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Co-prime of vectors

  1. Feb 13, 2012 #1
    Hi guys

    I have a question about the coprime of two vectors
    For two vectors (x1,x2) and (y1,y2).
    Given a,b with gcf (a,b)=1 .i.e. relatively prime.
    I do the linear combination of two vectors
    a(x1,x2)+b(y1,y2)=n(z1,z2) with some common factor n and gcf(z1,z2)=1.
    If n=1 for any a,b, two vectors are said co-prime.
    I wonder if any criteria to prove two vectors are coprime.
    For example, (2,3),(1,3) are not coprime b/c (2,3)+(1,3)=3(1,2).
    But (7,3),(2,1) are coprime b/c a(7,3)+b(2,1)=(7a+2b,3a+b) and gcf(7a+2b,3a+b)=gcf(a,3a+b)=gcf(a,b)=1.
    Also how to generalize it to vectors with n components?

    Thank you
  2. jcsd
  3. Feb 13, 2012 #2
    Hi, I havnt checked the details, but such problems screem for the use of the determinant formed by the x's and y's, and can then also be generalized immediately. I guess the condition is that this determinant has no prime factors, that is being 1 or -1. Any prime factor p would allow a nontrivial relation ax+by=0 over F_p, which then lifts to show a relation with gcd(z_1,z_2)=p. Tell me if that works out.
  4. Feb 13, 2012 #3
    Thanks Norwegian!
    I think you are almost right. But (4,-3) and (3,-4) are coprime but with det=7 will be a counter example. If we consider higher dimensions, are (2,-1,2,-1) and (-4,1,4,-1) co-prime? Do you know how to show it rigorously?
  5. Feb 13, 2012 #4
    (4,-3) + (3,-4) is divisible by 7, so they are not coprime. The sum of your other vectors is divisible by 2, so they are also not coprime. My guesses for generalizations: n vectors in n-space, determinant = 1 or -1. Two vectors in n-space, set of all 2x2 minors coprime, m<n vectors in n-space, all mxm minors coprime.
  6. Feb 13, 2012 #5
    I type the wrong vectors. I consider the two (-1,0,3,-1) and (-3,1,1,0). Are they coprime? I think you are right but I do not know how to prove it. Thank you!
  7. Feb 13, 2012 #6
    Yes, those vectors are coprime. You only need to look at the last two components.
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