Transformation of mixed Kronecker and Hadamard products

Question

Given $ w = \Big((A \otimes 1)\odot(1^T \otimes x)\Big)\ d $, express $ w $ in the form $ w = Q x $, where $ Q, A $ are matrices and $ d $ is a vector. I have tried to solve this by using the mixed properties of Hadamard Kronecker but it does not seem to help in decoupling $ x $ from the expression. Any property that would be useful?

Answer 1

The answer was rather simple. I leave the solution here in case someone needs it in the future $ Q = diag((A \otimes 1) d)$

Answer 2

First, note that $$\eqalign{ ({\tt1}^T\otimes x) &= ({\tt1}\otimes x^T)^T \\ &= ({\tt1}x^T)^T \\ &= x{\tt1}^T \\ }$$ Next, recall that the elementwise/Hadamard product of two vectors is commutative, and can be replaced by matrix multiplication with a diagonal matrix formed from either of the vectors $$f\odot g = g\odot f = Fg = Gf,\qquad F={\rm Diag}(f),\quad G={\rm Diag}(g)$$ Likewise, the Hadamard product of a rank-one matrix with a full-rank matrix can be replaced by multiplication with diagonal matrices formed from the rank-one factors $$\eqalign{ M\odot fg^T &= FMG \\ }$$ Finally, setting $$\eqalign{ M &= (A&\otimes&{\tt1}) \\ F &=\; X &= &{\rm Diag}(x) \\ G &=\; I &= &{\rm Diag}({\tt1}) \\ }$$ yields $$\eqalign{ w &= \Big((A\otimes{\tt1})\odot (x{\tt1}^T)\Big)\,d \\ &= \Big(X(A\otimes{\tt1})\Big)\,d \\ }$$ For typing convenience, define the variables $$q=(A\otimes{\tt1})\,d,\quad Q={\rm Diag}(q)$$ which can be used to simplify the given vector to $$w = Xq = Qx$$ This is identical to the result that you obtained.