# Flipping Summation of Kronecker Products

Question

Suppose $\mathbf A$ is an $n\times n$ matrix, and that $\mathbf B_i$ is an $m\times m$ matrix, for all $i\in\{1,\dots, n\}$. Is it possible to find $n\times n$ matrices $\mathbf U$ and $\mathbf V$, and $m\times m$ matrices $\mathbf X_i$, for all $i\in\{1,\dots, n\}$, such that

$$\left[\sum_{i}(\mathbf E_{ii}^n\otimes \mathbf B_{i})\right](\mathbf A\otimes \mathbf I_{m})\left[\sum_{i}(\mathbf E_{ii}^n\otimes \mathbf B_{i})\right]=(\mathbf U \otimes \mathbf I_{m})\left[\sum_{i}(\mathbf E_{ii}^n\otimes \mathbf X_{i})\right](\mathbf V\otimes \mathbf I_{m}),$$

where $\{\mathbf E_{ij}^n\}$ are the $n\times n$ matrix units?

Perhaps this is not possible under these general conditions, so basically I'm trying to find the minimal set of assumptions (i.e. necessary and sufficient) on the $\{\mathbf B_i\}$ matrices that would allow such a transformation.

Obs$_1$: The answers to this question can be helpful to generalize Assumption 2 bellow in the step that leads to equation $(2.2)$.

Obs$_2$: If a complete answer is not possible, I would be happy with an answer that simply relaxes one of the assumptions bellow.

 

How far I've been able to go, i.e. strong assumptions that work:

Assumption 1

Suppose that

$$\mathbf B_i \mathbf B_j = d_{ij} \mathbf D. \tag 1$$

To see how it works in this case, notice that

$$\left[\sum_{i}(\mathbf E_{ii}^n\otimes \mathbf B_{i})\right](\mathbf A\otimes \mathbf I_{m})\left[\sum_{i}(\mathbf E_{ii}^n\otimes \mathbf B_{i})\right]= \sum_{i,j}(\mathbf E_{ii}^n\mathbf A\mathbf E_{jj}^n\otimes \mathbf B_{i}\mathbf B_{j}),$$

and that under $(1)$,

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

so it is enough to set

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

Assumption 2

For all $i\in\{1,\dots, n\}$, let

$$\mathbf B_{i} = \operatorname{diag}(b_{i1},...,b_{im}),\tag{2.1}$$

and notice that

\begin{align} \left[\sum_{i=1}^n(\mathbf E_{ii}^n\otimes \mathbf B_{i})\right](\mathbf A\otimes I_{m})\left[\sum_{j=1}^n(\mathbf E_{jj}^n \otimes \mathbf B_{j})\right] &=\sum_{i=1}^n\sum_{j=1}^n(\mathbf E_{ii}^n \mathbf A \mathbf E_{jj}^n\otimes \mathbf B_{i}\mathbf B_{j})\\[1.5ex] &=\sum_{i=1}^n\sum_{j=1}^n(a_{ij}\mathbf E_{ij}^n \otimes\sum_{k=1}^{m}b_{ik}b_{jk}\mathbf E_{kk}^{m})\\[1.5ex] &=\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^{m}(a_{ij}b_{ik}b_{jk}\mathbf E_{ij}^n\otimes \mathbf E_{kk}^{m})\\[1.5ex] &=\sum_{k=1}^{m}\left(\left(\sum_{i=1}^n\sum_{j=1}^n a_{ij}b_{ik} b_{jk}\mathbf E_{ij}^n\right)\otimes \mathbf E_{kk}^{m}\right) \end{align}

Next, suppose $\sum_{i=1}^n\sum_{j=1}^n a_{ij}b_{ik}b_{jk} \mathbf E_{ij}^n$ has eigenvectors independent of $k$, so that its eigendecomposition can be written as

$$\sum_{i=1}^n\sum_{j=1}^n a_{ij}b_{ik}b_{jk} \mathbf E_{ij}^n = \mathbf Q \mathbf \Omega_{k}\mathbf Q^{-1} \tag{2.2}$$

with

$$\mathbf \Omega_{k} = \operatorname{diag}(\omega_{k1},...,\omega_{kn}).$$

Then

\begin{align} \sum_{k=1}^{m}\left(\left(\sum_{i=1}^n\sum_{j=1}^n a_{ij}b_{ik} b_{jk}\mathbf E_{ij}^n\right)\otimes \mathbf E_{kk}^{m}\right) &=\sum_{k=1}^{m}\left(\sum_{i=1}^n\omega_{ki}\mathbf Q\mathbf E_{ii}^n \mathbf Q^{-1}\otimes \mathbf E_{kk}^{m}\right)\\[1.5ex] &=(\mathbf Q\otimes \mathbf I_{m})\left[\sum_{k=1}^{m}\sum_{i=1}^n(\mathbf E_{ii}^n\otimes \omega_{ki}\mathbf E_{kk}^{m})\right](\mathbf Q^{-1}\otimes \mathbf I_{m})\\[1.5ex] &=(\mathbf Q\otimes \mathbf I_{m})\left[\sum_{i=1}^n(\mathbf E_{ii}^n\otimes \sum_{k=1}^{m}\omega_{ki}\mathbf E_{kk}^{m})\right](\mathbf Q^{-1}\otimes \mathbf I_{m}) \end{align}

so, we can set $$\mathbf U = \mathbf Q,\quad \mathbf X_{i} =\sum_{k=1}^{m}\omega_{ki}\mathbf E_{kk}^{m},\quad\text{and}\quad \mathbf V = \mathbf Q^{-1}.$$