TransWikia.com

Derivative of Inverse of sum of matrices

Mathematics Asked by Ronaldinho on February 1, 2021

Given is the function $f : mathbb{R}^p to mathbb{R}$ with

$$
f(x) = q(x)^{top} G^{-1} q(x)
$$

where $G = A + x_1 B_1 + ldots + x_p B_p$.

  • The matrices $A, B_1, ldots, B_p in mathbb{R}^{n times n}$ are all symmetric positive definit.

  • $q: mathbb{R}^n to mathbb{R}^n$ and the Jacobian $nabla q$ is known.

Is it possible to derive a closed form for $nabla f$? For me, the hard part is
$G^{-1}$. Any hints or suggestions are really appreciated!

One Answer

$$frac1{Delta x}big[(A+(x+Delta x)B)^{-1}-(A+xB)^{-1}big]=$$ $$frac1{Delta x}(A+xB)^{-1}(A+xB)big[(A+(x+Delta x)B)^{-1}-(A+xB)^{-1}big] (A+(x+Delta x)B) (A+(x+Delta x)B)^{-1}$$ $$=-(A+xB)^{-1}B (A+(x+Delta x)B)^{-1}rightarrow-(A+xB)^{-1}B(A+xB)^{-1}$$ as $Delta xrightarrow0$.

So that $partial f/partial x_j=2partial q/partial x_j^top G^{-1} q -q^top G^{-1}B_jG^{-1}q$.

Answered by Edwin Franks on February 1, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP