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Subspace and permutations

  1. Feb 5, 2007 #1
    1. The problem statement, all variables and given/known data
    choose x = (x1, x2, x3, x4) in R^4. It has 24 rearrangements like (x2, x1, x3, x4) and (x4, x3, x1, x2). Those 24 vectors including x itself span a subspace S. Find specific vectors x so that dimS is 0, 1, 3, 4

    3. The attempt at a solution
    So, i thought of it this way: 24 possiblities to arrange 4 numbers:
    4*3*2*1 = 4! = 24

    for dim S = 0 i got vector x = 0-vector;
    for dimS = 1 i got x = (1, 1, 1, 1) since whichever way you rearrange it, it's still same;
    for dim S = 4 i got x = (0, 0, 0, 1) or complement of this, since 1 can be carried through 4 times and new Lin indep vectors will result, which would be a basis and so dimS = 4;
    but for dimS = 3 i can't figure it out, I tried all possible combinations of (1, 1, 0, 0) and (1, 1, 1, 0), but it does not work out....
    it does not say that elements should be 0 or 1 necessarily, but i thought it was simple enough.....
    Thanks in advance.
     
  2. jcsd
  3. Feb 5, 2007 #2

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    Hint: It shouldn't be too hard to show that if any of the entries in x are distinct, then S contains all vectors in the plane perpendicular to (1,1,1,1).
     
  4. Feb 5, 2007 #3
    i don't get it..... yes, to check if they are really a basis i do arrange them into cols and make sure the rank is 3 for this case.
    but (1, 2, 3, 4) or (2, 3, 5, 7) do not work. Can you give a more explicit hint?
     
  5. Feb 5, 2007 #4

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    Forget about ranks of matrices. If two of the entries in x are different, say x_1 and x_2, then subtracting (x_1,x_2,x_3,x_4) from (x_2,x_1,x_3,x_4) gives a scalar multiple of (1,-1,0,0). Similarly, you can get (1,0,-1,0), etc. What subspace do these generate?
     
  6. Feb 5, 2007 #5
    ok, that works... do you mind explaining why that works? maybe geometrically .... because right now it is not very intuitive to me... thanks :)
     
  7. Feb 5, 2007 #6

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    Well, for any vector, the dot product with (1,1,1,1), which is just the sum of its entries, is unchanged by applying any permutation to it. This means all the permutations lie in the same plane perpendicular to (1,1,1,1), and so their differences lie in the 3 dimensional subspace perpendicular to (1,1,1,1) (ie, the plane perpendicular to (1,1,1,1) which passes through the origin). It's not hard to show that for any non-zero vector in this subspace, its permutations generate the entire subspace. Thus if x is any vector which isn't fixed by the permutations (ie, isn't a multiple of (1,1,1,1)), S contains this subspace.

    From there, S is either this subspace or all of R^4 depending on whether x is in the subspace itself: if it is in the subspace, so are all its permutations, so S is contained in, and so equal to, this subspace. If it isnt, it is linearly independent to any set of vectors in the subspace, and so generates a properly containing subspace, the only one of which is all of R^4.
     
    Last edited: Feb 5, 2007
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