I am trying to follow the document on http://web.mit.edu/people/raj/Acta05rmt.pdf, and I got a simple question that why for $f(A)=A^{-1}$ we have $df(E)=-A^{-1}EA^{-1}$ on page 5? (where $E$ is a small perturbation, and on page 4 the author says the $df$ is just the jacobian matrix. )
I got a method which seems correct: $df(A^{-1})=d(A^{-1})=-A^{-1}dfA^{-1}$, and substitute $E$ for $df$.
But my question is why another method would not work: I think the correct way to calculate jacobian is to find the derivative of $f(A+E)$ with respect to $E$, which I found somehow would not lead to the correct answer (a quick check is that the derivative of matrix wrt matrix would result in a squared size dimension matrix) ?
Thanks.
Following up : besides Raskolnikov's explane, there is also explanation here: http://web.mit.edu/18.325/www/handouts/handout2.pdf
Because of $(AB)^{-1}=B^{-1}A^{-1}$, we have
$$f(A+E)=(A+E)^{-1}=(\mathbb{I}+A^{-1}E)^{-1} A^{-1} \; .$$
Using a series development for the first factor
$$(\mathbb{I}+A^{-1}E)^{-1} = \mathbb{I}-A^{-1}E+\ldots$$
we obtain
$$f(A+E)=(A+E)^{-1}=A^{-1}-A^{-1}E A^{-1}+\ldots$$
thus
$$df(E)=f(A+E)-f(A)=-A^{-1}E A^{-1} \; .$$