1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving that a subspace must have a specific number of elements

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Let V = (F2)^3, the set of triples (x; y ; z) of numbers in F2, the fi eld with two
    elements. V is a vector space over F2.

    Prove that any subspace of V must have either 1, 2, 4, or 8 elements.

    2. Relevant equations

    F2 = {0,1}

    3. The attempt at a solution

    The only way that I can really think of approaching the problem is to say that some subspace U of V must have at least 1 element (the zero vector) and at most 8 elements, which would be V, the total set. From there I think I could somehow show that subsets with 3, 5, 6 or 7 elements are not subspaces, but I don't know how.
     
  2. jcsd
  3. Sep 27, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    A subspace must be an additive subgroup of the whole space. Do you know Lagrange's theorem about groups?
     
  4. Sep 27, 2011 #3
    We haven't learned it, so I'm assuming that the prof doesn't expect us to solve the question using it.
     
  5. Sep 27, 2011 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Fair enough. Suppose S is your subspace and not the whole space. So there is an element u that's not in S. Take the sets S and S+u. Can you show they don't intersect? That's a starting point.
     
  6. Sep 28, 2011 #5
    No, I don't know how I could do that :(
     
  7. Sep 28, 2011 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You could try. Pick an element of S, say s1. Now pick an element of S+u, say s2+u where s2 is in S. Why can't s1=s2+u? Remember u was chosen not to be in S.
     
  8. Sep 28, 2011 #7
    s2 + u cant equal s1 because s2 + some other element in S already equals s1, so that would imply u is in S, but u isn't in S. I don't think that's right...
     
  9. Sep 28, 2011 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think it's right. s1=s2+u. So u=s2-s1. s2-s1 is in S (because it's a subspace). u is assumed not to be in S. So u can't be equal to s2-s1. Hence S and S+u are nonintersecting sets. So S and S+u have the same number of elements, right? Do you see how this would lead to a proof that the number of elements in V is a multiple of the number of elements in S?
     
  10. Sep 28, 2011 #9
    Hmm... No, I can't say that I do.
     
  11. Sep 28, 2011 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Stop coming back so negative. Do you understand why S and S+u have the same number of elements and don't intersect? Given any element v of V, v is an element of S+v. The idea here is to show for any two elements v1 and v2 of V then S+v1 and S+v2 are either nonintersecting sets, or they are the same set. These are called 'cosets'. If you can fill in the details and prove that then you've shown V can be split into disjoint sets that all have the same number of elements. So the number of elements in V must be a multiple of the number of elements in S, right?
     
  12. Sep 28, 2011 #11
    If I've shown that S+v1 and S+v2 are either nonintersecting or the same (which you helped me do in the previous post), then how does it follow that V can be split into disjoint sets? All I sget from this is that S+v=S, or else S=v is not a subspace.
     
  13. Sep 28, 2011 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If S+v1 and S+v2 are either disjoint or equal (and I think you can prove that, they are equal if v1-v2 is in S and they are disjoint otherwise) then V is the union of a bunch of disjoint sets that all have the same number of elements, yes? Doesn't that mean the number of elements in V must be a multiple of the number of elements in S? I'm not sure how else to say this. We are just trying to prove Lagrange's theorem from scratch, since you say you don't already have it.
     
  14. Sep 28, 2011 #13
    Don't worry about it Dick, I'll discuss the problem with some of my buddies tomorrow, I'm sure we'll be able to work it out together. Thanks for trying to help
     
  15. Sep 28, 2011 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I'm sure you will work it out. It's not that hard a thing. I just don't seem to be explaining it well.
     
  16. Jun 5, 2012 #15
    Sorry to open up an old thread, but I've come across this problem in my studies and have some questions about it. I was wondering if it should be taken into consideration that a field of two elements treats addition and multiplication (though I'm only looking at addition for this problem) differently than the field of all real numbers - so that 1+1=0 in the field of two elements and 1 is its own inverse (if I'm not mistaken). Extending this, for a vector space over the field of two elements to make sense, wouldn't it be true that each vector must serve as its own inverse? (so v1=(-v1) for example).

    If this is the case, then in examining S={s1, s2, 0} it's easy to see that s1+s2 can't be defined, so a subspace of 3 elements is not feasible (unlike a subspace of 4). This argument can be extended to show that fields of 5, 6, and 7 elements aren't feasible either, but is my understanding of the problem correct?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook