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Homework Help: Determinant of a matrix with identity blocks

  1. Feb 15, 2010 #1
    Hi all,
    I'm studying my mathematics lesson, and there is an example I can't understand:

    Consider the matrix
    A=(0 In)
    (-In 0)
    with In the identity nxn
    We want to compute detA :

    We introduce the permutation
    p=(1 2 ... n n+1 ... 2n)
    (n n+1 ... 2n 1 ... n )


    s.t if we apply p to the columns of A, then we get
    B=(In 0)
    (0 -In)
    (so far so good)

    then we say: detB=(-1)^n (ok)
    the sign of p is (-1)^n (all right)

    then detA=(-1)^n*(-1)^n=1...(why???)


    Thanks in advance
     
    Last edited: Feb 16, 2010
  2. jcsd
  3. Feb 15, 2010 #2

    tiny-tim

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    Hi penguin007! :smile:

    PA = B, so detP*detA = detB :wink:
     
  4. Feb 15, 2010 #3

    vela

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    Re: matrix-determinant

    How'd you get the sign of p is [itex](-1)^n[/itex]? That's where the problem lies.
     
  5. Feb 16, 2010 #4
    Re: matrix-determinant

    Hi tiny-tim:
    I think I got it but I thought a map could be defined by a matrix only if it was linear?? Yet I can see p has a matrix (the one that reverses the columns 1 and n...) so:
    is p a linear map? or can some non linear maps have a matrix??

    vela:
    s=The sign of p is given by (-1)^(the number of inversions by p) I counted n-2 inversions then s=(-1)^n...


    Thanks again
     
  6. Feb 16, 2010 #5

    tiny-tim

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    Hi penguin007! :smile:
    P is linear :smile:

    what makes you think it isn't? :confused:
     
  7. Feb 16, 2010 #6

    vela

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    Re: matrix-determinant

    Oy, I misread what you wrote in this thread too. I think I'll just shut up now. :redface:
     
  8. Feb 16, 2010 #7
    Re: matrix-determinant

    tiny-tim:
    please correct me: p is a permutation; p is linear would mean, for instance:p(3)=p(1+2)=p(1)+p(2), but p(3)=n+2<>p(1)+p(2)...( I know there is a snag, but where??)

    vela:
    That's all right, these things happen to everyone.
     
  9. Feb 16, 2010 #8

    tiny-tim

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    oh i see …

    no, these Ps acts on n-tuples (or vectors).

    So eg P12(a,b,c,d,e,f) = (b,a,c,d,e,f),

    and you can check that P12(a+A,b,c,d,e,f) = (b,a,c,d,e,f) + (b,A,c,d,e,f) = P12(a,b,c,d,e,f) + P12(A,b,c,d,e,f) :wink:
     
  10. Feb 16, 2010 #9
    Re: matrix-determinant

    Thanks a lot tiny-tim for all these explanations, now I understand!
     
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