free vector space over a set∗
mathcam†
2013-03-21 15:54:25
In this entry we construct the free vector space over a set, or the vector space
generated by a set [?]. For a set X, we shall denote this vector space by C(X).
One application of this construction is given in [?], where the free vector space
is used to define the tensor product 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 : X → K | f −1 (K\{0}) is finite}.
In other words, C(X) consists of functions f : X → K that are non-zero only at
finitely many points in X. Here, we denote the identity element in K by 1, and
the zero element 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 x 7→ f (x) + g(x).
Similarly, if f ∈ C(X) and α ∈ K, then αf is the mapping x 7→ αf (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 a ∈ X, let us define the function ∆a ∈ C(X) by
1 when x = a,
∆a (x) =
0 otherwise.
These functions form a linearly independent basis for C(X), i.e.,
C(X)
=
span{∆a }a∈X .
(1)
Here, the space span{∆a }a∈X consists of all finite linear combinations of elements in {∆a }a∈X . It is clear that any element in span{∆a }a∈X is a member
in C(X). Let us check the other direction. Suppose f is a member in C(X).
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Then, let ξ1 , . . . , ξN be the distinct points in X where f is non-zero. We then
have
f
=
N
X
f (ξi )∆ξi ,
i=1
and we have established equality in equation ??.
To see that the set {∆a }a∈X is linearly independent, we need to show that
its any finite subset is linearly independent. Let {∆ξ1 , . . . , ∆ξN } be such a finite
PN
subset, and suppose
i=1 αi ∆ξi = 0 for some αi ∈ K. Since the points ξi
are pairwise distinct, it follows that αi = 0 for all i. This shows that the set
{∆a }a∈X is linearly independent.
Let us define the mapping ι : X → C(X), x 7→ ∆x . This mapping gives a
bijection between X and the basis vectors {∆a }a∈X . We can thus identify these
spaces. Then X becomes a linearly independent basis for C(X).
0.0.2
Universal property of ι : X → C(X)
The mapping ι : X → C(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 φ̄ such that the below diagram commutes:
X[r]φ [d]ι V C(X)[ur]φ̄
Proof. We define φ̄ as the linear mapping that maps the basis elements of C(X)
as φ̄(∆x ) = φ(x). Then, by definition, φ̄ is linear. For uniqueness, suppose that
there are linear mappings φ̄, σ̄ : C(X) → V such that φ = φ̄ ◦ ι = σ̄ ◦ ι. For
all x ∈ X, we then have φ̄(∆x ) = σ̄(∆x ). Thus φ̄ = σ̄ since both mappings are
linear and the coincide on the basis elements.
References
[1] W. Greub, Linear Algebra, Springer-Verlag, Fourth edition, 1975.
[2] I. Madsen, J. Tornehave, From Calculus to Cohomology, Cambridge University press, 1997.
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