I'm trying to calculate the derivative of $\mathrm{tr}((I+X^{-1})^{-1})$ with respect to $X$. By some sort of a chain rule, I believe this should be $X^{-1}(I+X^{-1})^{-2}X^{-1}$. However, I'm having a hard time finding a good reference for such a result. Any help would be greatly appreciated.
Matrix derivative of $\mathrm{tr}((I+X^{-1})^{-1})$
derivativesinversematricesmatrix-calculusscalar-fields
Best Answer
Alternative approach to Ben Grossmann's approach
We will use the following Frobenius product identity \begin{align} \operatorname{tr}\left(A^T B \right) := A:B . \end{align}
Further, we will use the differential of invertible (and assuming symmetric) matrix $X$ \begin{align} XX^{-1} = I \Longrightarrow dX X^{-1} + X dX^{-1} = 0 \Longleftrightarrow dX^{-1} = -X^{-1} dX X^{-1}. \end{align}
Let us define the following matrix with their differential \begin{align} M := \left(I + X^{-1} \right) \Longrightarrow dM = dX^{-1} = -X^{-1} dX X^{-1}. \end{align}
To this end, say $f := \operatorname{tr}\left( M^{-1} \right)$, then we find differential followed by the gradient. \begin{align} df &= d\operatorname{tr}\left( M^{-1} \right) = d\operatorname{tr}\left( I M^{-1} \right) \\ &= I : dM^{-1} \\ &= I : -M^{-1} dM M^{-1} \\ &= - M^{-2} : dM \\ &= - M^{-2} : -X^{-1} dX X^{-1} \\ &= X^{-1} M^{-2} X^{-1} : dX \end{align}
Then the gradient is \begin{align} \frac{\partial f}{\partial X} = X^{-1} M^{-2} X^{-1} = X^{-1} \left(I + X^{-1} \right)^{-2} X^{-1} . \end{align}