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Size of GLn

  1. Oct 12, 2008 #1
    What is the size of [itex]GL_n(\mathbb{Z}_2)[/itex]?
     
  2. jcsd
  3. Oct 13, 2008 #2

    morphism

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    Think in terms of linear independence.
     
  4. Oct 13, 2008 #3

    HallsofIvy

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    You could also think of this as n by n matrices whose entries must be either 1 or 0. How many entries are then in an n by n matrix? And if there are only two possible values for each?
     
  5. Oct 13, 2008 #4
    Well non-invertible matrices are not in GLn(Z_2), so it's NOT 2^(n^2). Rather it's that minus all the matrices with determinant 0, and I don't know how to count those.

    Is it true that an nxn matrix is invertible iff the column space has n dimensions?
     
  6. Oct 16, 2008 #5
    [tex]Z_2= \{0,1\}[/tex] and [tex]Z_2[/tex] has only one unit {1}.
    so matrix A which satisfy det A = 1 will be element of [tex]GL_n ( Z_2) [/tex]
     
  7. Oct 16, 2008 #6

    HallsofIvy

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    Yes, and the whole question is "how many of them are there"!

    In fact, a matrix A, in Z2 is invertible if det(A)= 1 and not invertible if det(A)= 0. Does that imply that exactly half of all n by n matrices in Z2[/sup] are invertible?
     
  8. Oct 16, 2008 #7

    morphism

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    Yes!


    No. In fact, this is only true if n=1.
     
  9. Oct 16, 2008 #8
    I still don't know how to count them. So you have [itex]2^n-1[/itex] choices for the first vector. [itex]2^n-2[/itex] for the second. But how many vectors are linearly independent wrt to the first two?
     
  10. Oct 17, 2008 #9

    morphism

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    How many linear combinations are there of the first two columns? Remember, there are only two elements in our field.
     
  11. Oct 17, 2008 #10
    I think it's
    (2^n -1)(2^n - 2) (2^n - 2^2 ) .... (2^n - 2^{n-1}).

    the reason is: you just have to make sure all the rows are linearly independent and non-zero.
    For the first row there are (2^n - 1) choices.

    For the second row there's (2^n-1) minus 1 (the number of rows already used).

    For the third row there's (2^n-1) minus 2 (the number of rows already used) minus another 1 for the number of sums of those rows.

    For the kth row, there's (2^n-1) minus (k choose 1) minus (k choose 2), etc., minus (k choose (k-1)). You can tell that none of these conditions are redundant because that would imply that some of the previous rows were linearly dependent.

    But
    k choose 1 + k choose 2 + ... + k choose (k-1) = 2^{k} -1.

    ?
     
    Last edited: Oct 17, 2008
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