# 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 . 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 $\mathbb{K}$, we define

 $\displaystyle C(X)$ $\displaystyle=$ $\displaystyle\{f:X\to\mathbb{K}\,\,|\,\,f^{-1}(\mathbb{K}\backslash\{0\})\,% \mbox{is finite}\}.$

In other words, $C(X)$ consists of functions $f:X\to\mathbb{K}$ that are non-zero only at finitely many points in $X$. Here, we denote the identity element  in $\mathbb{K}$ 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 $\alpha\in\mathbb{K}$, 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 $\Delta_{a}\in C(X)$ by

 $\displaystyle\Delta_{a}(x)$ $\displaystyle=$ $\displaystyle\left\{\begin{array}[]{ll}1&\mbox{when}\,x=a,\\ 0&\mbox{otherwise.}\\ \end{array}\right.$

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

 $\displaystyle C(X)$ $\displaystyle=$ $\displaystyle\mathop{\mathrm{span}}\{\Delta_{a}\}_{a\in X}.$ (1)

Here, the space $\mathop{\mathrm{span}}\{\Delta_{a}\}_{a\in X}$ consists of all finite linear combinations  of elements in $\{\Delta_{a}\}_{a\in X}$. It is clear that any element in $\mathop{\mathrm{span}}\{\Delta_{a}\}_{a\in X}$ is a member in $C(X)$. Let us check the other direction. Suppose $f$ is a member in $C(X)$. Then, let $\xi_{1},\ldots,Â \xi_{N}$ be the distinct points in $X$ where $f$ is non-zero. We then have

 $\displaystyle f$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{N}f(\xi_{i})\Delta_{\xi_{i}},$

and we have established equality in equation 1.

To see that the set $\{\Delta_{a}\}_{a\in X}$ is linearly independent, we need to show that its any finite subset is linearly independent. Let $\{\Delta_{\xi_{1}},\ldots,\Delta_{\xi_{N}}\}$ be such a finite subset, and suppose $\sum_{i=1}^{N}\alpha_{i}\Delta_{\xi_{i}}=0$ for some $\alpha_{i}\in\mathbb{K}$. Since the points $\xi_{i}$ are pairwise distinct, it follows that $\alpha_{i}=0$ for all $i$. This shows that the set $\{\Delta_{a}\}_{a\in X}$ is linearly independent.

Let us define the mapping $\iota:X\to C(X)$, $x\mapsto\Delta_{x}$. This mapping gives a bijection between $X$ and the basis vectors $\{\Delta_{a}\}_{a\in X}$. 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 $\phi$ is an arbitrary mapping from $X$ to a vector space $V$, then there exists a unique mapping $\bar{\phi}$ such that the below diagram commutes:

 $\xymatrix{X\ar[r]^{\phi}\ar[d]_{\iota}&V\\ C(X)\ar[ur]_{\bar{\phi}}&}$

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

## References

Title free vector space over a set FreeVectorSpaceOverASet 2013-03-22 13:34:34 2013-03-22 13:34:34 mathcam (2727) mathcam (2727) 8 mathcam (2727) Definition msc 15-00 vector space generated by a set TensorProductBasis