Derivative of $(XAX^T)^{-1}$

Question

Let $X$ is a rectangular matrix with full row rank, $A$ is positive definite.

How can I find the derivative of $$(XAX^T)^{-1}$$

with respect to $X$.

Answer 1

Given the matrix-valued function $$F=(XAX^T)^{-1}$$ its differential is easy to find. $$\eqalign{ I &= FF^{-1} \cr 0 &= F\,dF^{-1}+dF\,F^{-1} \cr dF &= -F\,dF^{-1}\,F \cr &= -F\,dX\,AX^TF - FXA\,dX^T\,F \cr }$$ A difficult issue is that the matrix-by-matrix gradient, i.e. $\frac{\partial F}{\partial X}$, is not a matrix but a 4th order tensor. One way to proceed is to vectorize all of the matrix quantities. $$\eqalign{ df &= -\Big((F^TXA^T\otimes F) + (F^T\otimes FXA)K\Big)\,dx \cr \frac{\partial f}{\partial x} &= -(F^TXA^T\otimes F) - (F^T\otimes FXA)K \cr }$$ where $K$ is the commutation matrix associated with the Kronecker product (represented by $\otimes$).