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It is well-known that, given real matrices $\mathbf{A}_1$, $\mathbf{A}_2$ and $\mathbf{B}$ with dimensions $m_1 \times n_1$, $m_2 \times n_2$ and $n_1 \times n_2$, respectively, then $$(\mathbf{A}_2 \otimes\mathbf{A}_1)\text{vec}(\mathbf{B}) = \text{vec}(\mathbf{A}_1\mathbf{B}\mathbf{A}_2^\top),$$ where $\text{vec}$ denotes the vectorization operator, $\otimes$ is the Kronecker product and $^\top$ is the transposition operation.

I wonder that similar relations can be deduced when more matrices $\mathbf{A}_i$, $i=1,\ldots, m$, are involved (and hence, $\mathbf{B}$ is a tensor with order $m$), that is to say, how can we express the product $$(\mathbf{A}_m \otimes \cdots \otimes\mathbf{A}_1)\text{vec}(\mathbf{B}) = \quad ?$$

Any help would be appreciated.

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$\def\v{\operatorname{vec}}\def\o{\otimes}\def\B{\,\color{red}{B}\,}$By reshaping $\v(B)$ we can create a matrix $\B$

Consider the case for $m=4$ $$\eqalign{ \v\Big((A_1) \B (A_4^T\o A_3^T\o A_2^T)\Big) &= (A_4\o A_3\o A_2\o A_1)\v(B) \\ \v\Big((A_2\o A_1) \B(A_4^T\o A_3^T)\Big) &= (A_4\o A_3\o A_2\o A_1)\v(B) \\ \v\Big((A_3\o A_2\o A_1) \B(A_4^T)\Big) &= (A_4\o A_3\o A_2\o A_1)\v(B) \\ }$$ Extrapolating, if the number of $A$ matrices is $m$, then there are $(m-1)$ locations where the $\color{red}{B}$ matrix can be placed, which will yield the same result upon vectorization. The $A$ matrices are cyclically permuted to accommodate the shifting $\color{red}{B}$ matrix.

Another approach is calculate the Kronecker factorization of the vector into constituent vectors whose dimensions are compatible with the $A$ matrices $$\eqalign{ &\v(B) = b_4\o b_3\o b_2\o b_1 \\ &(A_4\o A_3\o A_2\o A_1)\v(B) = (A_4b_4\o A_3b_3\o A_2b_2\o A_1b_1) \\ }$$ Although it may not be possible to find a monomial factor, it is always possible to find a sum of such factors, e.g. $$\eqalign{ \v(B) &= \sum_{k=1}^K b_{4k}\o b_{3k}\o b_{2k}\o b_{1k} \\ }$$

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