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Jul 7, 2019 at 8:09 vote accept livehhh
Jun 19, 2019 at 8:35 answer added Pietro Majer timeline score: 4
Jun 19, 2019 at 4:50 answer added pre-kidney timeline score: 4
Jun 18, 2019 at 13:15 comment added Federico Poloni it's not easy to find the derivative of a function of a matrix over a matrix Actually it's quite easy. All you need here is the formula $(X+H)^{-1} = X^{-1} - X^{-1}HX^{-1} + O(\|H\|^2)$, which follows from the Neumann series. Applying it twice you should get your derivative.
Jun 18, 2019 at 11:56 history edited YCor
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Jun 18, 2019 at 11:52 comment added Pietro Majer I understand that $A$ (and then the others matrices as well) is not assumed to be symmetric. Then what do you mean here by "$A$ is positive definite matrix" (and $X\ge0$) ? Is it $v^\top A v >0$ for nonzero vectors $v$? Or $A$ has positive real eigenvalues?
Jun 18, 2019 at 6:15 history asked livehhh CC BY-SA 4.0