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Hermitian of product of two matrices

  1. May 22, 2014 #1
    Hi,

    i was wondering how the following expression can be decomposed:

    Let A=B°C, where B, C are rectangular random matrices and (°) denotes Hadamard product sign. Also, let (.) (.)H denote Hermitian transposition.

    Then, AH *A how can be decomposed in terms of B and C ??

    For example, AH *A = BH*B ° CH*C, or something like that ??

    Thank you in advance
     
  2. jcsd
  3. Jul 2, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. Jul 2, 2014 #3
    The identity you propose is false.

    [tex]\left(A^H A\right)_{ij}= \sum_k (A^H)_{ik} A_{kj} = \sum_k A^*_{ki} A_{kj}[/tex]
    [tex]= \sum_k B^*_{ki}C^*_{ki} B_{kj}C_{kj} \neq \left[\left(B^H B \right) \circ \left(C^H C \right)\right]_{ij} = \left( \sum_k B^*_{ki}B_{kj} \right) \left(\sum_\ell C^*_{\ell i}C_{\ell j} \right) [/tex]

    From this I believe [itex]\left(B \circ C \right)^H \left(B \circ C \right) = \left(B^H B \right) \circ \left(C^H C \right) + \text{stuff}[/itex]

    However I'm not sure if there is a neat way of expressing "stuff". I wasn't able to find any identities using only the matrix product, Hadamard product, and the conjugate transpose. But, maybe there is one.
     
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