affine combination


Let V be a vector spaceMathworldPlanetmath over a division ring D. An affine combination of a finite setMathworldPlanetmath of vectors v1,,vnV is a linear combinationMathworldPlanetmath of the vectors


such that kiD subject to the condition k1++kn=1. In effect, an affine combination is a weighted average of the vectors in question.

For example, v=12v1+12v2 is an affine combination of v1 and v2 provided that the characteristic of D is not 2. v is known as the midpointMathworldPlanetmathPlanetmathPlanetmath of v1 and v2. More generally, if char(D) does not divide m, then


is an affine combination of the vi’s. v is the barycenter of v1,,vn.

Relations with Affine Subspaces

Assume now char(D)=0. Given v1,,vnV, we can form the set A of all affine combinations of the vi’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.


Suppose A is the set of affine combinations of v1,,vn. If n=1, then A is a singleton {v}, so A=0+v, where 0 is the null subspacePlanetmathPlanetmathPlanetmath of V. If n>1, we may pick a non-zero vector vA. Define S={a-vaA}. Then for any sS and dD, ds=d(a-v)=da+(1-d)v-v. Since da+(1-d)vA, dsS. If s1,s2S, then 12(s1+s2)=12((a1-v)+(a2-v))=12(a1+a2)-vS, since 12(a1+a2)A. So 12(s1+s2)S. Therefore, s1+s2=2(12(s1+s2))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 dim(S)=dim(A)=n, S has a basis {s1,,sn}. For each i=1,,n, define vi=nsi+v. Given aA, we have

a = s+v=k1s1++knsn+v
= k1n(v1-v)++knn(vn-v)+v
= k1nv1++knnvn+(1-k1n--knn)v.

From this calculation, it is evident that a is an affine combination of v1,,vn, 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 translateMathworldPlanetmath of a one-dimensional subspace S (a line through 0). Every element in A has the form dv+(1-d)w, dD. Inspecting the first part of the proof in the previous propositionPlanetmathPlanetmath, 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.


  • If one of v1,,vn 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={k1v1++knvnviV,kiD,ki=1}, the subset


    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 similarMathworldPlanetmathPlanetmathPlanetmath concept of a spanning setPlanetmathPlanetmath of an affine subspace. A minimalPlanetmathPlanetmath spanning set M of an affine subspace is said to be affinely independent. We have the following three equivalentMathworldPlanetmathPlanetmathPlanetmath characterization of an affinely independent subset M of a finite dimensional affine subspace:

  1. 1.

    M={v1,,vn} 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 vM, N={vi-vvvi} is linearly independentMathworldPlanetmath.


We will proceed as follows: (1) implies (2) implies (3) implies (1).

(1) implies (2). If aA has two distinct representations k1v1++knvn=a=r1v1++rnvn, we may assume, say k1r1. So k1-r1 is invertiblePlanetmathPlanetmathPlanetmath with inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath tD. Then




So for any bA, we have


The sum of the coefficients is easily seen to be 1, which implies that {v2,,vn} is a spanning set of A that is smaller than M, a contradictionMathworldPlanetmathPlanetmath.

(2) implies (3). Pick v=v1. Suppose 0=s2(v2-v1)++sn(vn-v1). Expand and we have 0=(-s2--sn)v1+s2v2++snvn. So (1-s2--sn)v1+s2v2++snvn=v1A. By assumptionPlanetmathPlanetmath, there is exactly one way to express v1, so we conclude that s2==sn=0.

(3) implies (1). If M were not minimal, then some vM could be expressed as an affine combination of the remaining vectors in M. So suppose v1=k2v2++knvn. Since ki=1, we can rewrite this as 0=k2(v2-v1)++kn(vn-v1). Since not all ki=0, N={v2-v1,,vn-v1} is not linearly independent. ∎


  • If {v1,,vn} is affinely independent set spanning A, then 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 vM, such that N={w-vvwM} 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 infiniteMathworldPlanetmath 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 Aff(S). Every vector in 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