Recently, I have been having trouble understanding the concept of a "free vector space." Several google searches have resulted in some very vague explanations. All I have to go off of is a friend's lecture notes which denote some definitions and goes over some of the consequences of having a free vector space. These are the definitions I was given:

  • 1) Definition of the Free Vector Space

"Let $S$ be a set, and $K$ be a field. The free vector space generated by $S$, $K$$\langle{S}\rangle$, is the subspace of $Fun(S,K)$ (the set of all functions from $S$ to $K$) spanned by the characteristic functions $\chi_s$, for $s \in S$."

Just to clarify, the characteristic function is defined as:

$$\chi_s(x) = \begin{cases} 0, & \text{if $x$ $\neq$ s} \\ 1, & \text{if $x$ = s} \end{cases} $$

Since any function in $Fun(S,K)$ can be expressed as a linear combination of characteristic functions, this also means that any element within the free vector space is expressed as a linear combination of characteristic functions. That is, for any $k \in K \langle S \rangle $, $k = a_{s_1} \chi_{s_1} + ... + a_{s_n} \chi_{s_n}$, for some $s_1, ..., s_n \in S$.

It is clear to me that these characteristic functions are linearly independent; that is, you cannot write any one of them in terms of the others.

  • 2) "Freeness" of the Free Vector Space

"Let $S$ be a set, $K$ a field, and $V$ a $K$-vector space. Then any function $\psi \colon S \to V$ extends uniquely to a linear transformation $T_{\psi} \colon K \langle S \rangle \to V$."

The notes go on to mention that the reason why we use the term "free" is because we are "free" to define a linear transformation out of $K \langle S \rangle$ by just choosing any vectors for $S$ to be sent to.

  • 3) Alternative Definition of the Free Vector Space

A second definition of the free vector space is also mentioned, though informally. It states that another way to think of $ K \langle S \rangle$ is to think of it as the span of $S$, where we assume that $S$ is its basis.

A final note is that the reason why definitions 1) and 3) are equivalent (and ultimately why we are able to find basis vectors for a vector space) is because the characteristic functions are isomorphic to the elements of S. That is, for each $s \in S , \, \, \, \chi_s \longleftrightarrow s$ $\, \, \,$ (he says it is a "relabeling").

Within all of this, there are a few things I understand. Definition 3) is very intuitive to me - treating the elements of $S$ as basis elements, and treating $K \langle S \rangle$ as its span seems relatively natural - much more so than having to first consider a (potentially very very large) set of functions. This leads me to my first question:

If we are interested in a vector space whose elements are some object, then why are we able to reasonably compare them to functions between those objects and some field?

I understand that using the first definition is helpful because it is more directly related to linear independence, but why are we able to take the first definition as the "main" definition, and then do some strange math to say that these functions are "essentially the same as" the elements of $S$ themselves? I assume that this has to do with the fact that $\chi_s$ seems to be a very reasonable "stand in" for $s$, within $Fun(S,K)$, but I still do not understand why this leads to the elements of $S$ being linearly independent (we are only directly saying things about their related characteristic functions).

The generality of the statement is also somewhat confusing. $S$ can be any set, so if I were to consider the set containing the 3 primary colors, they would be fundamentally different from characteristic functions. How can I reasonably "relabel" the colors "red, blue, and yellow", to "$\chi_{red}$, $\chi_{blue}$, and $\chi_{yellow}$?" There is a section that states that it's ok to do, since a linear combination from either set would have the same structure (a sum of vectors which have been multiplied by scalars), but it still seems problematic. I can plug things into functions, but I can't "plug" anything into the colors red, blue, and yellow. It seems like relabeling would be a problem, since I wouldn't be able to do the same things with $S$ as I could with its related characteristic functions.

My second question has to do with the definition of freeness, as well as the note the writer made:

Just what does the statement "we are free to define a linear transformation out of $K \langle S \rangle$ by choosing vectors for $S$ to be sent to" mean?

More precisely, how are we able to do this? It seems very far-fetched for us to be able to extend ANY function from $S$ to $V$. Does this statement mean that we can define the transformation however we want?

Thank you for your help - I hope you can offer some insight.

  • $\begingroup$ If you are familiar with the language of category theory, this is simply that the forgetful functor ${_k}\mathsf{Vect} \to \mathsf{Set}$ admits a left adjoint, called the free vector space functor. $\endgroup$ – TheGeekGreek May 13 '18 at 14:09

The statement "we are free to define a linear transformation out of $K \langle S \rangle$ by choosing vectors for $S$ to be sent to", can be seen as follows: let $V$ be any vector space and assume we have a function $c: S \to V$, that assigns to each $s \in S$ some element in $V$, totally arbitrarily.

Then we can define a linear function $F_c: K \langle S\rangle \to V$ by $$F_c\left(\sum_{i=1}^n k_i \chi_{s_i}\right) = \sum_{i=1}^n k_i c(s_i)$$

where we define the image of an element of the span of all of the $\chi_s$ (so we have finitely many $\{s_1,\ldots,s_n\}$ from $S$ and finitely many field elements $k_i$ as well, and the right hand sum is taken in $V$. So basically we send $\chi_s$ to $c(s)$ and extend by linearity, as usual.

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