naima
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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.naima said: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?
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.naima said:Do you agree that the second is a n^2 * n^2 with n^2 "1".in it?
Yes, tensors can viewed this way:naima said: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"
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.naima said:In this http://isites.harvard.edu/fs/docs/icb.topic1533461.files/Chois%20Theorem-Bill.pdf