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I have not yet taught Linear Algebra, but I teach Computer Graphics regularly, which uses linear algebra at many junctures, and uses concepts such as the cross product. I have often been disappointed to learn that even students who took Linear Algebra and did well, have little (or no) familiarity with the cross product in $\mathbb{R}^3$.

Q. Does the cross product properly belong to linear algebra, or is it a geometric diversion from the main thrust of linear algebra?

My sense is that mathematics departments prefer to keep linear algebra abstract, to shy away from geometry in favor of formal rigor, often the first such rigorous course for students. Whereas in my experience, I never fully "grokked" linear algebra until I saw orthogonal transformations (etc.) geometrically. But that is my personal bias.

I think it can be instructive to explain (at some level) that the cross product can only exist in its familiar form in in $\mathbb{R}^3$ and in $\mathbb{R}^7$:

Massey, W. S. "Cross products of vectors in higher dimensional Euclidean spaces." American Mathematical Monthly (1983): 697-701. (JSTOR link)

(My question is prompted by the recent question on determinants.)

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    $\begingroup$ I would hazard that the geometry of the dot and cross product is covered thoroughly in multivariable/vector calculus, and mathematics departments eschew redundancy. $\endgroup$ – user1527 Apr 2 '14 at 0:06
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    $\begingroup$ Students in engineering and the physical sciences will also get the cross product in their physics course. $\endgroup$ – Ben Crowell Apr 2 '14 at 4:53
  • $\begingroup$ Tangentially related: matheducators.stackexchange.com/questions/55/… $\endgroup$ – Willie Wong Apr 2 '14 at 8:13
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    $\begingroup$ Sorry: here you go. math.stackexchange.com/a/606720/34287 $\endgroup$ – Steven Gubkin Apr 3 '14 at 2:02
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    $\begingroup$ Also the cross product can exist in more than 3 dimensions: it just is a (n-1) ary operation, not a binary operation. I talk about that (from the same perspective as the above link) here: math.stackexchange.com/a/674130/34287 $\endgroup$ – Steven Gubkin Apr 3 '14 at 2:04
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The problem is that there are multiple main thrusts of linear algebra.

  1. For many students (especially pure math majors), linear algebra is their first introduction to abstract algebra. When a course is designed with this in mind, rigor and formalism takes higher precedence as we want as much of the material to be reusable in the context of modules over rings and also multilinear algebra

  2. On the other hand, even in the world of pure mathematics, linear algebra is fundamental in the development of differential geometry, functional analysis ("infinite dimensional linear algebra"), and differential equations (including dynamical systems, stability analysis, and PDEs). With these applications in mind, it would only make sense to introduce to the students appropriate geometric intuition behind linear transformations.

  3. Outside of pure mathematics, linear algebra is often viewed as synonymous with "stuff having to do with matrices". For these applications the crucial points of understanding are how to manipulate matrices to facilitate analysis and computation (think PCA and SVD, computationally extracting information from the matrix without inverting it).

If you are somehow able to teach a class addressing all of the above, more the power to you. But the trends that I have been familiar with is to have linear algebra classes designed with different applications in mind: at the very least often times math majors end up taking a different version of the class compare to the engineers and economists.


Going back to your question about cross-products: using a strict definition of linear algebra as the study of linear transformations of vector spaces, then cross products do not belong to linear algebra. It requires necessarily some multilinear input, be it a quadratic form (an inner product, so we can talk about orthogonality), or a volume form (to convert from two-vectors to one-vector by Hodge dual). On the other hand, these topics may already be covered in a linear algebra class (which usually do not restrict itself to the strict definition anyway), and in this case cross product may be mentioned as an application.

But if the cross product is mentioned, care must be given to the fact that it behaves "weirdly" under coordinate transformations. (The result of a cross product is a pseudovector.) One of the themes of linear algebra (at least for the first two "thrusts" that I describe) is the possibility of thinking about things in a manner that is coordinate independent (which should certainly be part of any "geometric intuition"). The formula used to compute the cross product, as it is usually taught, is certainly not coordinate independent. (I have seen students trying to compute the cross product in a spherical coordinate system.) There are a lot of subtle issues involved that may bring the discussion in conflict with some of the earlier themes of the class, and care must be given to resolve them.

