PlanetMath: affine combination

For example, $v=\frac{1}{2}v_1+\frac{1}{2}v_2$ is an affine combination of

and

provided that the characteristic of

is not

is known as the midpoint of

and

. More generally, if $\operatorname{char}(D)$ does not divide

, then

Relations with Affine Subspaces

Proof. Suppose

is the set of affine combinations of $v_1,\ldots,v_n$ . If

, then

is a singleton $\lbrace v\rbrace$ , so

, where 0 is the null subspace of

. If

, 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 $d\in D$ ,

. 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

is a vector subspace of

and that

is an affine subspace.

Conversely, let be a finite dimensional affine subspace. Write , where is a subspace of . Since $\operatorname{dim}(S)= \operatorname{dim}(A)=n$ , has a basis $\lbrace s_1,\ldots,s_n\rbrace$ . For each $i=1,\ldots, n$ , define . 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

is an affine combination of $v_1,\ldots,v_n$ , and

. $\qedsymbol$

When

is the set of affine combinations of two distinct vectors

, we see that

is a line, in the sense that

, a translate of a one-dimensional subspace

(a line through 0). Every element in

has the form

, $d\in D$ . 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:

Note, however, that the

in the above corollary is not assumed to be finite dimensional.

Affine Independence

Since every element in a finite dimensional affine subspace

is an affine combination of a finite set of vectors in

, we have the similar concept of a spanning set of an affine subspace. A minimal spanning set

of an affine subspace is said to be affinely independent. We have the following three equivalent characterization of an affinely independent subset

of a finite dimensional affine subspace:

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 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

that is smaller than

, a contradiction.

(2) implies (3). Pick . 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 , so we conclude that $s_2=\cdots=s_n=0$ .

(3) implies (1). If were not minimal, then some $v\in M$ could be expressed as an affine combination of the remaining vectors in . 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 , $N=\lbrace v_2-v_1,\ldots, v_n-v_1\rbrace$ is not linearly independent. $\qedsymbol$

Definition

Relations with Affine Subspaces

Affine Independence