Derivative of matrix inverse?

Question

Let $b$ be a column vector, $A$ a symmetric square matrix and $X$ a square matrix. Define the scalar $$ f(X) = b^T(I-X)^{-1}A((I-X)^{-1})^Tb. $$ I would like to find an expression for $\frac{dF}{dX}$. Note that this should be a matrix.

The question derivative of inverse matrix by itself provides an answer for a simpler expression, but I am not able to adapt the derivations to the current setup.

Answer 1

Introduce the matrix variable $$\eqalign{ \def\c#1{\color{red}{#1}} \def\qiq{\quad\implies\quad} Y &= (I-X)^{-1} \qiq \c{dY=Y\:dX\;Y} \\ }$$ the sym() function $${\rm sym}(A) = \frac12(A+A^T) \qquad\qquad\qquad\;$$ and the matrix inner product $(:)$ $$\eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; {\rm Tr}(A^TB) \\ A:A &= \|A\|^2_F \qquad \{ {\rm Frobenius\;norm} \} \\\\ }$$ Use the above notation to write the function, then calculate its differential and gradient $$\eqalign{ f &= bb^T:YAY^T \\ df &= bb^T:2\:{\rm sym}(dY\,AY^T) \\ &= 2bb^T:(\c{dY}\,AY^T) \\ &= 2bb^TYA:(\c{Y\,dX\:Y}) \\ &= 2Y^Tbb^TYAY^T:dX \\ \frac{\partial f}{\partial X} &= 2Y^Tbb^TYAY^T \\ }$$

Answer 2

The differential of $X\mapsto X^{-1}$ is the map $H\mapsto -X^{-1}HX^{-1}.$ Proof: $$(X+H)^{-1}-X^{-1}=X^{-1}((I+HX^{-1})^{-1}-I)=X^{-1}(I-HX^{-1}-o(H)-I)$$