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$\ I ⊗ I ⊗ A = Z ⊗ I$

$A$ is a known matrix. $I$ is the identity matrix. $A$ and $I$ are n by n matrices. $⊗$ is the Kronecker product. Is there a way to find $Z$ in terms of $A$ and $I$?

Thanks in advance.

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A general approach how you can solve equations involving Kronecker products for one of its factors is vectorization: since $A \otimes B$ is linear in $A$ for given $B$ and vice versa, you can rewrite it as ${\rm vec}\{A \otimes B\} = \tilde{B} \cdot {\rm vec}\{A\} = \tilde{A} \cdot {\rm vec}\{B\}$. This would allow you to solve for the vectorized version of your desired matrix in standard ways (e.g., using Least Squares).

Whether your equation even has a solution will of course depend on the column space of the matrix $\tilde{A}$ or $\tilde{B}$, respectively.

That said, in your particular problem you are trying to find $Z$ such that $I \otimes A = Z \otimes I$ (note that the first two identity matrices can be combined into one since the Kronecker product is associative and $I \otimes I = I$). Solving this will be tough: $I \otimes A$ is a block-diagonal matrix with $A$ appearing as repeated blocks. $Z \otimes I$ is a matrix containing identity matrices stacked horizontally and vertically and scaled by the elements of $Z$. I would claim that this equation only has a solution when $Z$ and $A$ are diagonal. In this case, you don't need the matrices and their vectorizations, just extract the diagonals and equate those.

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