Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Affine space

  • Last Updated : 21 Sep, 2021

Affine spaces

Affine space

Affine space is the set E with vector space \vec{E} and a transitive and free action of the additive \vec{E} on set E. The elements of space A are called points. The vector space \vec{E} that is associated with affine space is known as free vectors and the action +: E * \vec{E} \rightarrow E satisfying the following conditions:

  1. Right Identity: a + 0 = a  \, \forall \, a \,\epsilon \,A
  2. Associativity: (a + u) + v = a + (u+v) \, \forall a \, \epsilon \, E \,\, \forall \, \, u, v \, \epsilon \, \vec{E}
  3. For any two points a,b \epsilon E, there is unique u such that :

a + u =b

Attention reader! Don’t stop learning now. Get hold of all the important Machine Learning Concepts with the Machine Learning Foundation Course at a student-friendly price and become industry ready.

where u \epsilon \vec{E} and can be denoted ab or \vec{ab} or sometimes b-a. Therefore , we can write above equation

b= a + ab


. Consider the subset L of A^2       consisting of all points (x, y) satisfying the equation:

x+y - 1=0

L is the line having slope -1 passing through the point (1,0) and (0,1). The line L can be an affine space by defining the action +: L * R \rightarrow L of R on L defined such that every point (x, 1-x) on L and any u \epsilon R.

(x, 1-x) + u = (x+u, 1-x-u)

Now, for any two points a =(a_1, 1- a_1) and b = (b_1, 1-b_1) on L, the unique vector u \epsilon R such that b = a+u is u = b_1 – a_1. Note that vector space R isomorphic to the line of equation x + y = 0 passing through the origin. 

Chasles’s Identity

Given any three points a,b,c \epsilon     , since c = a + ac, b = a + ab, and c = b + bc, we get 

c = b+ bc = a + ab +bc

By applying the above properties 2 and 3, 

ab +bc  = ac

THe above equation is known as chasles Identity. Since

 a = a + aa 

and by using property 1 we,get 

a = a+ 0

Thus, by using property 3, we get:


Replacing a in place of c in Chasles Identity, we get:

ba =-ab

Now, For 4 points a,b,c,d \epsilon E. The chasles identity can be given as:

ad+bc = ad+ dc = ac

Affine combinations/ Barycentres

Similar to linear combination in linear algebra, the corresponding concept in affine geometry is that of an affine combination, also called a barycenter

Consider 2-dimensional space as an affine space, with origin O= (0,0) and basis vectors (1,0) and (0,1). Given any two points a =(a1, a2) and b =(b1, b2) there can be a natural combination such that \lambda a+ \mu b or:

(\lambda a_1 + \mu b_1, \lambda a_2 + \mu b_2)

when a = (-1, -1) and b = (2, 2), thus a+b can be given as: c = (1,1). 

Now, consider the new coordinate system with respect to the origin c = (1, 1). Now, the co-ordinates of a = (-2, -2), the co-ordinates of b are (1, 1), and the point of d = (-1, -1). However, point d is identical to the origin O = (0, 0) of the first coordinate system. 

Thus, a + b corresponds to two different points depending on which coordinate system is used for its computation. This means that we need extra conditions are required for affine computations. It turns out the scalars sum up 1. This helps us define the Barycentres

For any family of points (a_i)_{i\epsilon I} in  E, for any family of scalars such that \sum_{i \epsilon I } \lambda_i =1    and for any a \epsilon E   , the point

a + \sum_{i \epsilon I} \lambda_i a a_i    is called the barycentres of the points a_i assigned the weights \lambda_i    and is denoted by:

\sum_{i \epsilon E} \lambda_i a_i

Barycentres is conveniently denoted by the notations (a, \lambda)   , whereas a \epsilon E    is a point and \lambda \epsilon R    is called a scalar.

Affine Subspace

V is the affine subspace of \vec{E}  in the direction of \vec{V}  

Given an affine space \left \langle E, \vec{E}, +  \right \rangle   , a subset V of E is an affine subspace of \left \langle E, \vec{E}, +  \right \rangle   , if for every family of weighted points ((ai , λi))_{i \epsilon I}    in V such that \sum_{i \epsilon I} \lambda_i = 1     , the barycenter \sum_{i \epsilon I } \lambda_{i} a_i    belongs to V .


My Personal Notes arrow_drop_up
Recommended Articles
Page :