free vector space over a set

In this entry we construct the free vector space over a set, or the vector space 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 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

$C(X)$ $=$ $\mathrm{\{}f:X\to 𝕂|f^{-1}(𝕂\backslash \{0\})\text{is finite}\}.$

In other words, $C(X)$ consists of functions $f:X\to 𝕂$ 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 $x\mapsto f(x)+g(x)$. Similarly, if $f\in C(X)$ and $\alpha \in 𝕂$, then $\alpha f$ is the mapping $x\mapsto \alpha f(x)$. It is not difficult to see that these operations are well defined, i.e., both $f+g$ and $\alpha f$ are again functions in $C(X)$.

0.0.1 Basis for $C(X)$

If $a\in X$, let us define the function ${\mathrm{\Delta }}_a\in C(X)$ by

${\mathrm{\Delta }}_a(x)$ $=$

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

$C(X)$ $=$ $span{\{{\mathrm{\Delta }}_a\}}_{a\in X}.$ (1)

Here, the space $span{\{{\mathrm{\Delta }}_a\}}_{a\in X}$ consists of all finite linear combinations of elements in ${\{{\mathrm{\Delta }}_a\}}_{a\in X}$. It is clear that any element in $span{\{{\mathrm{\Delta }}_a\}}_{a\in X}$ is a member in $C(X)$. Let us check the other direction. Suppose $f$ is a member in $C(X)$. Then, let ${\xi }_1,\mathrm{\dots },Ã\mathrm{‚}Â\mathrm{ }{\xi }_N$ be the distinct points in $X$ where $f$ is non-zero. We then have

$f$ $=$ ${\displaystyle \sum _{i=1}^N}f({\xi }_i){\mathrm{\Delta }}_{{\xi }_i},$

and we have established equality in equation 1.

To see that the set ${\{{\mathrm{\Delta }}_a\}}_{a\in X}$ is linearly independent, we need to show that its any finite subset is linearly independent. Let $\{{\mathrm{\Delta }}_{{\xi }_1},\mathrm{\dots },{\mathrm{\Delta }}_{{\xi }_N}\}$ be such a finite subset, and suppose ${\sum }_{i=1}^N{\alpha }_i{\mathrm{\Delta }}_{{\xi }_i}=0$ for some ${\alpha }_i\in 𝕂$. Since the points ${\xi }_i$ are pairwise distinct, it follows that ${\alpha }_i=0$ for all $i$. This shows that the set ${\{{\mathrm{\Delta }}_a\}}_{a\in X}$ is linearly independent.

Let us define the mapping $\iota :X\to C(X)$, $x\mapsto {\mathrm{\Delta }}_x$. This mapping gives a bijection between $X$ and the basis vectors ${\{{\mathrm{\Delta }}_a\}}_{a\in X}$. We can thus identify these spaces. Then $X$ becomes a linearly independent basis for $C(X)$.

0.0.2 Universal property of $\iota :X\to C(X)$

The mapping $\iota :X\to C(X)$ is universal in the following sense. If $\varphi$ is an arbitrary mapping from $X$ to a vector space $V$, then there exists a unique mapping $\overline{\varphi }$ such that the below diagram commutes:

Proof. We define $\overline{\varphi }$ as the linear mapping that maps the basis elements of $C(X)$ as $\overline{\varphi }({\mathrm{\Delta }}_x)=\varphi (x)$. Then, by definition, $\overline{\varphi }$ is linear. For uniqueness, suppose that there are linear mappings $\overline{\varphi },\overline{\sigma }:C(X)\to V$ such that $\varphi =\overline{\varphi }\circ \iota =\overline{\sigma }\circ \iota$. For all $x\in X$, we then have $\overline{\varphi }({\mathrm{\Delta }}_x)=\overline{\sigma }({\mathrm{\Delta }}_x)$. Thus $\overline{\varphi }=\overline{\sigma }$ since both mappings are linear and the coincide on the basis elements.$\mathrm{\square }$

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