14
$\begingroup$

I've recently been thinking about why my peers and other people I've helped learn vector spaces had trouble intuitively understanding the concept, and it occurred to me that non-numerical (i.e. nothing like $\langle 3,2,3 \rangle$ or obvious addition/multiplication operations) examples could reinforce intuition. For example, a huge problem was understanding that a vector space is simply a set of vectors with two operations that follow 10 axioms, and that a zero vector isn't necessarily all zeroes, and so on.

Does anyone have any great examples of vector spaces (and the vectors and operations in them, of course) that are non-numerical, and thus can't lead to those trying to prove their validity to being stuck in ruts (like assuming the zero vector is all zeroes, that the inverse vector is the negative scalar multiple, etc.)? Pictures, letters, and any others would certainly be interesting!

Note: I know that there are questions about vector spaces with unusual (and only partially valid) characteristics or out of the ordinary operations, but I'm looking for examples that have minimal numbers involved, to remove all automatic assumptions that are involved in them.

  • 0
    @Eugene: for any set $S$ the set of subsets of $S$ forms a vector space over $\mathbb{F}_2$ where addition is given by symmetric difference. In particular if $S = \mathbb{R}^2$ elements of this vector space include pictures and shapes. But this is still, in some sense, a "numerical" example, since one is really just talking about functions $S \to \mathbb{F}_2$. So I didn't include it in my answer.2011-05-08

6 Answers 6

14

A simple example is to take $\mathbb{R}^n$ but to fix a vector $w$ and modify scalar multiplication to $a \otimes v = a (v - w) + w$ and addition to $u \oplus v = u + v - w$. This is just the usual vector space structure on $\mathbb{R}^n$, but shifted by $w$, and in my experience many students have a lot of trouble with these kinds of examples; they have never really learned to think in a translation-invariant way. If nothing else, this example should quickly diagnose the problem you mention about the zero vector (which is of course $w$ here).

Perhaps a more "non-numerical" example is to take the space of solutions to a linear homogeneous differential equation or recurrence relation, such as y'' - 3y' - 2y = 0 or $a_{n+3} = a_{n+1} + a_n$. While the zero vector is in some sense "all zeroes" in these examples, I like them because it's not immediately obvious how to write down a basis for these spaces (or, having done so, it's not obvious that you've chosen a useful one).

7

I like function spaces for this example (pick your favorite kind). It is easy to see that it is a vector space, but a basis might be out of reach. Restricting to polynomials gives you a nice and obvious basis though.

3

I think that I first realized that vector spaces can be really strange creatures when I took a course in representation theory.

Consider $G$ a a set $\mathbb C[G] = \{\sum_{g\in G} \alpha_g g\mid \alpha_g\in\mathbb C\}$

This is a collection of formal sums. $G$ is a set and has no summation defined on, its elements are not numbers so there is really no "numerical result" of this sort of $\displaystyle\sum_{g\in G}\alpha_g g$.

The addition is defined naturally as: $\sum_{g\in G} \alpha_gg + \sum_{g\in G} \beta_gg = \sum_{g\in G} (\alpha_g+\beta_g)g$

And the scalar multiplication is very similar too.

This can be thought at as all the functions from $G$ into $\mathbb C$, again not a numerical vector space.

When $G$ is finite, this is isomorphic to $\mathbb C^{|G|}$ which is just a collection of $n$-tuples, which you essentially try to avoid. However the fact remains valid, vector spaces are "simple" creatures, and under the assumption they have a basis you can always just assume they are tuples (of the right length) and return to the "known" domain from which you are attempting to escape.

If $G$ is also a group, then this construction is called Group ring and it has an additional structure defined which is very useful for representation theory applications.

  • 0
    @Theo: I am dazed and confused, and you clearly have more idea about that bit than I do. Feel free to correct and edit things into my answer. I have to get to a Bocalist meeting in Minehead... :-)2011-05-08
2

When I tried to explain bases to my friend I used the set of all furniture in a room. I simplified it by saying it was just chairs and tables. The basis vectors would be a table and a chair. Multiplication would be like a(2 chairs) = 2a chairs and addition would be 2 tables and 3 tables = (2 + 3) tables.

  • 0
    Yeah, fair enough. I guess $\mathbb{Z}^+ \cup \{0\}$ isn't exactly a field.2011-05-08
1

In this Wikipedia article they mention a vector space of words of fixed length n, it can also be provided with a metric (the hamming distance) to make it a normed vector space.

  • 0
    This is still a vector space coming with a distinguished basis (the words with one nonzero element); in particular, the zero vector is all zeroes.2011-05-08
1

What about we take all colours and think of it as a vector space (with basis say, red, blue and green)? The scalars would be related to the intensity of that light and the zero vector would be white. Think of -(blue) as orange.

  • 0
    This is sort of a weird construction because you seem to be taking scalar multiplication in a [subtractive colour model](https://en.wikipedia.org/wiki/Subtractive_color) but vector addition (blue + orange = white) in [additive colour](https://en.wikipedia.org/wiki/Additive_color).2013-06-18