When I taught linear algebra, I did not talk about the cross product in any great detail (it was not on the centrally-set syllabus; a student brought it up when we talked about computing determinants for 3-by-3 matrices). Looking at the above, it would be hard to fit it into an already tight schedule: to do justice to the subject it will have to be at least one full lecture just on its properties.

I suspect somehow that this is the heart of your problem: many of your students must have seen the cross product before. But the amount of geometry related to the cross product and its properties for a computer graphics course is likely a bit more involved than the cursory discussion the cross product is afforded in any other introductory level mathematics class.


As a side remark: the first time I saw cross products was not in a math class at all! This was in a high school physics class where it naturally came up in the discussion of torque and angular momentum. (Yes, American high school.) But I wouldn't claim to have a good understanding of the cross product until having studied multilinear algebra and differential geometry.

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  • $\begingroup$ Thorough & thoughtful answer! I especially benefitted from your remarks on coordinate independence. $\endgroup$ – Joseph O'Rourke Apr 2 '14 at 12:15
  • $\begingroup$ Absolutely. So to answer the question in the title: "Should" is a No, but "Could" is a Yes since if it fits in with whatever bits of linear algebra are already in the course then it can be a way to connect it to something they might already know. But there's no point introducing it without that surrounding structure. $\endgroup$ – Loop Space Apr 2 '14 at 14:17
  • $\begingroup$ An interesting and informative answer! In my answer I was thinking of #3. Many people are first exposed to linear algebra as high school seniors.There was a section in our high school Alegebra 3 & 4 textook. Students would ask "why are we learning matrix multiplication?" and they'd quickly forget it while resolving to avoid linear algebra in the future. For an introduction to linear algebra at this level I believe a geometric approach is very helpful. $\endgroup$ – HopDavid Apr 2 '14 at 16:52
  • $\begingroup$ @HopDavid: Ah, interesting point. I sort of know Joseph O'Rourke and so when reading the question immediately assumed the discussion at the level of a college course in linear algebra. For the very first exposure at the high school level, you are likely right that the usual presentations leave something to be desired. $\endgroup$ – Willie Wong Apr 3 '14 at 7:38
  • $\begingroup$ Linear independence - orthogonality. Determinants - cross-product. Linear algebra - geometry. They inform each other. $\endgroup$ – Tom Copeland Mar 1 '15 at 21:53
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One of the main reasons why the cross product is seldomly (to my experience) taught in linear algebra is not only the fact that this particular vector product only works well in $\mathbb{R}^3$ (a specific vector space which isn't the main focus of linear algebra), but that the arsenal of linear algebra provides already a tool to find vectors which are orthogonal to a given set of vectors:

The Gram-Schmidt algorithm.

If you take $n-1$ vectors $v_1, v_2, \dots, v_{n-1} \in \mathbb{R}^n$ and want to know a vector which is perpendicular to each of these, i.e. the normal vector on the hyperplane spanned by them, then you find some vector $v_n$ such that it completes the set to a basis. Then you can start the Gram-Schmidt procedure: Subtract orthogonal projections, normalize, repeat. You end up with an orthonormal basis $w_1, \dots, w_n$ of $\mathbb{R}^n$ such that $$\text{span}\{v_1, \dots, v_i\}=\text{span}\{w_1, \dots, w_i\}, i=1, \dots, n.$$ In particular, $w_n$ is orthogonal to all $w_1,\dots,w_{n-1}$, and, thus to $v_1,\dots,v_{n-1}$.

From a computational point of view, this might not be very effective, but from a mathematical point of view, there is a tool which does what the cross product does, and it's constructive and I think has an easier geometric interpretation than the cross product.

As a final note, there are a number of generalizations from the cross product in different areas of abstract algebra (I am no algebraist, so I cannot tell how popular or important these are).

