0

As far as I understand, a vector space is a set + two operations (addition and multiplication by a number) + a set of coefficients. The set is closed with respect to operations, operations satisfy axioms, the set of coefficients is a field.

Well, for example, a set of directed line segments on a plane with the traditional operations of addition and multiplication by a number, and the coefficients from $\mathbb{Q}$ are a vector space, isn't it?

If this is a vector space, then the question about the basis arises. After all, an ordinary pair of non-parallel vectors will not be a basis here. It turns out that we got some kind of infinite-dimensional space for nothing?

In general, debunk my delusions, please...

CatMario
  • 398
  • 1
    Just to be clear, are you describing $\mathbb{Q}^2$? – angryavian Nov 23 '21 at 19:06
  • I don't understand your disagreement in the last paragraph. Can you clarify? – user317176 Nov 23 '21 at 19:09
  • If I understand your question correctly, then you are describing the vector space $\mathbb{Q}^2$ over the scalar field $\mathbb{Q}$. If my understanding is correct, then the vector space is $2$-dimensional, and the basis has cardinality $2$. I am not sure why you think it has an infinite basis. What makes you think a pair of non-parallel vectors will not be a basis here? – Angel Nov 23 '21 at 19:10
  • I encourage you to think about this: I assume you are familiar with the vector space $\mathbb{R}^2$ with respect to $\mathbb{R}$.. And I assume you agree that this vector space is indeed $2$ dimensional. Now, what makes you think that this changes so drastically when you replace $\mathbb{R}$ with $\mathbb{Q}$? Try thinking about that and analyzing your own thought process. – Angel Nov 23 '21 at 19:13
  • @angryavian I mean the vector space $_{\mathbb{Q}}\mathbb{R}^2$. – CatMario Nov 23 '21 at 19:21
  • How do you add two directed line segments (if they are not "based at the origin", which is a condition you omit)? – Arturo Magidin Nov 23 '21 at 19:26
  • @ArturoMagidin but what has the origin to do with it? The vectors are free, they are not attached to the points... and while there is no basis, no one is talking about the origin, and the space of directed segments is already being built. – CatMario Nov 23 '21 at 19:38
  • So I'm asking: how do you add two such "vectors"? If I give you the line segment from $(1,1)$ to $(1,2)$, and then the line segment from $(0,1)$ to $(2,-1)$, what is their sum? Where does it "start" and where does it "end"? The reason I bring up the origin is that, usually, we do not consider the set of "directed line segments on the plane" to be a vector space. We consider the set of directed line segments that being at the origin to be a vector space. Remember that to describe a vector space you need to give not only what the vectors are, but also how to add and scale them. – Arturo Magidin Nov 23 '21 at 19:38
  • "... and while there is no basis..." First, I am not convinced you have correctly described a vector space. And second, if it is a vector space, then you have not established that "there is no basis". – Arturo Magidin Nov 23 '21 at 19:41
  • "that being at the origin" should be "that begin at the origin", but it is too late to edit. – Arturo Magidin Nov 23 '21 at 19:47
  • If you mean the usual vector space $\mathbb{R}^2$ (ordered pairs of the form $(a,b)$ with $a,b\in\mathbb{R}$, coordinatewise addition and scalar multiplication), but with coefficients restricted to $\mathbb{Q}$, you are correct that the dimension is inifnite; whether a basis exists or not depends on your set theory. Having an infinite dimension is not a problem in and of itself: the vector space of all polynomials $\mathbb{R}[x]$ has infinite dimension (and basis $1,x,x^2,\ldots,x^n,\ldots$). Not sure what "for nothing" means, though. – Arturo Magidin Nov 23 '21 at 19:49
  • @ArturoMagidin Why will we consider vectors emerging from only one point? We have vectors on a plane. We define the concept of equality of vectors and then there are no problems at all. – CatMario Nov 23 '21 at 19:52
  • @ArturoMagidin "The reason I bring up the origin is that, usually, we do not consider the set of "directed line segments on the plane" to be a vector space." - perhaps you are used to this approach. I am used to determining the coordinates of points through the coordinates of vectors. I'm guessing it's just a figure of speech?) – CatMario Nov 23 '21 at 19:55
  • 1
    I'm okay with you using whatever approach you want, but then you need to be precise. I gave you two directed line segments. How do you add them? How do you define "equality of vectors"? The government doesn't like it when I read minds without a warrant, so I need you to say it explicitly and not just say "It's what I'm thinking". – Arturo Magidin Nov 23 '21 at 19:55
  • 1
    Note that if you consider, for example, the line segment from $(1,1)$ to $(1,2)$ to be "the same vector" as the line segment from $(0,1)$ to $(0,2)$, then you are not considering the set of directed line segments: in that set, the two line segments are different. You are actually considering a quotient set, the partition induced by an equivalence relation on the set of line segments. That's a very different object from "the set of directed line segments." That's not a figure of speech, but a rather important mathematical distinction. – Arturo Magidin Nov 23 '21 at 19:59
  • "And second, if it is a vector space, then you have not established that "there is no basis"." - I'm used to the fact that first the concept of a vector space is defined and only then it is said about its basis. – CatMario Nov 23 '21 at 20:04
  • @ArturoMagidin Directional line segments are translated in parallel so that they have one starting point ... and are added, for example, according to the parallelogram rule ... right? – CatMario Nov 23 '21 at 20:10
  • And I don’t understand what you’re getting at... want me to write a couple of pages from the textbook here?

