| Part 2: ============= From the comment on that video: >It is more fun to proof that the set of a 2 by 2 matrices with everywhere the same value x with x not equal to 0 is a group. The determinant of such matrices is 0 but it is still a group. This is only surprising if you don't understand 2x2 matrices as operators on a plane, in which case the exercise is a cruel perversion (why on Earth would anyone want to consider such matrices, or check that they form a group, with identity that's not the identity matrix?! And how would one come up with this to begin with?!). "Even though the determinant is zero" is a symptom of a conceptual gap. Of course the determinant being zero has nothing to do with these matrices forming a group! You can embed GL(2) into GL(3) by filling the rest of the entries with zero, and of course this will still be a group: because it acts on the XY plane in just the same way as before, and matrix multiplication still gives their composition because we defined it to work this way. And of course matrices of the form [x x; x x] form a group. A better question would be, why wouldn't they? Take the following kindergarten-accessible definition of a group: actions that you can undo, repeat, and combine. Let's hand-wave the above exercise with this definition. What does [t t; t t] do? Let's take t=1. [1 1; 1 1] takes a point in [a, b] in a plane, and sends it to a point [a+b, a+b]. Doesn't seem like you can undo that, because you don't know whether [3, 3] came from [1, 2] or [2, 1]. Bummer. Well no surprise, the operation [1 1; 1 1] smashes the whole plane into a single line spanned by [1; 1], aka y = x (the image is spanned by column vectors). We might as well ignore anything off this line, 'cause we can't tell points away from the line apart after applying [1 1; 1 1]. What does [1 1; 1 1] do on its image, the line y = x? It sends [a; a] to [2a, 2a] on the same line. So [1 1; 1 1] acts like multiplying by 2. That's certainly something you can undo. The same works for other matrices; [x x; x x] acts like multiplying by 2x on that line. You can undo that as long as x is not 0. And you can repeat/combine these operations because who's gonna stop you? There is nothing left to prove. This "hand-wavy" argument, of course, is something that gives much more understanding than a "formal" proof from the "definition". That proof is "fun" because it is surprising - and it is surprising because it doesn't make sense. And I would argue that it's much easier to make a mistake there - and conclude that it's not a group because such-and-such axiom doesn't hold. The hand-wavy argument, though, ultimately comes from (or would lead to) an understanding that matrices act on their eigenspaces - and little more needs to be said (given that all of those matrices share an eigenspace). Furthermore, this gives an example of a group representation for a group of nonzero real numbers with operation defined by xy = 2xy. Of course, such a definition is utterly confusing; why would anyone come up with such a thing, other than to torture people? Why would* one want to redefine the product of real numbers to be something else?! Seeing people work it out on Math StackExchange [3] is painful. The people giving the answers are confident that a * b := 2(a+b) is both is and isn't a group! This alone should tell you that at some point, rigor becomes a hindrance. This is that point. I say, the correct answer is that if someone gives you a group without a thing that it acts on, ask for your money back. Of course that thing isn't a group, but what breaks if we just say it is? Since nobody is giving me a refund on this, let's see how it would act on itself. An element a would act by sending b to (2a) + 2b, so we have translation and scaling. Can we undo this? Sure, we just need to shift back by (-2a) and scale down by 1/2. But scaling down isn't an option here, so tough luck. It's not the only problem, of course; but the student is left utterly confused (again, see comments in [3]!) by this exercise, whose point seems to be that "arbitrarily messing with definitions sometimes works and sometimes doesn't". But it feels* like this should be a group. Let's fix it.
The rule "a * x -> 2(a+x)" is kosher; we can take it to be the action of a on the real line. What does the composition look like? well, a * b * x = a * (2b + 2x) = 2a + 4b + 4x That "4x" there tells us that the group generated by these actions is larger than just the generating set. Nothing in our generating set can multiply by 4 (again, that would be a way to see that the rule doesn't define a group). The exploration can then go on further to examining which subgroup of the affine transformations of the real line this generates. It's interesting! I trust you that rigor being forced on you could have improved your mathematical reasoning. But in that case, you are exceptional - or there was more to it than "do it this way just because". The most common case, in my experience, is represented in the [1][2][3] (particularly in the comments): it makes people confused, wrong, and lost. I'd rather have them never seen a definition of a group than go through that kind of brain damage. >I agree that you cannot teach math in a purely bourbakist style. I prefer a "visual" style like that inspired by the books of Arnold, Strang, Needham, and I am the sole teacher in my lab that seriously uses the word "amplitwist" to refer to the complex derivative :) In that case, they might be crying tears of joy or grief over all the years they were taught otherwise :) [1] https://www.youtube.com/watch?v=q_JqHQPbmUk [2] https://math.stackexchange.com/questions/919040/proving-a-gr... [3] https://math.stackexchange.com/questions/1108349/prove-that-... |