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  • $\begingroup$ If you want to use the Gram-Schmidt algorithm to find a vector perpendicular to two other vectors, you first have to find a vector which is not in the span of those two vectors. Probably a "random" vector will not be in the span, but this still is not "algorithmic". The cross product is algorithmic. $\endgroup$ – Steven Gubkin Jun 17 '14 at 12:11
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    $\begingroup$ It can be algorithmic if you point out that one of the three standard basis vectors will always work. Just loop through them until you find one that's linearly independent (with probability 1, of course, it will be the first one). $\endgroup$ – Ryan Reich Jun 17 '14 at 18:10
  • $\begingroup$ That is true! I guess it only looses the continuity of the process. $\endgroup$ – Steven Gubkin Jun 18 '14 at 6:22
  • $\begingroup$ The importance of the cross product is not that it gives you a way to find a vector perpendicular to two other vectors. The main reasons students need to know it is its physics applications. $\endgroup$ – Ben Crowell Aug 13 '18 at 22:41
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Our eyes and visual cortex are fabulous tools for modeling scenarios, processing information, and communication. It's a disservice to students to ditch the visual approach because it's a "geometric diversion"

2 space and 3 space are familiar and can be visualized. Presenting coordinates as vertices of polygons or polyhedra can give a visually oriented student a handle on what the numbers can represent. Showing that matrices can scale, rotate, flip or skew these objects can engage the interest of a student that might otherwise see linear algebra as abstract concepts with little application.

To answer your question, if vector representation of 3-space is used, it's a good idea to talk about cross products. Not only is the cross product a fun and useful tool for making stuff, but it also gives an opportunity to talk about determinants.

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    $\begingroup$ Welcome to this site! I empathize with your view, and remember "relearning" a lot of linear algebra/vector calc years later after getting a more sophisticated view of everything. However, your answer here might be better if it (i) was less "opinion" based (I had ... I learned... etc.) (ii) addressed the original question about whether the cross product, in particular, is an essential component of linear algebra or merely a "geometric diversion". I would edit your post for you, but don't want to suffocate your opinion, so go right ahead! $\endgroup$ – Brendan W. Sullivan Apr 2 '14 at 5:00
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Where I am working now (EPFL, Switzerland), first year students have a dedicated two semester geometry course in parallel with the usual calculus, analysis, and linear algebra sequence. I have not taught these courses but they seem to do a lot of plane and space geometry using the various tools at hand (inner product, cross products, determinants, angles...). They also cover some elementary spherical geometry and geometry of curves and surfaces.

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The cross product is an important element of "determinants," which is a topic normally covered in linear algebra.

As such, it belongs in most courses of linear algebra. Except one that is "light" on determinants (and heavier in vector spaces and mappings).

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  • $\begingroup$ You confuse determinant (that is a function of a matrix) with triple product (that is a function of three vectors). Moreover, one doesn’t need cross product to define the (scalar-valued) triple product. $\endgroup$ – Incnis Mrsi Aug 14 '15 at 17:37
  • $\begingroup$ @IncnisMrsi He does not even mention the triple product, so I am not sure what your comment is all about. In any case, there are clear links here, and all he is saying that this is enough to justify their inclusion in a linear algebra course. In fact the map $V \mapsto Det(X,Y,V)$ where $Det$ is being viewed as a multilinear function on vectors has the cross product $X \times Y$ as its representative. This is a natural way to characterize the cross product. $\endgroup$ – Steven Gubkin Aug 14 '15 at 18:55
  • $\begingroup$ @Steven Gubkin: the thing you refer to with “a multilinear function on vectors” is the triple product. Determinant is a function of 1 matrix, not of 3 vectors, again. $\endgroup$ – Incnis Mrsi Aug 15 '15 at 0:42
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    $\begingroup$ @Steven Gubkin: Yes, these “clear links here” becomes clear only after remembering that there is such thing as the standard basis, that the author ignored. Ī̲ call the 3-linear form on Euclidean 3-vectors “triple” because it is conventionally called so. Linear algebra is GL-invariant, vector calculus is SO-invariant, and one may not conflate different theories looking on some algebraic expressions from here and there that seem similar. $\endgroup$ – Incnis Mrsi Aug 15 '15 at 7:47
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    $\begingroup$ @Steven Gubkin: Ī̲ don’t know how to call the 4-product in English. But if you even use yourself the word multilinear, then which problems could you see with terms “3-linear”, “4-linear”, and “n-linear”? Cross product belongs to vector calculus and multilinear algebra, not to linear algebra, period. BTW, your upvote provided to this confusing and vague half-true “answer” is as silly as my “hostile little criticisms” are rude. $\endgroup$ – Incnis Mrsi Aug 16 '15 at 1:14

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