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affine combination
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(Definition)
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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 $$k_1v_1+\cdots+k_nv_n$$ 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 $$v=\frac{1}{m}(v_1+\cdots+v_m)$$ is an affine combination of the $v_i$ s. $v$ is the barycenter of $v_1,\ldots,v_n$
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 \begin{eqnarray*} a &=& s+v=k_1s_1+\cdots+k_ns_n+v \\ &=& \frac{k_1}{n}(v_1-v)+\cdots+\frac{k_n}{n}(v_n-v)+v \\ &=& \frac{k_1}{n}v_1+\cdots+\frac{k_n}{n}v_n+(1-\frac{k_1}{n}-\cdots- \frac{k_n}{n})v. \end{eqnarray*}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 $A=\lbrace k_1v_1+\cdots +k_nv_n \mid v_i\in V, k_i\in D, \sum k_i=1 \rbrace$ the subset $$\lbrace k_1v_1+\cdots+ k_nv_n\in A\mid k_i=0\rbrace$$ is also an affine subspace.
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:
- $M=\lbrace v_1,\ldots,v_n\rbrace$ is affinely independent.
- every element in $A$ can be written as an affine combination of elements in $M$ in a unique fashion.
- 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 $$v_1=t(r_2-k_2)v_2+\cdots+t(r_n-k_n)v_n.$$ 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 $$b=s_1v_1+\cdots+s_nv_n=s_1(t(r_2-k_2)v_2+\cdots+t(r_n-k_n)v_n)+\cdots+s_nv_n.$$ 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. 
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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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, vector subspace, non-zero vector, subspace, null, singleton, conversely, affine subspace, finite dimensional, barycenter, divide, midpoint, characteristic, weighted average, linear combination, vectors, finite set, division ring, vector space
There are 4 references to this entry.
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 8009 times total.
Classification:
| AMS MSC: | 51A15 (Geometry :: Linear incidence geometry :: Structures with parallelism) |
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Pending Errata and Addenda
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