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Introduction

Generally most students are introduced to the concept of Vector as something that has both a "magnitude and direction" and Scalars as something that only has a "magnitude and no direction". Usually this definition of Vectors and Scalars comes straight out of a high-school Physics class, as most students learn about Vectors and Scalars far earlier in Physics than in Mathematics.

However once you get to first and second year undergrad Mathematics courses at Universities, you realize that Vectors and Scalars are really abstract mathematical objects. Once you study further, you realize that there are higher mathematical objects such as Tensors which are generalizations of Vectors and Scalars.

The "Physics Definition" of a Vector

We've all had the "definition" of a Vector drummed into us that : "A vector is a quantity that has both magnitude and direction", but in a Pure Mathematical sense, that is not the case. A Vector is not a "quantity" it's a mathematical object. Likewise a Scalar is not a "quantity" either, but also a mathematical object (e.g a real number).

The Pure Maths Definition of a Vector

I argue that saying a Vector is a "quantity that has both magnitude and direction" takes away the Pure Mathematical intuition behind a Vector. I argue that using this "definition" of a Vector is just a shorthand to help us familiarize ourselves with the concept of Vectors, and to find physical intuition for it. Saying that it has magnitude and direction is just a physical interpretation of it, i.e. what we are doing in this definition is giving "physical context" to an abstract Mathematical object.

How do you define the "direction" of a an abstract Mathematical object, such as a Vector in a Pure Mathematical way? The concept of "direction" is given context by the physical world, and hence not a purely mathematical concept. The closest thing we have to "direction" in pure mathematics is associated with a number or axis being either positive or negative.

(I realize in Vector Calculus that there are concepts like "Directional Derivatives", and the like, but again I argue that calling it that is just a shorthand to work with Vectors without getting too much into their true roots in purely mathematical sense (i.e. in terms of tuples of scalars), as you will see below)

Correcting the Definition of a Vector

I've included a short description of Tuples and a link for further reading to further emphasize my point.

Tuples : A tuple is a finite ordered list of elements. In mathematics, an n-tuple is a sequence (or ordered list) of n elements, where n is a non-negative integer. : https://en.wikipedia.org/wiki/Tuple

Now I know that you could plot a Euclidean vector on a Cartesian plane, but saying it has a "direction" is just our visual/physical interpretation of the vector, the vector itself is just a tuple of Scalars, likewise a Tensor is just a tuple of Vectors (which in turn is just a tuple of Scalars).

For example a vector in $$\mathbb{R^{n}}$$ is just a n-tuple of Scalars (Real Numbers)

\begin{equation} \vec{V}=\begin{bmatrix} \mathbb{R_1} \\ \mathbb{R_2} \\ \mathbb{R_3}\\ ... \\ \mathbb{R_n} \end{bmatrix} \end{equation}

I find that thinking about Vectors the Pure Maths way, helps to bridge the gap and allow an easy transition from learning about numbers (Scalars), to Vectors to Tensors and higher Mathematical Objects, as it makes full sense once you realize that these mathematical objects are nothing more than ordered collections/sequence of numbers (scalars). Moving from an understanding of what a Vector is to what a Tensor is becomes so much easier.

For example a second-order Tensor can be represented by the following tuple of Vectors

\begin{equation} T=\begin{bmatrix} V_{1_1} & V_{1_2} & V_{1_3}\\ V_{2_1} & V_{2_2} & V_{2_3} \\ V_{3_1} & V_{3_2} & V_{3_3}\\ \end{bmatrix} \end{equation}

Wrapping Up

Is my intuition correct? Is a Vector just a tuple of Scalars (e.g Real Numbers). And likewise is a Tensor just a tuple of Vectors? And finally is what I've said that there is no true concept of "direction" in Pure Mathematics (i.e. we can only "represent" direction in its physical sense by interpreting abstract mathematical objects in a certain way)?

Would you agree that thinking about Vectors in their true abstract purely mathematical definition makes them so much more "powerful" and useful?

Suggestions

If you have spotted any gaps in my understanding, or if you have any criticism to offer on this way of thinking about Vectors, or if I'm completely flat out wrong at any point please comment below.

