This is somewhat related to my previous question, but in a slightly different form and with different assumptions.
Say I have the following expression:
$\frac{\delta E}{\delta \mathbf{R}} = 2\Theta \mathbf{R}^T - 2\Phi$
where $\Phi, \Theta \text{ and } \mathbf{R}$ are all matrices.
Now, given the additional constraint:
$\mathbf{R} = \mathbf{B}(\mathbf{I} \otimes \mathbf{K})$
($\mathbf{I}$ is an identity matrix, $\mathbf{B}$ is also a matrix)
How does one find $\frac{\delta E}{\delta \mathbf{K}}$?
Reverting to the differential expression $$\eqalign{ dE &= 2\,(\Theta R^T-\Phi):dR \cr &= 2\,(\Theta R^T-\Phi):B(I\otimes dK) \cr &= 2\,B^T(\Theta R^T-\Phi):I\otimes dK \cr &= M:I\otimes dK \cr\cr }$$ Now you need the Kronecker factorization of $M$ $$\eqalign{ M &= \sum_{j=1}^r A_j\otimes Z_j \cr }$$ where the $A_j$ matrices are shaped like $I$, and the $Z_j$ like $K$.
Then apply the Frobenius-Kronecker mixed product rule $$\eqalign{ (A\otimes Z):(E\otimes F) &= (A:E)\,(Z:F) \cr }$$ to get $$\eqalign{ dE &= \sum_{i=j}^r A_j\otimes Z_j : I\otimes dK \cr &= \sum_{i=j}^r (A_j:I)\,Z_j:dK \cr &= \sum_{i=j}^r {\rm tr}(A_j)\,Z_j:dK \cr\cr \frac{\partial E}{\partial K} &= \sum_{j=1}^r\,Z_j\,{\rm tr}(A_j) \cr\cr }$$ For help with the Kronecker factorization search for articles by van Loan and Pitsianis.