Derivative involving the trace of a Kronecker product 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! 
 A: 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.
