affine combination

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

,

…

,

v

n

∈

V

is a

linear combination

of the vectors

k

1

⁢

v

1

+

⋯

+

k

n

⁢

v

n

such that

k

i

∈

D

subject to the condition

k

1

+

⋯

+

k

n

=

1

. In effect, an affine combination is a weighted average of the vectors in question.

For example,

v

=

1

2

⁢

v

1

+

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

char

⁡

(

D

)

does not divide

m

, then

v

=

1

m

⁢

(

v

1

+

⋯

+

v

m

)

is an affine combination of the

v

i

’s.

v

is the

barycenter

of

v

1

,

…

,

v

n

.

Relations with Affine Subspaces

Assume now

char

⁡

(

D

)

=

0

. Given

v

1

,

…

,

v

n

∈

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

,

…

,

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

∈

A

. Define

S

=

{

a

-

v

∣

a

∈

A

}

. Then for any

s

∈

S

and

d

∈

D

,

d

⁢

s

=

d

⁢

(

a

-

v

)

=

d

⁢

a

+

(

1

-

d

)

⁢

v

-

v

. Since

d

⁢

a

+

(

1

-

d

)

⁢

v

∈

A

,

d

⁢

s

∈

S

. If

s

1

,

s

2

∈

S

, then

1

2

⁢

(

s

1

+

s

2

)

=

1

2

⁢

(

(

a

1

-

v

)

+

(

a

2

-

v

)

)

=

1

2

⁢

(

a

1

+

a

2

)

-

v

∈

S

, since

1

2

⁢

(

a

1

+

a

2

)

∈

A

. So

1

2

⁢

(

s

1

+

s

2

)

∈

S

. Therefore,

s

1

+

s

2

=

2

⁢

(

1

2

⁢

(

s

1

+

s

2

)

)

∈

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

{

s

1

,

…

,

s

n

}

. For
each

i

=

1

,

…

,

n

, define

v

i

=

n

⁢

s

i

+

v

. Given

a

∈

A

, we have

a

=

s

+

v

=

k

1

⁢

s

1

+

⋯

+

k

n

⁢

s

n

+

v

=

k

1

n

⁢

(

v

1

-

v

)

+

⋯

+

k

n

n

⁢

(

v

n

-

v

)

+

v

=

k

1

n

⁢

v

1

+

⋯

+

k

n

n

⁢

v

n

+

(

1

-

k

1

n

-

⋯

-

k

n

n

)

⁢

v

.

From this calculation, it is evident that

a

is an affine combination of

v

1

,

…

,

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

d

⁢

v

+

(

1

-

d

)

⁢

w

,

d

∈

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

,

…

,

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

+

⋯

+

k

n

⁢

v

n

∣

v

i

∈

V

,

k

i

∈

D

,

∑

k

i

=

1

}

, the subset

{

k

1

⁢

v

1

+

⋯

+

k

n

⁢

v

n

∈

A

∣

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.

M

=

{

v

1

,

…

,

v

n

}

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

∈

M

,

N

=

{

v

i

-

v

∣

v

≠

v

i

}

is

linearly independent

.

Proof.

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

(1) implies (2). If

a

∈

A

has two distinct

representations

k

1

⁢

v

1

+

⋯

+

k

n

⁢

v

n

=

a

=

r

1

⁢

v

1

+

⋯

+

r

n

⁢

v

n

, we may assume, say

k

1

≠

r

1

. So

k

1

-

r

1

is

invertible

with

inverse

t

∈

D

. Then

v

1

=

t

⁢

(

r

2

-

k

2

)

⁢

v

2

+

⋯

+

t

⁢

(

r

n

-

k

n

)

⁢

v

n

.

Furthermore,

∑

i

=

2

n

t

⁢

(

r

i

-

k

i

)

=

t

⁢

(

∑

i

=

2

n

r

i

-

∑

i

=

2

n

k

i

)

=

t

⁢

(

1

-

r

1

-

1

+

k

1

)

=

1

.

So for any

b

∈

A

, we have

b

=

s

1

⁢

v

1

+

⋯

+

s

n

⁢

v

n

=

s

1

⁢

(

t

⁢

(

r

2

-

k

2

)

⁢

v

2

+

⋯

+

t

⁢

(

r

n

-

k

n

)

⁢

v

n

)

+

⋯

+

s

n

⁢

v

n

.

The sum of the

coefficients

is easily seen to be 1, which implies that

{

v

2

,

…

,

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

)

+

⋯

+

s

n

⁢

(

v

n

-

v

1

)

. Expand and we have

0

=

(

-

s

2

-

⋯

-

s

n

)

⁢

v

1

+

s

2

⁢

v

2

+

⋯

+

s

n

⁢

v

n

. So

(

1

-

s

2

-

⋯

-

s

n

)

⁢

v

1

+

s

2

⁢

v

2

+

⋯

+

s

n

⁢

v

n

=

v

1

∈

A

. By

assumption

, there is exactly one way to express

v

1

, so we conclude that

s

2

=

⋯

=

s

n

=

0

.

(3) implies (1). If

M

were not minimal, then some

v

∈

M

could be expressed as an affine combination of the remaining vectors in

M

. So suppose

v

1

=

k

2

⁢

v

2

+

⋯

+

k

n

⁢

v

n

. Since

∑

k

i

=

1

, we can rewrite this as

0

=

k

2

⁢

(

v

2

-

v

1

)

+

⋯

+

k

n

⁢

(

v

n

-

v

1

)

. Since not all

k

i

=

0

,

N

=

{

v

2

-

v

1

,

…

,

v

n

-

v

1

}

is not linearly independent.
∎

Remarks

.

•

If

{

v

1

,

…

,

v

n

}

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

v

∈

M

, such that

N

=

{

w

-

v

∣

v

≠

w

∈

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

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

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