Flipping Summation of Kronecker Products

Question

Suppose A\mathbf A is an n×nn\times n matrix and that Bi\mathbf B_i is an m×mm\times m matrix, for all i∈{1,…,n}i\in\{1,\dots, n\}. I am trying to find n×nn\times n matrices U\mathbf U and V\mathbf V and m×mm\times m matrices Xi\mathbf X_i, for all i∈{1,…,n}i\in\{1,\dots, n\}, such that

[∑i(Eii⊗Bi)](A⊗Im)[∑i(Eii⊗Bi)]=(U⊗Im)[∑i(Eii⊗Xi)](V⊗Im),\left[\sum_{i}(\mathbf E_{ii}\otimes \mathbf B_{i})\right](\mathbf A\otimes \mathbf I_{m})\left[\sum_{i}(\mathbf E_{ii}\otimes \mathbf B_{i})\right]=(\mathbf U \otimes \mathbf I_{m})\left[\sum_{i}(\mathbf E_{ii}\otimes \mathbf X_{i})\right](\mathbf V\otimes \mathbf I_{m}),

where {Eij}\{\mathbf E_{ij}\} are the n×nn\times n matrix units.

I don’t think this is possible under these general conditions, so basically I’m trying to find the minimal set of assumptions on the {Bi}\{\mathbf B_i\} matrices that would allow such a transformation.

A strong assumption that works

Suppose that

BiBj=bijD. \mathbf B_i \mathbf B_j = b_{ij} \mathbf D. \tag 1

To see how it works in this case, notice that

[∑i(Eii⊗Bi)](A⊗Im)[∑i(Eii⊗Bi)]=∑i,j(EiiAEjj⊗BiBj),\left[\sum_{i}(\mathbf E_{ii}\otimes \mathbf B_{i})\right](\mathbf A\otimes \mathbf I_{m})\left[\sum_{i}(\mathbf E_{ii}\otimes \mathbf B_{i})\right]= \sum_{i,j}(\mathbf E_{ii}\mathbf A\mathbf E_{jj}\otimes \mathbf B_{i}\mathbf B_{j}),

and that under (1)(1),

\sum_{i,j}(\mathbf E_{ii}\mathbf A\mathbf E_{jj}\otimes \mathbf B_{i}\mathbf B_{j})=\sum_{i,j}(a_{ij}b_{ij}\mathbf E_{ij}\otimes \mathbf D), \sum_{i,j}(\mathbf E_{ii}\mathbf A\mathbf E_{jj}\otimes \mathbf B_{i}\mathbf B_{j})=\sum_{i,j}(a_{ij}b_{ij}\mathbf E_{ij}\otimes \mathbf D),

so it is enough to set

\mathbf U = \sum_{i,j}a_{ij}b_{ij}\mathbf E_{ij}, \quad \mathbf X_i = \mathbf D, \quad \text{and} \quad \mathbf V = \mathbf I_n. \mathbf U = \sum_{i,j}a_{ij}b_{ij}\mathbf E_{ij}, \quad \mathbf X_i = \mathbf D, \quad \text{and} \quad \mathbf V = \mathbf I_n.

It seems to me that condition (1)(1) is too restrictive though.

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