Hermitian of product of two matrices

[nikozm]

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, A^H *A how can be decomposed in terms of B and C ??

For example, A^H *A = B^H*B ° C^H*C, or something like that ??

Thank you in advance

 

[Greg Bernhardt]

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?

 

[MisterX]

The identity you propose is false.

\left(A^H A\right)_{ij}= \sum_k (A^H)_{ik} A_{kj} = \sum_k A^*_{ki} A_{kj}
= \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)

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

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.