Given a subset $S$, is there any universal construction (property) of $L(S)$, the linear span of $S$?

Question

Given a subset $S$ of a vector space $V$, is there any universal construction (property) of $L(S)$, the linear span of $S$ (like the way we can construct the fraction field of an integral domain)?

Answer 1

$L(S)$ is the image of the map $F(S)\to V$ induced by the universal property from the inclusion $S\hookrightarrow V$ ($F(S)$ being the free vector space spanned by $S$). the image of a map itself is defined as the kernel of the cokernel, which both are described by universal properties.
for example, the kernel of $f:V\to W$ is an object $V'$ with a map $i:V'\to V$ (nessecarily monomorphism) such that for every $g:U\to V$ with $fg=0$, $g$ factors through $i$. the cokernel is defined dually.
I don't know if it's possible to encapsulate all these three construction in one.

Answer 2

The span of $S$ is the smallest subspace of $V$ that contains $S$ as a subset.

If you want something that look more categorical, we can say that whenever $W$ is a vector space and $f:S\to W$ is a map that respects linear combinations in the sense that $a_1s_1+\cdots+a_ns_n=0$ implies $a_1f(s_1)+\cdots+a_nf(s_n)=0$, then $f$ extends so a linear transformation $L(S)\to W$ in exactly one way. And $L(S)$ is the only subspace of $V$ that has this property.

Answer 3

If $S$ is linearly independent, then $L(S)$ is the free vector space on $S$, i.e. we have the universal property $\hom(L(S),W) \cong \hom(S,|W|)$ for vector spaces $W$; here $|W|$ denotes the underlying set. If $S$ is arbitrary, you cannot really specify a universal property of $L(S)$ inside the category of vector spaces directly. We know from linear algebra that $S$ contains a linearly independent subset $S'$ which still generates $L(S)$, and $L(S)=L(S')$ is the free vector space on $S'$. For a universal property inside the partial order of subspaces of $V$ (which you may regard as a thin category), see the answer by Henning Makholm.