# 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 non-zero vector $v\in A$. Define $S=\{a-v\mid a\in A\}$. Then for any $s\in S$ and $d\in D$, $ds=d(a-v)=da+(1-d)v-v$. Since $da+(1-d)v\in A$, $ds\in S$. 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 one-dimensional subspace $S$ (a line through 0). Every element in $A$ has the form $dv+(1-d)w$, $d\in D$. 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.

• If one of $v_{1},\ldots,v_{n}$ 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.

• 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. 1.

$M=\{v_{1},\ldots,v_{n}\}$ is affinely independent.

2. 2.

every element in $A$ can be written as an affine combination of elements in $M$ in a unique fashion.

3. 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,

 $\sum_{i=2}^{n}t(r_{i}-k_{i})=t(\sum_{i=2}^{n}r_{i}-\sum_{i=2}^{n}k_{i})=t(1-r_% {1}-1+k_{1})=1.$

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 $0=(-s_{2}-\cdots-s_{n})v_{1}+s_{2}v_{2}+\cdots+s_{n}v_{n}$. So $(1-s_{2}-\cdots-s_{n})v_{1}+s_{2}v_{2}+\cdots+s_{n}v_{n}=v_{1}\in A$. 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 $\operatorname{dim}(A)=n-1$.

• 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 $N=\{w-v\mid v\neq w\in M\}$ 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 non-zero.

• 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 $S$.

 Title affine combination Canonical name AffineCombination Date of creation 2013-03-22 16:00:13 Last modified on 2013-03-22 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