I realize that there is a similar question asked here (albeit from a different viewpoint), which has a good answer similar to my line of reasoning, except the answer given describes Vectors by working down from the definition of Vector Spaces, whereas I try to describe Vectors by working up from the definition of Scalars, tuples and Real Numbers : https://math.stackexchange.com/a/1395270/266135

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  • $\begingroup$ Any vector space can be identified with a set of tuples, even over fields other than the real numbers, but in general there are many ways to do this and fixing a particular way is arbitrary. $\endgroup$ – Matt Samuel Apr 13 '16 at 23:21
  • $\begingroup$ @MattSamuel for that you need a fairly liberal definition of tuple and as far as I can see also axiom of choice. $\endgroup$ – DRF Apr 13 '16 at 23:25
  • $\begingroup$ @DRF it's not very liberal in my opinion. I consider it as a function from an arbitrary index set into the field. The axiom of choice allows you to construct sets. If a vector space has a basis, then it has a basis. It doesn't magically disappear when you remove an axiom, it's just that you no longer have that particular method for proving that a linearly independent set exists that spans the space. If the collection exists, then it always exists, but it may not be a set. $\endgroup$ – Matt Samuel Apr 13 '16 at 23:29
  • $\begingroup$ @Matt Samuel, could you recommend any good introductory books, or texts/Wikipedia articles to read further, especially on abstract vector spaces? I'm a first year undergrad, majoring in Pure Mathematics, so currently this is where my knowledge breaks down and I lose the ability to debate further on this. $\endgroup$ – Perturbative Apr 13 '16 at 23:37
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    $\begingroup$ I believe the canonical reference you're looking for is Axler's Linear Algebra Done Right. It's got to be time to create a meme for that, by now... Also, Winitzki's Linear Algebra via Exterior Products focuses heavily on abstract vector spaces, and is freely available on his website. $\endgroup$ – pjs36 Apr 14 '16 at 0:08
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There are several misconceptions in the OP about both mathematicians' and physicists' use of the word "vector", and even about what scalars and tensors are. To keep this a concise overview I'll be linking to fuller explanations.

Firstly, anything you've heard about magnitude and direction was just an attempt to help schoolchildren avoid certain fallacies without having to explain the entire concept of a vector space to them. The aim is to make sure they understand that, for example, a particle's momentum points a certain way but its amount of energy doesn't.

In general, vectors are not tuples. Admittedly some sets of tuples satisfy the axioms of a vector space if you define arithmetic the usual way, but vectors are so much more general than that case, as examples discussed above show. What is true in general is that, if a vector space $V$ has a basis of the form $\left\{e_i|i\in I \right\}$, then each vector in $V$ is expressible as a linear combination of the $e_i$. Depending on the details, this "linear combination" might be a sum or an integral. Armed with this, the coefficients used can provide a tuple representation of vectors (although in some cases you need infinitely many numbers), but the vector is an independent object. The map is not the territory. In fact, making a terrain look different by creating a new map that's rotated relative to an old one is a special case of what you'll sometimes here called a basis change. Since you're familiar with $\mathbb{R}^n$, I'll give a simple example. The vectors $\left(\begin{array}{c} 1\\ 0 \end{array}\right),\,\left(\begin{array}{c} 0\\ 1 \end{array}\right)$ comprise a basis of $\mathbb{R}^2$, but I can rotate a 2D map by an angle $\theta$ because $\left(\begin{array}{c} \cos\theta\\ \sin\theta \end{array}\right),\,\left(\begin{array}{c} \sin\theta\\ -\cos\theta \end{array}\right)$ comprise a basis too.

I should also point out in passing that, while in some contexts the word "basis" simply means a choice of $\left\{e_i|i\in I \right\}$ for which this can be done, the proper definition requires that the linear combination need only use finitely many of the $e_i$. Many vector spaces of interest that do not have finite dimension nonetheless meet some additional technical conditions that allow the less strict meaning of "basis" to be useful. However, the famous statement that two bases of a vector space have the same cardinality refers to the finite-combinations-only definition.

So that's what mathematicians mean by vector spaces. A vector space is always "over" a field of scalars. Just as a vector is defined as an element of a vector space which in turn has a long definition, a scalar is defined as an element of a field which in turn has a long definition.

Or is it? Let's talk about what physicists really mean when they discuss vectors. On the one hand, they know about all the mathematics I mentioned above. On the other hand, they also want to describe nature in terms of quantities that transform in certain convenient ways when we switch coordinate systems, to exemplify "symmetries". This leads them to define "vector" in a stricter way. For example, one thing schoolchildren aren't told is that, although angular momentum has a magnitude and direction, it's not a vector because of the way it transforms under reflections. The distinction in $\mathbb{R}^3$ between vectors and axial vectors takes some explaining. The confusion is understandable. Position and momentum are "in" $\mathbb{R}^3$ and are vectors; angular momentum is "in" $\mathbb{R}^3$ is an axial vector. The reason is simply that none of these things are really "in" a famous set of tuples, because they're not tuples at all; they're quantities that admit a tuple representation. That's one similarity axial vectors have with "true" vectors.

