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I have been looking for a matrix multiplier that is similar to a tensor product. The best way I can define the product is with the following example:

Suppose $A=\left[\begin{array}{cc} 1 & 2\\ 3 & 4 \end{array}\right]$, and $B=\left[\begin{array}{cc} 0 & 5\\ 6 & 7 \end{array}\right]$.

Does their exist a multiplier (call it $\#$) such that

$A\#B=\left[\begin{array}{cc} 1\left[\begin{array}{cc} 0 & 5\end{array}\right] & 2\left[\begin{array}{cc} 0 & 5\end{array}\right]\\ 3\left[\begin{array}{cc} 6 & 7\end{array}\right] & 4\left[\begin{array}{cc} 6 & 7\end{array}\right] \end{array}\right]=\left[\begin{array}{cccc} 0 & 5 & 0 & 10\\ 18 & 21 & 24 & 28 \end{array}\right]$.

Does such a multiplier exist? If not, how could I use existing operators to attain my desired outcome?

Pareto
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  • I am trying to work out the details, but I think it is a Tracy–Singh product when the second matrix is partitioned by rows, instead of the better known partition by columns giving rise to the Khatri–Rao product. – Miguel Mar 29 '18 at 18:08
  • Similar to, but certainly not the same as, the Kronecker product. For a discussion of related products, see https://en.wikipedia.org/wiki/Kronecker_product – marty cohen Mar 29 '18 at 18:09
  • @Miguel I believe you are correct. I'm working on the details now. – Pareto Mar 29 '18 at 18:27

1 Answers1

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Can someone verify?

Suppose $A=\left[\begin{array}{cc} 1 & 2\\ 3 & 4 \end{array}\right],$and $B=\left[\begin{array}{cc} 0 & 5\\ 6 & 7 \end{array}\right]$.

By definintion of the Khatri-Rao product: $A*B=\left(A_{ij}\otimes B_{ij}\right)_{ij}$.

$A^{T}=\left[\begin{array}{cc} 1 & 3\\ 2 & 4 \end{array}\right],$and $B^{T}=\left[\begin{array}{cc} 0 & 6\\ 5 & 7 \end{array}\right]$

\begin{eqnarray*} \left(A^{T}*B^{T}\right)^{T} & = & \left[A_{1}\otimes B_{1}|A_{2}\otimes B_{2}\right]^{T}\\ & = & \left[\begin{array}{cc} 1\cdot0 & 3\cdot6\\ 1\cdot5 & 3\cdot7\\ 2\cdot0 & 4\cdot5\\ 2\cdot5 & 4\cdot7 \end{array}\right]^{T}\\ & = & \left[\begin{array}{cc} 1\cdot0 & 3\cdot6\\ 1\cdot5 & 3\cdot7\\ 2\cdot0 & 4\cdot6\\ 2\cdot5 & 4\cdot7 \end{array}\right]^{T}\\ & = & \left[\begin{array}{cc} 0 & 18\\ 5 & 21\\ 0 & 24\\ 10 & 28 \end{array}\right]^{T}\\ & = & \left[\begin{array}{cccc} 0 & 5 & 0 & 10\\ 18 & 21 & 24 & 28 \end{array}\right] \end{eqnarray*}

This seems related: Macedo, Hugo Daniel, and José Nuno Oliveira. 2015. A linear algebra approach to OLAP. Formal Aspects of Computing 27, (2) (03): 283-307

Pareto
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