free vector space over a set

In this entry we construct the free vector space over a set, or the vector spaceMathworldPlanetmath generated by a set [1]. For a set X, we shall denote this vector space by C(X). One application of this construction is given in [2], where the free vector space is used to define the tensor productPlanetmathPlanetmathPlanetmath for modules.

To define the vector space C(X), let us first define C(X) as a set. For a set X and a field 𝕂, we define

C(X) = {f:X𝕂|f-1(𝕂\{0})is finite}.

In other words, C(X) consists of functions f:X𝕂 that are non-zero only at finitely many points in X. Here, we denote the identity elementMathworldPlanetmath in 𝕂 by 1, and the zero elementMathworldPlanetmath by 0. The vector space structure for C(X) is defined as follows. If f and g are functions in C(X), then f+g is the mapping xf(x)+g(x). Similarly, if fC(X) and α𝕂, then αf is the mapping xαf(x). It is not difficult to see that these operations are well defined, i.e., both f+g and αf are again functions in C(X).

0.0.1 Basis for C(X)

If aX, let us define the function ΔaC(X) by

Δa(x) = {1whenx=a,0otherwise.

These functions form a linearly independentMathworldPlanetmath basis for C(X), i.e.,

C(X) = span{Δa}aX. (1)

Here, the space span{Δa}aX consists of all finite linear combinationsMathworldPlanetmath of elements in {Δa}aX. It is clear that any element in span{Δa}aX is a member in C(X). Let us check the other direction. Suppose f is a member in C(X). Then, let ξ1,, ξN be the distinct points in X where f is non-zero. We then have

f = i=1Nf(ξi)Δξi,

and we have established equality in equation 1.

To see that the set {Δa}aX is linearly independent, we need to show that its any finite subset is linearly independent. Let {Δξ1,,ΔξN} be such a finite subset, and suppose i=1NαiΔξi=0 for some αi𝕂. Since the points ξi are pairwise distinct, it follows that αi=0 for all i. This shows that the set {Δa}aX is linearly independent.

Let us define the mapping ι:XC(X), xΔx. This mapping gives a bijection between X and the basis vectors {Δa}aX. We can thus identify these spaces. Then X becomes a linearly independent basis for C(X).

0.0.2 Universal property of ι:XC(X)

The mapping ι:XC(X) is universal in the following sense. If ϕ is an arbitrary mapping from X to a vector space V, then there exists a unique mapping ϕ¯ such that the below diagram commutes:


Proof. We define ϕ¯ as the linear mapping that maps the basis elements of C(X) as ϕ¯(Δx)=ϕ(x). Then, by definition, ϕ¯ is linear. For uniqueness, suppose that there are linear mappings ϕ¯,σ¯:C(X)V such that ϕ=ϕ¯ι=σ¯ι. For all xX, we then have ϕ¯(Δx)=σ¯(Δx). Thus ϕ¯=σ¯ since both mappings are linear and the coincide on the basis elements.


  • 1 W. Greub, Linear AlgebraMathworldPlanetmath, Springer-Verlag, Fourth edition, 1975.
  • 2 I. Madsen, J. Tornehave, From Calculus to CohomologyPlanetmathPlanetmath, Cambridge University press, 1997.
Title free vector space over a set
Canonical name FreeVectorSpaceOverASet
Date of creation 2013-03-22 13:34:34
Last modified on 2013-03-22 13:34:34
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 8
Author mathcam (2727)
Entry type Definition
Classification msc 15-00
Synonym vector space generated by a set
Related topic TensorProductBasis