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I was reading some online notes on vector spaces and one authors insisted on turning a set $\mathbb{X}$ into a vector space. I thought it was quite insane but maybe I am not seeing the point.

The example went along the line of something like:

Given a set $\mathbb{X} = \mathbb{R^2}$, let $x = (u_1,u_2), u_1,u_2\in \mathbb{R}$

Define $(u_1,u_2)$ + $(v_1,v_2)$ = $(u_1+v_1, u_2+v_2)$, $\alpha (u_1,u_2) = (\alpha u_1, \alpha u_2)$

Then $\mathbb{X} = \mathbb{R^2}$ becomes a vector space over $\mathbb{R}$

... Is it really necessary to construct a vector space out of the set of $\mathbb{R^2}$? Can't we simply just use $\mathbb{R^2}$ as a vector space without forcing ourselves to define operations on $\mathbb{R^2}$ everytime we wish to treat it as a vector space?

Is this too much rigor or is a standard practice?

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    $\begingroup$ If we're going to add together two elements of $\mathbb R^2$, then we have to define what we mean by that. $\endgroup$ – littleO Aug 7 '15 at 7:36
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    $\begingroup$ Wehre in the online notes does this example occur? Relatively soon after the intorduction of the concept of a vector space? I.e., was $\mathbb R^ 2$ ever introduced as a vector space before that? $\endgroup$ – Hagen von Eitzen Aug 7 '15 at 11:43
  • $\begingroup$ @FemaleTank : As someone who moved from physics into more pure mathematics, I can tell you that the applied / physics way of thinking, where we don't define everything and just assume we are all talking about the same thing, can sometimes lead to serious problems. You yourself bring up the issue of norms (in your comment below Aloizio's answer). A single space can admit many different norms, and sometimes the one that you know best is not the one that you should use, depending on the problem. $\endgroup$ – MathsByTheSea Aug 7 '15 at 14:54
  • $\begingroup$ How is this a vector space, it's not a vector space without those operations. So of course you need to know them. $\endgroup$ – Alec Teal Aug 7 '15 at 15:26
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When people say $\mathbb{R}^2$, it is commonly implicit all its "canonical" structure (as a vector space, topological space, metric space etc).

But you must know what this canonical structure is!

Imagine you enter a room where everyone knows Bob. You also know Bob. But everyone also knows Bob's dog, which everyone simply assumes everybody else knows: it is a simple dog, after all! It so happens that you've never been introduced to Bob's dog, nor seen in photos etc. You only know that it is a dog. You can even say: Oh, okay, Bob has a dog. But you don't know its color, or if it eats a lot: you don't know its characteristics/structure. What will you do if you need to buy a present for Bob's dog? No matter how simple it is, you still need to know Bob's dog to know its "structure". After you've been introduced, you can enter the "club", and talk about Bob, with an underlying, well-known dog.

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    $\begingroup$ It is just a huge leap from applied math to analysis type of math where you need to define everything that I am not comfortable with. Another example would be, if I was given two points and told to compute ||u-v||, I wouldn't hesitate to use the 2 norm, but in analysis || || is standard notation for all types of norms ...should I ask my professor what norm he is talking about or would he assume I am smart enough to figure this all out. Should I ask him whether he wants it to be a metric space or a normed space? A complete one? It just feels that math is out of my control at this point $\endgroup$ – Shamisen Expert Aug 7 '15 at 7:43
  • $\begingroup$ It is like an "assumption fest". Never ending amount of assumption before you actually get to something interesting. $\endgroup$ – Shamisen Expert Aug 7 '15 at 7:45
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    $\begingroup$ @FemaleTank Unless otherwise specified, use the "canonical structure", but it is never wrong to ask for clarification. $\endgroup$ – user223391 Aug 7 '15 at 7:46
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    $\begingroup$ Haha... take it easy! It is not out of your control. Maybe you are a little overwhelmed with a lot of possible structures. Knowing "what norm he is talking about" most of times is not a matter of "being smart", but a matter of context. You'll get used to that. But keep in mind that if you are seeing things like this, although it may seem like pedantism at first sight, it is actually for a good cause: there will be times where the structure you will want is not at all clear. or times where you will have to deal with multiple structures (even on the same space) at once. $\endgroup$ – Aloizio Macedo Aug 7 '15 at 7:48
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    $\begingroup$ Perhaps introductory courses will stress rigor a tad strongly, but we need to make sure students are capable of being rigorous. It is a learned skill that requires practice for most. Though, your issue with functional norms is perhaps more an issue with generalization than it is with rigor. Also, "assumptions" are not the same thing as definitions, explanations and clarifications, so don't call them that. $\endgroup$ – whacka Aug 7 '15 at 7:55
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The notation $\Bbb R^2$ already refers to the vector space you're referring to, so there's no need to define the vector space structure every time you use the notation. But we do need to define everything, even common notation, at least once in our lives so that terms even have meaning. Whether or not you need to rewrite the definition again in another context depends on your purposes, the audience, and the how standard your terms or notations are. In other words, just like words themselves.

