# Hermitian of product of two matrices

1. May 22, 2014

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

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

2. Jul 2, 2014

### 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?

3. Jul 2, 2014

### 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.