# Derivative with respect to vectorized inverse Kronecker product

I am trying to derive the gradient of a function I wish to optimize, and wish to obtain the following derivative: $$\frac{\partial}{\partial \pmb{x}} \left(\pmb{I} - \pmb{X} \otimes \pmb{X} \right)^{-1} \pmb{y}$$ with $$\pmb{x} = \mathrm{vec}(\pmb{X})$$, $$\pmb{X}$$ being a square asymetric matrix and $$\pmb{y}$$ a vector that is not a function of $$\pmb{x}$$, and $$\otimes$$ the Kronecker product. My thought was to first write: $$\left( \pmb{y}^{\top} \otimes \pmb{I} \right) \mathrm{vec}\left( \left(\pmb{I} - \pmb{X} \otimes \pmb{X} \right)^{-1}\right)$$ next to let $$\pmb{f} = \mathrm{vec}\left( \left(\pmb{I} - \pmb{X} \otimes \pmb{X} \right)^{-1}\right)$$ and then to express the differential of $$\pmb{f}$$. I got to: $$d\pmb{f} = \left(\left(\pmb{I} - \pmb{X} \otimes \pmb{X} \right)^{-\top} \otimes \left(\pmb{I} - \pmb{X} \otimes \pmb{X} \right)^{-1}\right) \left( \mathrm{vec}\left( (d\pmb{X}) \otimes \pmb{X} \right) + \mathrm{vec}\left( \pmb{X} \otimes (d\pmb{X})\right) \right)$$ in which $$-\top$$ is short for the transpose of an inverse. This seems close to the answer, but not quite there yet. I guess I am getting lost in trying to express $$\mathrm{vec}\left( (d\pmb{X}) \otimes \pmb{X} \right)$$ in terms of $$d\pmb{x}$$.

Edit: continuing this, I recognized there must be some permutation matrix $$\pmb{P}$$ such that: $$\pmb{P}\mathrm{vec}( (d\pmb{x})\pmb{x}^{\top} ) = \mathrm{vec}((d\pmb{X}) \otimes \pmb{X})$$ which I can use to further derive: \begin{align} d\pmb{f} &= \left(\left(\pmb{I} - \pmb{B} \otimes \pmb{B} \right)^{-\top} \otimes \left(\pmb{I} - \pmb{B} \otimes \pmb{B} \right)^{-1}\right)\pmb{P}\left((\pmb{b} \otimes \pmb{I}) + (\pmb{I} \otimes \pmb{b})\right)d\pmb{b} \\ \frac{\partial \pmb{f}}{\partial \pmb{b}} &= \left(\left(\pmb{I} - \pmb{B} \otimes \pmb{B} \right)^{-\top} \otimes \left(\pmb{I} - \pmb{B} \otimes \pmb{B} \right)^{-1}\right) \pmb{P}\left((\pmb{b} \otimes \pmb{I}) + (\pmb{I} \otimes \pmb{b})\right). \end{align} Which seems plausible. Thus, all that seems to be needed is an expression for $$\pmb{P}$$. I guess that will take a similar form as this answer, but I am not sure about it.

Let $$X\in {\mathbb R}^{n\times n}$$ and $$E$$ be the identity matrix of the same size.
Let's also denote the $$k^{th}$$ column of $$X$$ by $$x_k$$.
Define the matrices \eqalign{ A &= (E\otimes E - X\otimes X),\quad M &= \pmatrix{E\otimes x_1\cr E\otimes x_2\cr\vdots\cr E\otimes x_n} \cr } Calculate the differential of $$A$$. \eqalign{ dA &= -(X\otimes dX+dX\otimes X) \cr da &= {\rm vec}(dA) = -(M\otimes E+E\otimes M)\,dx \cr } Now we can answer the question. \eqalign{ w &= A^{-1}y \cr dw &= dA^{-1}y \cr &= -A^{-1}\,dA\,A^{-1}y \cr &= -{\rm vec}(A^{-1}\,dA\,w) \cr &= -(w^T\otimes A^{-1})\,da \cr &= (w^T\otimes A^{-1})\,(M\otimes E+E\otimes M)\,dx \cr \frac{\partial w}{\partial x} &= (w^T\otimes A^{-1})\,(M\otimes E+E\otimes M) \cr }