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Difficult Problem

  1. Sep 29, 2007 #1
    Here's a difficult problem in an area I'm not at all familiar with.

    A product * is defined on R[tex]^{4}[/tex] in the following way:

    (a, b, c, d)*(a', b', c', d') = (cd' - c'd, ac' - a'c + cb' - c'b, a'd - ad' + bd' - b'd, c'd - cd')

    Find all subsets S of R[tex]^{4}[/tex] that satisfies the following two conditions:

    1: For all x,y [tex]\in[/tex] S and all a,b [tex]\in[/tex] R, ax + by [tex]\in[/tex] S

    2: For all x [tex]\in[/tex] S and all y [tex]\in[/tex] R[tex]^{4}[/tex], x * y [tex]\in[/tex] S

    Condtion 1 seems to me to say we're on a 2D-plane in 4-space but the strange symmetry of the product must restrict this in some way. Condition 2 seems to be about linearity within S somehow. I'm really new at this and would really appreciate quite a lot of help with it.
     
    Last edited: Sep 30, 2007
  2. jcsd
  3. Sep 30, 2007 #2
    the conditions remind me of the properties of an ideal.
     
  4. Sep 30, 2007 #3

    HallsofIvy

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    Why not just go ahead an DO it? Let [itex]x= (x_1, x_2, x_3, x_4)[/itex], let [itex] y= (y_1,y_2,y_3,y_4)[/itex], go ahead and multiply out the conditions and see what you get.
     
  5. Sep 30, 2007 #4
    I just realised a typo in Condition 1. it shoud read ...and all a,b in R (not R4).
     
  6. Sep 30, 2007 #5
    I take it an ideal is a concept in linear algebra...?
     
  7. Sep 30, 2007 #6

    matt grime

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    It is completely immaterial (and the answer is no, by the way).

    The important thing here is what Halls wrote. Just do it. It is just a calculation to find the conditions, nothing complicated.
     
  8. Sep 30, 2007 #7
    Missing something...?


    OK, so I've tried your advice. But setting x and y to x1... and y1... and doing the multiplication didn't really give anything more than the original definition of the product. What am I missing here? Please remember that I'm liable to miss simple observations simply because I hardly understand what I'm after.

    E.g. I cannot see how this excersize helps me put any restrictions on S.

    However, I followed another path. Every product should have an identity element, right?

    So putting x * I = x, I came up with a nice 4th order system of equations, the first one being [itex] x_3i_4 - x_4i_3 = x_1[/itex] etc.

    Please help. And not just with the details, but conceptually too!

    I crunched it into my TI-Nspire which actually produced a complete solution on the form

    {i1 =..., i2=..., i3=... i4=... AND x1=..., x2=...} OR {three more solutions}

    The interesting bit here is the x1= bit. This is, as far as I understand, the restrictions that define the subsets S. DO PLEASE correct me if I'm missing something here.

    Since this is 4-space I can let one coordinate be time and visualise the solutions and also check them and they do make sense algebraically but there are some points I need to clear up.

    1. Uniqueness of Identity element: I thought that an Identity element satisfies x*I = I*x = x. However, since in this case x*y = -y*x (I managed to prove that, at least) isn't x*I = -I*x??? What is the correct way out of this?
    2. Effects on S of this uniqueness: So I put I * x = x into the TI-Nspire to check and it came up with some sign changes in two of the solutions. But these sign changes only affect the i=... bit. NOT the x=... bit. But should these solutions be disregarded anyway because of the identity element not being "stable"? Or are these solutions still valid?
     
  9. Sep 30, 2007 #8
    Some findings...

    Well to be perfectly true I did find something when I did the multiplication. If You let z = ax + by, I the tried to multiply z with x to see if that would yield anything and I found that z*x = by * x and z*y = ax*y.

    It is as if the product takes just the part "orthogonal" to the vector to find the product, rather like torque (or as opposed to work).

    But I can't see how this gives me any bearings on S. It seems to be valid for all x,y.
     
  10. Sep 30, 2007 #9

    HallsofIvy

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    ?? Since this is on R4, why would z= ax+ by?
     
  11. Sep 30, 2007 #10

    matt grime

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  12. Sep 30, 2007 #11

    matt grime

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    Please, stop thinking like that.
     
  13. Sep 30, 2007 #12
    ...?

    I like to visualise. Do offer an alternative before discarding other peoples mode of thinking. By thinking like that I can visualize what x1 = -x4 means. Admittedly rotations are difficult :-) but I find I can understand problems better if I can visualise them as well as numbercrunch them. E.g. in this particular case I can visualise where the solutions intersect each other and how they relate to each other.

    If my thinking is flawed, do please point it out to me, but then tell me how to do it better and be specific.
     
  14. Sep 30, 2007 #13
    Condition 1

    Condition 1 specifically states that for all x,y in S and a,b in R ax+by should be in S. As far as I understand this is a condition that states that the subsets S have to be linear subspaces like straight lines, flat planes etc. My solutions so far seem to be just that but there is also a class of straight lines that together form a curved surface which need more investigating. I haven't got round to seeing how my solutions hold up to the conditions 1&2 yet and I won't have time for some 48 hrs or so. So if you can help do please do so. I'm going to bed now.
     
  15. Oct 1, 2007 #14

    matt grime

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    I didn't say 'don't visualize', though if you can visualize four space you're doing better than almost anyone else. I said you shouldn't assert 'one co-ordinate is time' since (a) it doesn't help, (b) it doesn't change anything, and (c) it isn't true.

    You just have to get conditions that (x_1,x_2,x_3,x_4) satisfies to be in such an S. Why haven't you written out what the conditions imply? Yes, 1 just states it is a subspace. 2. allows you to write other things out as well. Why not pick some y's , like (1,0,0,0) and see what happens? It appears that you want the solution to magically appear, when you should just be calculating things and seeing what happens.
     
  16. Oct 1, 2007 #15
    I do admit magic is wonderful, but I don't shirk hard work - If I know what to do. My problem here is I hardly know where to begin. What, for instance do you mean by picking some different y's and seeing what happens? I guess you mean doing the multiplication but for what X? Numeric example or generic. So this produces new vectors. Am I then to guess what S looks like from seeing which values of X produces new vectors that are in the same subspace as the original X for any Y? Or what?
     
  17. Oct 1, 2007 #16

    matt grime

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    I said you might want to try multiplying x in S by some ys in R^4 like (1,0,0,0).... Or you might want write down the matrix that represents the linear transformation x-->y*x for a generic y. There are lots of things to try, and only by trying them will you have some ideas as to what's going on. I don't understand what you mean by 'I don't know where to begin' - the only thing you can do is play around until the light appears at the end of the tunnel. That's how you do maths. If you don't know the method needed, then discover it.
     
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