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    $\begingroup$ "The notation $\mathbb{R}^2$ already refers to the vector space" I'm not sure that's really true. For example, in contexts where the vector space structure isn't even being considered, I'd argue that $\mathbb{R}^2$ just stands for the naked set of points $\{(x,y)\mid x,y\in \mathbb{R}\}$. It wouldn't seem sensible to use $\mathbb{R}^2$ to stand for any vector space other than the canonical one, but I don't think you can claim it always stands for the vector space. $\endgroup$ – David Richerby Aug 7 '15 at 10:15
  • $\begingroup$ Point taken, but maybe splitting hairs. That it refers to one thing doesn't mean it can't refer to other things in other contexts too, so there's no issue. $\endgroup$ – whacka Aug 7 '15 at 20:55
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    $\begingroup$ Your phrasing can be read as meaning that it only refers to the vector space, as you can see me assuming in the last sentence of my comment. $\endgroup$ – David Richerby Aug 7 '15 at 21:11
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So most of us encounter $\mathbb R^2$ as the cartesian plane in which we create axes, sketch functions, explore geometry, encounter vectors etc. It is a familiar object.

Then we meet, in algebra, the abstract concept of a vector space. We understand the idea - one thing we want to do is to take what we've learned in $\mathbb R^2$ and other examples which are familiar to us, and to use that knowledge in a more general context. We perhaps want to explore more dimensions, and need to know whether there are things which are special about two dimensions which won't apply in three. We encounter the idea of complex vector spaces, and someone perhaps tells us that the whole idea has proved fruitful in abstract algebra.

So we make a definition in abstract. But does the definition in fact relate to our original and most familiar example? Will it help us to make the conceptual leap from the concrete to the abstract? Let's check that it works ... and then we're ready to go.

As others have said, we don't need to do it every time - but we do need to do it at least once.

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There is never too much rigor in math. ;)

A vector space is a set of vectors over a set of scalars, with two operations: vector addition and scalar multiplication. If you don't have operations, you don't have a vector space.

And you will discuss different operations on $\Bbb{R}^2$ other than the standard addition and multiplication, so yes it is necessary to be explicit with what operations you are talking about.

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To use language effectively, you have to consider your audience...

In the context of an introductory upper-division course on linear algebra, it's good to be explicit about the vector space operations. After all, they do have to be defined, and there's value in emphasizing that point.

In a professional context, it would be insane. The reader knows what you mean if you call $\mathbb R^2$ a vector space. Acting as if they didn't would be an insult and a waste of time.

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    $\begingroup$ I think my linear algebra professor wrote "$\mathbb{R^2}$ is a vector space" on the first day of college and that was the end of my obsession with defining $+$ and $\times$ on that set. Unsuprisingly it seems as crazy to me as having to prove "1+1 = 2" using Russell's set theory while doing your student loan debt $\endgroup$ – Shamisen Expert Aug 7 '15 at 7:59
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    $\begingroup$ @Female Tank: Russell didn't have a set theory. He developed with Whitehead type theory. $\endgroup$ – Asaf Karagila Aug 7 '15 at 9:16
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This is Mark Bennet's answer, with a different emphasis. I am also operating under the assumption, like many others, that you saw this toward the beginning of a set of linear algebra, abstract algebra, or analysis notes. I agree with you that the check that $\Bbb R^2$ is a vector space is generally boring, obvious, and tedious, but I posit that, in this context, it still has value.


One reason that we go through the rigor of showing that $\Bbb R^2$ is a vector space is to provide some evidence that a vector space "is what you think it is". Or, if you don't think anything of it, to give you some idea of what you should think it is

A vector space is, after all, an object with two operations satisfying nine axioms, and an additional object called a field which is an object with two operations satisfying ten axioms. (Numbers varying depending on how you count, of course). But it's very easy to read this list and then come out at the end without having any clue what the heck a vector space is.

A very good reason that we go through the (boring, tedious, and obvious) effort of showing that $\Bbb R^2$ is a vector space is because we need some indication that these axioms have any sort of relationship to any object we care anything about. Most likely, the author is assuming you know what $\Bbb R^2$ is, and how to add vectors, and so on, and is using that past knowledge to suggest that vector spaces in general might be something that you should care about as well.

