$$ 
Furthermore,
$$\sum _{i=2}^{n}t({r}_{i}{k}_{i})=t(\sum _{i=2}^{n}{r}_{i}\sum _{i=2}^{n}{k}_{i})=t(1{r}_{1}1+{k}_{1})=1.$$ 
So for any $b\in A$, we have
$$b={s}_{1}{v}_{1}+\mathrm{\cdots}+{s}_{n}{v}_{n}={s}_{1}(t({r}_{2}{k}_{2}){v}_{2}+\mathrm{\cdots}+t({r}_{n}{k}_{n}){v}_{n})+\mathrm{\cdots}+{s}_{n}{v}_{n}.$$ 
The sum of the coefficients is easily seen to be 1, which implies that $\{{v}_{2},\mathrm{\dots},{v}_{n}\}$ is a spanning set of $A$ that is smaller than $M$, a contradiction^{}.
(2) implies (3). Pick $v={v}_{1}$. Suppose $0={s}_{2}({v}_{2}{v}_{1})+\mathrm{\cdots}+{s}_{n}({v}_{n}{v}_{1})$. Expand and we have $0=({s}_{2}\mathrm{\cdots}{s}_{n}){v}_{1}+{s}_{2}{v}_{2}+\mathrm{\cdots}+{s}_{n}{v}_{n}$. So $(1{s}_{2}\mathrm{\cdots}{s}_{n}){v}_{1}+{s}_{2}{v}_{2}+\mathrm{\cdots}+{s}_{n}{v}_{n}={v}_{1}\in A$. By assumption^{}, there is exactly one way to express ${v}_{1}$, so we conclude that ${s}_{2}=\mathrm{\cdots}={s}_{n}=0$.
(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 ${v}_{1}={k}_{2}{v}_{2}+\mathrm{\cdots}+{k}_{n}{v}_{n}$. Since $\sum {k}_{i}=1$, we can rewrite this as $0={k}_{2}({v}_{2}{v}_{1})+\mathrm{\cdots}+{k}_{n}({v}_{n}{v}_{1})$. Since not all ${k}_{i}=0$, $N=\{{v}_{2}{v}_{1},\mathrm{\dots},{v}_{n}{v}_{1}\}$ is not linearly independent. ∎
Remarks.

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If $\{{v}_{1},\mathrm{\dots},{v}_{n}\}$ is affinely independent set spanning $A$, then $\mathrm{dim}(A)=n1$.

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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 $N=\{wv\mid v\ne w\in M\}$ 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.

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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 $\mathrm{Aff}(S)$. Every vector in $\mathrm{Aff}(S)$ can be written as an affine combination of vectors in $S$.
Title  affine combination 
Canonical name  AffineCombination 
Date of creation  20130322 16:00:13 
Last modified on  20130322 16:00:13 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  19 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 51A15 
Synonym  affine independent 
Related topic  AffineGeometry 
Related topic  AffineTransformation 
Related topic  ConvexCombination 
Defines  affine independence 
Defines  affinely independent 
Defines  affine hull 