Derivative involving the trace of a Kronecker product

Question

I'm stuck trying to solve a derivative that looks like this:

$$\frac{\partial}{\partial X} \mbox{Tr} \{ A(X^{-1} \otimes I_{n} )B \},$$

where A is a $N\times 2n$ matrix, B is a $2n \times N$ matrix, and X is a $2 \times 2$ matrix.

Thank you!

Answer 1

Assume that we know the Kronecker factorization $$\eqalign{ A^TB^T &= Y\otimes Z \cr }$$ where $(Y,Z)$ are shaped like $(X,I)$ respectively.

Then write the function in terms of the Frobenius (:) Inner Product $$\eqalign{ f &= A^TB^T:X^{-1}\otimes I \cr &= Y\otimes Z:X^{-1}\otimes I \cr &= (Z:I)\otimes(Y:X^{-1}) \cr &= {\rm tr}(Z)\,Y:X^{-1} \cr\cr }$$ Calculation of the differential and gradient are straight-forward $$\eqalign{ df &= -{\rm tr}(Z)\,Y:X^{-1}\,dX\,X^{-1} \cr &= -X^{-T}\Big({\rm tr}(Z)\,Y\Big)X^{-T}:dX \cr\cr \frac{\partial f}{\partial X} &= -X^{-T}\Big(Y\,{\rm tr}(Z)\Big)\,X^{-T} \cr\cr }$$ In general, the Kronecker factorization will have more than one term $$\eqalign{ A^TB^T &= \sum_{k=1}^r Y_k\otimes Z_k \cr }$$ which modifies the result slightly $$\eqalign{ \frac{\partial f}{\partial X} &= -X^{-T}\,\Bigg(\sum_{k=1}^r Y_k\,{\rm tr}(Z_k)\Bigg)\,X^{-T} \cr }$$ For details of the factorization, search for "Kronecker Product Approximation" and for papers by Van Loan & Pitsianis.