Transformation of mixed Kronecker and Hadamard products 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?
 A: The answer was rather simple. I leave the solution here in case someone needs it in the future $ Q = diag((A \otimes 1) d)$
A: 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.
