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Hadamard product to Matrix product transformation

  1. Jul 16, 2007 #1
    I've been trying to find an answer to this for some time now, if anyone can provide a definitive answer, they'll have the satisfaction of knowing that several University mathematics professors haven't been able to come up with one...

    I am looking for a way to transform from a Hadamard product (element-wise matrix multiplication) into a classic matrix product. Before I get flamed for posing a supposedly stupid question, the answer is related to quantum networks - (see -

    http://archive.numdam.org/ARCHIVE/AIF/AIF_1999__49_3/AIF_1999__49_3_927_0/AIF_1999__49_3_927_0.pdf" [Broken]

    p. 951).

    So more specifically - Let us start with a matrix Z, Z is an element of Real. Element wise multiplication is given by .*, regular multiplication is * . A ranges between 0 and positive infinity, B ranges between -infinity and positive infinity.

    A.*Z = B*Z

    Is there a function that maps A to B?

    To provide more information, think about B. If we exponentiate it to 0, we get the identity matrix (I) . If we exponentiate it by -1 (assuming it does have an inverse) then we get Z. On the left hand side of the equation - if A is uniform (i.e. all matrix elements are identical), then we get Z. If A is the identity matrix, then we get (I). I'm chiefly interested in what happens in the inbetween states (i.e. between no change and orthogonalization). Transitioning B is straightforward (exponentiate to p where p ranges between 0 and -1). Transitioning A is less so. I get the feeling that the answer has something to do with rotation, but it's n-dimensional...

    Any advice (or references) would be appreciated. Thanks!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jul 16, 2007 #2
    To map between A and B, I guess you can use A = (BZ)./Z, and B = (A.*Z)Z^-1. The function is probably dependent on the structure of Z, and therefore Z can't be separated out.

    If that's the case, then what is the function that transitions from identity (I) through Z and then to uniform (all values are 1/n)?
  4. Jun 7, 2010 #3

    I was wondering if you found an answer to your question. I stumbled on a similar problem.

  5. Feb 26, 2011 #4
    The formula A.*Z=B*Z is not possible in general. Here is a counter example using a matrix rank inequality.

    We have rank(A.*Z)=rank(B*Z)<=min(rank(B),rank(Z)).

    Let A=[1 -1;1 1]' and Z=[1 1;1 1]', then rank(A)=2 and rank(Z)=1. Pick whatever B you like,

    then 2=rank(A)=rank(A.*Z)<=min(rank(B),rank(Z))=1 which is impossible

    and we have a counter example.
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