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B What is this notation?

  1. Aug 5, 2016 #1

    naima

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    I am reading the proof of the Choi's theorem in his own paper.
    he first introduces ##E_{ij} ## as the nn null matrix but with a 1 at i,j.
    Then he uses ##(E_{ij})_{1<=i,j <=n}##
    he says that thi is a positive matrix. What is talking about?
    Is it a matrix of matrices?
     
    Last edited: Aug 5, 2016
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  3. Aug 5, 2016 #2

    jedishrfu

    Staff: Mentor

    I think the notation means that 1<=i<=n and 1<=j<=n ie for an n x n matrix the i and j indices can have integer values between 1 and n.
     
  4. Aug 5, 2016 #3

    fresh_42

    Staff: Mentor

    I read this as ##E_{kl} = (\delta_{ki}\delta_{jl})_{i,j}## and ##(E_{ij})_{1<=i,j <=n}## as either an all one matrix or more likely the same as ##E_{ij}## with only the ranges of ##i## and ##j## added, as he considers non square matrices as well. A standard basis vector if you like.
     
  5. Aug 5, 2016 #4

    naima

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    There is no problem with the first definition. it is a n*n matrix with one "1".
    Do you agree that the second is a n^2 * n^2 with n^2 "1".in it?
     
  6. Aug 5, 2016 #5

    fresh_42

    Staff: Mentor

    This would really surprise me. I think it is more like an ill-fated version of ##A=(a_{ij})_{1≤i≤n,1≤j≤n}##. However, I wouldn't bet on it.
    What could be a reason to arrange the ##E_{ij}## like this, as a matrix of matrices?
     
  7. Aug 5, 2016 #6

    naima

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    I do not feel comfortable with the proof of the Choi's theorem.
    But read the top of page 2. we have ##M_n(M_m)## which is a tensor product.
    Choi says that this space contains
    "n X n block matrices with m x m matrices as entries"
     
  8. Aug 5, 2016 #7

    fresh_42

    Staff: Mentor

    Yes, tensors can viewed this way:
    ## c \otimes d = cd## with scalars ##c,d## is a scalar.
    ##c \otimes v## with ##c## a scalar and ##v## a vector is a vector.
    ##v \otimes w## with ##v,w## vectors is a matrix (of rank 1).
    ##v \otimes A## with a vector ##v## and a matrix ##A## is a stack of weighted copies of ##A##.
    ##A \otimes B## with matrices ##A,B## are a four-dimensional array of coordinates.
    ... and so on ...
    The rest of tensor spaces are linear combinations of those.

    It may be right that ##(E_{i,j})_{i,j}## is a tensor product or otherwise arranged array of matrices. I don't want to rule it out. I simply haven't seen a construction like this noted in coordinates. It would be a linear function of linear functions, and all in coordinates.
     
  9. Aug 6, 2016 #8

    naima

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  10. Aug 6, 2016 #9

    fresh_42

    Staff: Mentor

    I don't know whether it is always like that. Rui Li's notations are new to me. E.g. I see ##A\otimes B## as a four-dimensional array, but this can't be put on paper. So he writes ##A \otimes B = (A_{ij}B)_{ij}##. This makes certainly sense though.
    But I haven't heard of a generalized or partial trace as in ##A.3##. However, I'm not a physicist and a little bit allergic to coordinates, because they often hide the principle behind. And furthermore: it isn't important whether it is always the case or not, because Rui Li defines his notations and is consistent.
     
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