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I would like to know if there is a convenient mathematical expression for the following.

I want to construct a (n x n) matrix $\mathbf{C}$ for which i know that each column $\mathbf{C}_i$ is equal to $(\mathbf{A}_i \circ \mathbf{B})\mathbf{1}$. Matrix $\mathbf{B}$ is (n x n) whereas $\mathbf{1}$ is a (n x 1) vector of ones. Further, $\mathbf{A}$ is 3-dimensional (n x n x n) where i consider $\mathbf{A}_i$ to be the (n x n) submatrix at level $i$ of $\mathbf{A}$. Equivalently, if $\mathbf{A}$ would be viewed as a (1 x n) vector of 2d matrices, $\mathbf{A}_i$ would be the (n x n) matrix at position $(1,i)$ of that vector. To conclude, the $\circ$ operator indicates the Schur (element-wise) product.

Ideally, but i'm sure wrongfully, it would be nice to write $\mathbf{C} = (\mathbf{A} \circ \mathbf{B})\mathbf{U}$, with $\mathbf{U}$ being the (n x n) matrix of ones, but i think columns of $\mathbf{U}$ will not just correctly "hop over" to each level $i$ submatrix of $\mathbf{A}$ in constructing $\mathbf{C}$.

Is there a correct way for expressing $\mathbf{C}$, maybe via tensor algebra?

dixity
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I assume that "adheres to" means the same as "is equal to".

Let $\text{vec}$ denote the vectorization operator and $\text{diag}(M_1,\dots,M_k)$ the direct sum of matrices. We can write $$ \operatorname{vec}(\mathbf C) = \operatorname{diag}(\mathbf A_1 \circ \mathbf B, \dots, \mathbf A_n \circ \mathbf B) \mathbf 1_{n^2 \times 1} $$

Ben Grossmann
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