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Coprime of vectors 
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#1
Feb1312, 09:53 AM

P: 13

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 coprime. 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
Feb1312, 11:09 AM

P: 144

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.



#3
Feb1312, 12:45 PM

P: 13

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) coprime? Do you know how to show it rigorously? 


#4
Feb1312, 01:09 PM

P: 144

Coprime of vectors
(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 nspace, determinant = 1 or 1. Two vectors in nspace, set of all 2x2 minors coprime, m<n vectors in nspace, all mxm minors coprime.



#5
Feb1312, 04:14 PM

P: 13

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!



#6
Feb1312, 09:20 PM

P: 144

Yes, those vectors are coprime. You only need to look at the last two components.



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