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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.

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The positive real numbers are a vector space over the reals with addition $x\oplus y = xy$ and scalar multiplication $\alpha\odot x = x^{\alpha}$. But this is still "numbers". And of course, every vector space is isomorphic to either $F^X$ or $F^{(X)}$ for some set $X$ and some field $F$... –  Arturo Magidin May 8 '11 at 18:43
    
@Arturo yes, that last part is true, but this is aimed more at the elementary level (probably not even involving vector spaces over any field other than the reals) –  Eugene Bulkin May 8 '11 at 18:46
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@Eugene: I guess my attempted point was that no matter how "non-numerical" a vector space might "look", it can always be "thought of" as a set of "numerical vectors" (when viewed from a certain angle)... –  Arturo Magidin May 8 '11 at 18:47
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@Arturo: sure, but I think Eugene is looking for examples where that angle isn't the first thing a student would try. For example, a construction of a vector space where it's not obvious how to specify a basis. –  Qiaochu Yuan May 8 '11 at 19:21
    
yeah. I'm especially interested to see if anyone can come up with a vector space where the vectors are something really strange, like pictures or shapes or something weird like that! also, for some reason I can't mark this as community wiki; can a moderator do so, please? –  Eugene Bulkin May 8 '11 at 22:24
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6 Answers 6

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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).

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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.

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This is sort of a weird construction because you seem to be taking scalar multiplication in a subtractive colour model but vector addition (blue + orange = white) in additive colour. –  Rahul Jun 18 '13 at 0:42
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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.

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It might be better to just let $G$ be a set: the group structure is quite irrelevant for this, so it can only distract from your point. –  Mariano Suárez-Alvarez May 8 '11 at 22:35
    
@Mariano: The need for finite-ness is also redundant. I shall edit. –  Asaf Karagila May 8 '11 at 22:38
    
I don't understand your last paragraph. What is less straightforward? You take the vector space generated by $G$ and define the multiplication by extending $(g,h) \mapsto gh$ from the basis bilinearly to all of $\mathbb{C}[G]$. And the thing is called group ring or group algebra regardless of whether $G$ is finite or not and it is also useful and interesting for infinite groups. –  t.b. May 8 '11 at 22:50
    
@Theo: Taking "just all formal sums" is slightly more straightforward (in my view) than starting to deal with a support (e.g. compact, finite, countable, etc...) for. I can't completely recall the definitions in the infinite case, although I'm certain I have studied some of the basic LCA case. In my mind, this is a proof of less-straightforward-ness. But who am I? A set theorist in a world of algebraic analysis :-) –  Asaf Karagila May 8 '11 at 22:54
    
:D Yeah, you poor oppressed set-theorist [obligatory Python-quote: "Help help, I'm being oppressed! - Now you see the violence inherent in the system"]. But you're just taking the group elements as a basis, hence you're only taking finite combinations of the basis elements. The correspondence between representations of $G$ and $\mathbb{C}[G]$-modules holds verbatim. We're not talking about Haar measure, $L^{1}(G)$ and the like. Ok, I see a possible confusion: You want to say the group algebra is not $\prod_{G} \mathbb{C} = \mathbb{C}^{|G|}$ but rather $\bigoplus_{G} \mathbb{C}$... –  t.b. May 8 '11 at 23:06
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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.

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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.

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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. –  Qiaochu Yuan May 8 '11 at 19:31
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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.

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Well, this is more of an example of a $\mathbb{Z}$-module (abelian group) than a vector space, unless we are going to talk about $\pi$ chairs + $\sqrt{2}$ tables. Also, this reminds me of a great joke on MO. –  Zev Chonoles May 8 '11 at 18:54
    
Yeah, fair enough. I guess $\mathbb{Z}^+ \cup \{0\}$ isn't exactly a field. –  badatmath May 8 '11 at 19:00
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