I have seen this operation in a vectorization operation of $D=AX^{T}B$ i.e. $vec(D)=\left( A\circ B\right) vec(X),$ \begin{equation*} A\circ B= \begin{pmatrix} A_{1}B_{1}^{T} & A_{2}B_{1}^{T} & \cdots & A_{n}B_{1}^{T} \\ A_{1}B_{2}^{T} & A_{2}B_{2}^{T} & \cdots & A_{n}B_{2}^{T} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1}B_{n}^{T} & A_{2}B_{n}^{T} & \cdots & A_{n}B_{n}^{T} \end{pmatrix}, \end{equation*} where $A_{j}$, $B_{j}$, $j=1,2,\cdots ,N,$ denoting the column vector of $A$ and $B,$ respectively.
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Where have you seen this? I can't see what it accomplishes that cannot be done just as well by the Kronecker product and a commutation matrix. – Mårten W Apr 30 '19 at 08:55
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I have seen this in this work https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1529639, I have tried with Kronecker product but it did not work for me. – core May 01 '19 at 00:39
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Quite interesting. They introduce it in the paragraph between equation (11) and equation (12), but they don't mention any name or reference for it. As I said before, it seems somewhat pointless, as $(A\circ B)=(A\otimes B)K^\top$ for a suitable commutation matrix $K$. – Mårten W May 01 '19 at 10:04
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You are right, this solves my question, thanks for the observation. Now I have to compute which form has the commutation matrix K. – core May 01 '19 at 15:48