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How many vectors in span({v})

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Let v be an element of F2p \ {(0,0)}. How many vectors does Span({v}) have? How many 1-dimensional vector subspaces does F2p have?

    F2p is the two-dimensional field (a,b) where each a, b are elements of Fp, where p is a prime number.

    3. The attempt at a solution

    I know the total number of elements in the vector space F2p \ {(0,0)} is p2 - 1. I also thought that this was the number of vectors in Span({v}) but I've told by a couple people that is not so. I started with the definition of span but I just couldn't see the rest unfold.

    Thank you ahead of time for your help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2009 #2

    Office_Shredder

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    An element in span(v) must be of the form av for some a in the field. How many choices of a do you have? Do all of these choices yield a different vector, or can av=bv if a=/=b?
     
  4. Sep 16, 2009 #3

    Dick

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    Take an example. Let F=Z_3, the field with three elements {0,1,2}. Yes, there are 8 elements in (Z_3)^2-{0,0}. How many elements are in span((1,1))? That's (1,1)*x for all x in Z_3. Does that help you to see things unfold?
     
  5. Sep 16, 2009 #4
    Would (1,1)*x have 3 elements? (1,1,0), (1,1,1), (1,1,2)? Or am I reading this incorrectly?
     
  6. Sep 16, 2009 #5

    Dick

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    Well, yeah. Would have three elements. But they would be (0,0)=(1,1)*0, (1,1)=(1,1)*1 and (2,2)=(1,1)*2.
     
  7. Sep 16, 2009 #6
    Okay now I see what you meant sorry about that.
     
  8. Sep 16, 2009 #7
    "a" I believe can be any number in F.
     
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