    My vector is a "stick with an arrow", addition - the parallelogram rule or the triangle rule. If you want to consider the isomorphic space R^2 instead of "sticks with arrows", then the coordinates by points are determined by the difference between the end and the beginning. The sum is the addition of coordinates. Did I answer your questions?

     – CatMario Nov 23 '21 at 20:13
  • 3
    So then your vector space does not consist of the set of all directed line segements, because the line segment from $(0,0)$ to $(1,1)$ is "the same" as the line segment from $(1,1)$ to $(2,2)$. And so, you are taking a quotient set modulo an equivalence relation. And thus, you can take a single representative from each equivalence class instead. And the standard representative is the vector that begins at $(0,0)$, and is described exclusively by their terminal points... and so you end up with the usual description. – Arturo Magidin Nov 23 '21 at 20:13
  • 1
    The sum is not "addition of coordinates" until you translate them so that they all begin at the origin. Yet you objected vocally to my talking about the vectors all beginning at the origin. I don't want you to copy pages from the book, I just wanted you to be clear and explicit. Note: Your vector space does not consist of all directed line segments, it consists of equivalence classes of directed line segments. And so they can be represented as the set of vectors starting at the origin, exactly as I said and what you objected to. – Arturo Magidin Nov 23 '21 at 20:15
  • So, now that I know exactly what your vector space is, which is essentially the set of ordered pairs $(a,b)$ with $a,b\in\mathbb{R}$ with coordinatewise addition and scalar multiplication, if you restrict the scalars to $\mathbb{Q}$ then you do not have any misunderstandings: this vector space is infinite dimensional over $\mathbb{Q}$. In fact, the real numbers, by themselves are infinite dimensional over $\mathbb{Q}$. For example, see here. – Arturo Magidin Nov 23 '21 at 20:18

2 Answers2

2

The vector space $\mathbb{R}^2$ of ordered pairs of real numbers, considered as a vector space over $\mathbb{Q}$, is infinite dimensional. You do not have any misunderstandings or delusions there. It is not difficult to exhibit an infinite linearly independent subset, since in fact we can do that for $\mathbb{R}$ considered as a vector space over $\mathbb{Q}$; so just take the ordered pairs of the form $(\sqrt{p},0)$ with $p$ ranging over all primes to exhibit an infinite $\mathbb{Q}$-linearly independent set of vectors. This suffices to show that this vector space does not have a finite basis.

Whether this vector space has a basis at all, though, depends on your set theory. In standard set theory, the Axiom of Choice guarantees that every vector space has a basis. In fact, the Axiom if Choice is equivalent to the statement that "Every vector space has a basis".

But there are models of set theory without the Axiom of Choice in which $\mathbb{R}$ does not have a basis over $\mathbb{Q}$. In those models, $\mathbb{R}^2$ does not have a basis over $\mathbb{Q}$ either, even though you can exhibit infinitely many linearly independent vectors.

Again, there is nothing to debunk: one can prove in Zermel-Fraenkel Set Theory that $\mathbb{R}^2$ is not finite dimensional over $\mathbb{Q}$.

Arturo Magidin
  • 417,286
1

Let $\mathbb{K}$ a general field (such as $\mathbb{R}$ or $\mathbb{Q}$). We define vector space $\mathbb{V}$ and we write $(\mathbb{K}, +,\cdot)$ a structure where are defined the following operations:

$+:\, \mathbb{V}\cdot\mathbb{V}\to\mathbb{V}$ that is known as vector addition: $\forall v, w \in \mathbb{V}\,\, v+w$.

$\cdot\,: \mathbb{K}\cdot\mathbb{V}\to \mathbb{V}$, known as scalar multiplication: $\forall v\in \mathbb{V} \forall k\in \mathbb{K}\,\, k\cdot v$.

$\mathbb{Q}$ is a a field, but $\mathbb{N}$ is not. (because it's not close with sum). Thus, $(\mathbb{Q}, +,\cdot)$ is a vector space.

Matteo
  • 6,661