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affine combination (Definition)

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
$\displaystyle k_1v_1+\cdots+k_nv_n$
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

$\displaystyle v=\frac{1}{m}(v_1+\cdots+v_m)$
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 $ \lbrace v\rbrace$, 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=\lbrace a-v\mid a\in A\rbrace$. 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 $ \lbrace s_1,\ldots,s_n\rbrace$. For each $ i=1,\ldots, n$, define $ v_i=ns_i+v$. Given $ a\in A$, we have

$\displaystyle a$ $\displaystyle =$ $\displaystyle s+v=k_1s_1+\cdots+k_ns_n+v$  
  $\displaystyle =$ $\displaystyle \frac{k_1}{n}(v_1-v)+\cdots+\frac{k_n}{n}(v_n-v)+v$  
  $\displaystyle =$ $\displaystyle \frac{k_1}{n}v_1+\cdots+\frac{k_n}{n}v_n+(1-\frac{k_1}{n}-\cdots- \frac{k_n}{n})v.$  

From this calculation, it is evident that $ a$ is an affine combination of $ v_1,\ldots,v_n$, and $ v$. $ \qedsymbol$

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=\lbrace k_1v_1+\cdots +k_nv_n \mid v_i\in V, k_i\in D, \sum k_i=1 \rbrace$, the subset
    $\displaystyle \lbrace k_1v_1+\cdots+ k_nv_n\in A\mid k_i=0\rbrace$
    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=\lbrace v_1,\ldots,v_n\rbrace$ 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=\lbrace v_i-v\mid v\neq v_i\rbrace$ 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_1v_1+\cdots+k_nv_n=a= r_1v_1+\cdots+r_nv_n$, we may assume, say $ k_1\neq r_1$. So $ k_1-r_1$ is invertible with inverse $ t\in D$. Then

$\displaystyle v_1=t(r_2-k_2)v_2+\cdots+t(r_n-k_n)v_n.$
Furthermore,
$\displaystyle \sum_{i=2}^n t(r_i-k_i)=t(\sum_{i=2}^n r_i-\sum_{i=2}^n k_i)=t(1-r_1-1+k_1)=1.$
So for any $ b\in A$, we have
$\displaystyle b=s_1v_1+\cdots+s_nv_n=s_1(t(r_2-k_2)v_2+\cdots+t(r_n-k_n)v_n)+\cdots+s_nv_n.$
The sum of the coefficients is easily seen to be 1, which implies that $ \lbrace v_2,\ldots,v_n\rbrace$ 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_2v_2+\cdots+s_nv_n$. So $ (1-s_2-\cdots-s_n)v_1+s_2v_2+\cdots+s_nv_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_2v_2+\cdots+k_nv_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=\lbrace v_2-v_1,\ldots, v_n-v_1\rbrace$ is not linearly independent. $ \qedsymbol$

Remarks.

  • If $ \lbrace v_1,\ldots,v_n\rbrace$ 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=\lbrace w-v\mid v\neq w\in M\rbrace$ 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$.



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See Also: affine geometry, affine transformation, convex combination

Other names:  affine independent
Also defines:  affine independence, affinely independent, affine hull
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Cross-references: number, infinite, operation, finite, spanning, expand, contradiction, coefficients, sum, inverse, invertible, representations, implies, linearly independent, characterization, equivalent, minimal, spanning set, similar, subset, contains, zero vector, iff, argument, proposition, proof, translate, line, basis, non-zero vector, subspace, null, singleton, affine subspace, finite dimensional, barycenter, divide, midpoint, characteristic, weighted average, linear combination, vectors, finite set, division ring, vector space
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This is version 16 of affine combination, born on 2006-06-14, modified 2006-10-20.
Object id is 8033, canonical name is AffineCombination.
Accessed 5840 times total.

Classification:
AMS MSC51A15 (Geometry :: Linear incidence geometry :: Structures with parallelism)

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Help with proof by joshsamani on 2008-04-30 04:08:48
I need to prove two things

1) If x_0,...x_n are n+1 affine-independent 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 affine-independent 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!
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Homework Problem Help by joshsamani on 2008-04-27 17:24:06
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 affine-independent 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.
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