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$\mathbb{R}^2$ simply means the cartesian product of $\mathbb{R}\times\mathbb{R}$, which is the set of all ordered pairs $(a,b)$ with $a,b\in\mathbb{R}$. This is not a vector space, this is not a metric space, itis nothing but a set of ordered pairs. You must always define what operations means for it to be meaningful. What does $a+b$ mean for $a,b\in\mathbb{R}^2$? Unelss defined we don't know, you can't make assumptions that are unwarrented but must clearly define it which we do most often by utilizing previous structures where it is already defined. That is why you must define it, I can for example define a vector space like $$(a,b)+(c,d)=(2a+2c,b+d)$$ and $$a(b,d)=(2ab,ad)$$ and so on, it'll be peculiar but it'll work and we don't know it unless stated.

Edit: As someone pointed out this is not a proper vector space as it violates one of the properties but it was made on the top of my head so I didn't check.

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    $\begingroup$ You're correct on the example is poor, I just made some example on the top of my head. As for the rest you are however it is wrong. We need to define it because if you've never encountered it, how would you know? If I gave you $\mathbb{A}^2$ you'd have no clue what operations mean or anything. I have seen vector spaces and modules being very strange made out of very peculiar objects where the definitions are not necciserly intuitive so we ALWAYS need to define it. However for $\mathbb{R}^n$ the standard is implied unless specified but it's otherwise nothing but a set of ordered n-tuplets. $\endgroup$ – Zelos Malum Aug 7 '15 at 8:38
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    $\begingroup$ If $\mathbb{A}$ had any sort of algebraic structure, operations would assume to carry over componentwise unless otherwise stated (or if we had to define new operations). For example if $\mathbb{K}$ is a field, there would be no problem with someone saying $\mathbb{K}^{d}$ is a vector space. Of course there are nonstandard ones, in which case we should say what we are doing and show it is a vector space. $\endgroup$ – Morgan Rodgers Aug 7 '15 at 8:42
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    $\begingroup$ Those are assumptions you're making and in a rigorous setting you CANNOT make them. In informal cases there are usually no issue and I have no real problem with it either but it's important to know that unless specified you may have people making lots of unwarrented assumptions that may conflict. They might not agree amongst themselves and even less with you as the speaker or author. $\endgroup$ – Zelos Malum Aug 7 '15 at 8:44
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    $\begingroup$ I'll grant you the example is complete and utter shite. However if you ARE in a rigorous setting and you're that sloppy you're more likely to get people against you and think of you as unprofessional or even incompetent. Making mistakes like mine is fine and in rigorous settings it is appriciated to be corrected, which I am here. $\endgroup$ – Zelos Malum Aug 7 '15 at 8:50
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Stating it in this way is EXTREMELY pedantic. Okay, $\mathbb{R}^{2}$ is technically a set (Cartesian product), but when there is already an algebraic structure, addition and scalar multiplication can just be assumed to be componentwise unless we are explicitly told otherwise. If you really wanted to make this clear, we could just say "$\mathbb{R}^{2}$ is a vector space under componentwise addition and scalar multiplication".

The only possible reason to do this.... maybe if you JUST defined a vector space to someone who had never heard the term before, and then were immediately proceeding with an example.

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    $\begingroup$ I think that's the point, we begin by defining $\mathbb{R}^2$ to be the Cartesian product, with no prior extra structure. Then we say, oh look, we can make it a vector space by giving it these addition and scalar multiplication rules. One doesn't "know" that there is "already an algebraic structure" until one explicitly shows what that structure is, which I think is the reason the author in the OP's question did things the way (s)he did. $\endgroup$ – Josh Chen Aug 7 '15 at 8:49
  • $\begingroup$ $\mathbb{R}$ already has an algebraic structure. It carries over. $\endgroup$ – Morgan Rodgers Aug 7 '15 at 8:52
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    $\begingroup$ Well, of course, but what does it mean to say it "carries over"? In this case, precisely that we can do things componentwise. But for a first-timer seeing these things, we should show this explicitly. $\endgroup$ – Josh Chen Aug 7 '15 at 8:54
  • $\begingroup$ I do say that in my answer. If you had IMMEDIATELY seen the term "vector space" for the first time, sure, point out the details for $\mathbb{R}^{2}$. Otherwise, "carries over" is pretty standardly componentwise for a Cartesian product. $\endgroup$ – Morgan Rodgers Aug 7 '15 at 8:56

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