With the development of differential geometry, we realised there is a more elegant way to talk about all this. Instead of distinguishing between true vectors and axial vectors, we can distinguish between contravariant and covariant vectors, provided our "both types count" definition of vector means "rank one tensor". quantity $T^{\alpha_1\cdots\alpha_p}_{\beta_1\cdots\beta_q}$ with $p,\,q$ non-negative integers is called a tensor of rank $p+q$ and order (or type) $\left(p,\,q\right)$ iff a coordinate transformation of spacetime from $x^\mu$ to $x^{'\nu}$ obeys $$T^{'\alpha_1\cdots\alpha_p}_{\beta_1\cdots\beta_q}=\sum_{\gamma_1\cdots\gamma_p \delta_1\cdots\delta_q}\frac{\partial x^{'\alpha_1}}{\partial x^{\gamma_1}}\cdots\frac{\partial x^{'\alpha_p}}{\partial x^{\gamma_p}}\frac{\partial x^{\delta_1}}{\partial x^{'\beta_1}}\cdots\frac{\partial x^{\delta_q}}{\partial x^{'\beta_q}}T^{\gamma_1\cdots\gamma_p}_{\delta_1\cdots\delta_q}.$$(We never actually write the summation sign; we take for granted that any index that appears twice, once as a subscript and once as a superscript, is summed over all possible values. In relativity, there is one such value for each spacetime dimension.) A tensor of rank $0$ is a scalar, and is unchanged under coordinate transformations. A tensor of positive rank is called covariant if $p=0$, contravariant if $q=0$ and mixed otherwise. Mixed tensors have $p\geq 1$ and $q\geq 1$, so have rank $\geq 2$.

Something that looks like a tensor by virtue of its indices may not transform the right way to actually be a tensor. (Of course, if there are no indices at all something would "look like a scalar", but might not be one.) Here are three important examples.

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  • $\begingroup$ By the way, the existence of bases in the strict sense for any vector space is provable using the axiom of choice. $\endgroup$ – J.G. Apr 14 '16 at 8:16
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A vector is probably not just a tuple of scalars unless your definition of "tuple" is very very broad (and you also probably need to assume some extras like AC). In general a vector space is quite a bit more abstract.

A first easy example of a vector space where the vectors are not really "tuples" for most definition of what a "tuple" is would be functions from $[0,1]$ to $\mathbb{R}$ (or from $\mathbb{R}$ to $\mathbb{R}$). This can easily be shown to be a vector space but does not really consist of tuples.

You also get $m\times n$ matrices over an arbitrary field $K$ which will be a vector space. You can take polynomials in multiple unknowns. All of these can be shown to have the structure of vector spaces but are quite far from looking much like tuples.

The one thing that does sort of allow you to think of Vector spaces as "tuples" of elements of the base field you never mention. That is the fact that if the axiom of choice holds then every vector space has a basis (note this is actually and iff so every vector space having a basis is a pretty strong statement). If you then allow your tuples to have arbitrary cardinalities you could really think of vectors as tuples, but for many vector spaces that sort of approach will lost a lot of oomph. For matrices you really have to in some way reorder them, for polynomials the tuples can be really strange I think. Since you can obviously take the trivial unitary monomials approach but your basis might be a lot more complex then that aswell. And for real functions you can't really work with the representation since there is no constructible basis.

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  • $\begingroup$ I was gonna comment something of this sort! That (ordinary) tuples cant capture the essence of functions was what first popped into my head. Well written answer! $\endgroup$ – Nap D. Lover Apr 13 '16 at 23:24
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    $\begingroup$ The functions from $\left[0,\,1\right]$ to some subset $S$ of $\mathbb{R}$ comprise a vector space if $S=\mathbb{R}$, but not if $S=\left[0,\,1\right]$ (because you don't get closure under multiplication by arbitrary scalars) or even if $S=\left[0,\,\infty\right)$ (because you don't get additive inverses - or, to put it another way, you don't get closure under multiplication by $-1$.) $\endgroup$ – J.G. Apr 14 '16 at 7:12
  • $\begingroup$ @J.G. Woops ofcourse sorry my bad. $\endgroup$ – DRF Apr 14 '16 at 7:18

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