I am cross posting this question to the mathermatics stack exchange. please find it either at this link, https://math.stackexchange.com/q/2952989/430980, or below:

I have a question concerning the derivatives of approximate matrix inverses. I have a system, $$Ax = b$$ which I solve approximately (and with an iterative method) with: $$\Delta x = x - \tilde{A}^{-1}b$$ I would like to take the derivative of this process, i.e. find $$\frac{d\Delta x}{dD} = \frac{dx}{dD} - \frac{d\tilde{A}^{-1}}{dD}b - \tilde{A}^{-1}\frac{db}{dD}$$ I know that if I had an ideal inverse, my expression would be without tildes as I'd have the exact result: $$\frac{d\Delta x}{dD} = \frac{dx}{dD} - {A}^{-1}\frac{dA}{dD}{A}^{-1}b - \tilde{A}^{-1}\frac{db}{dD}$$ But for this, I cannot start with the typical assumption that $$AA^{-1} = I$$ and instead have $$A\tilde{A}^{-1} = C \neq I$$ Does anyone have any ideas for resources or techniques for such problems? Thank you.

Edit: To add some additional context, I'm using a Newton method in a CFD solver, so my linear system is $$\frac{\partial R}{\partial u}\Delta u= -R(u)$$ I'm attempting to differentiate the process. I solve the linear system using point Jacobi, and I break up the matrix into diagonal and off diagonal, so it looks like so: $$D \Delta u^{k+1}= -R(u) - \frac{\partial R(u) }{\partial u}\Delta u^{k}$$

  • $\begingroup$ It depends how you build $\tilde{A}^{-1}$. If $C$ is known, then you have $\tilde{A}^{-1} = A^{-1} C$ and you can differentiate the right hand side. $\endgroup$ – user7440 Oct 12 '18 at 21:25
  • $\begingroup$ C is not known. I am using an iterative method. $\endgroup$ – EMP Oct 13 '18 at 0:21
  • $\begingroup$ It seems to be like you might not have enough information: you have a bunch of unknown matrices, with no explicit known relationships between them, and you want to differentiate one of the unknown matrices. Somewhere in there would have to be a way to relate the unknowns to the input data, but I don't see that in the question. Could you be more explicit about how exactly $\tilde A$ is computed, for example? $\endgroup$ – Kirill Oct 13 '18 at 10:36
  • $\begingroup$ I added an edit with some additional information. Hope that is helpful. Thank you $\endgroup$ – EMP Oct 13 '18 at 13:30
  • $\begingroup$ This problem seems similar to the works of Nemili et al., see papers here researchgate.net/profile/Anil_Nemili2 (Anil is a friend and collaborator of mine) $\endgroup$ – cpraveen Oct 13 '18 at 14:04

Call the approximate inverse $V$, then its products with $A$ are $$AV = I+E_{a},\quad VA = I+E_{v}$$ where $E_k$ and $E_a$ are "small" random error matrices which characterize your iterative method.

The error matrices are related since $$VAV=V+VE_a=V+E_vV\implies VE_a=E_vV$$ $$AVA=A+E_aA=A+AE_v\implies AE_v=E_aA$$

Pick one of the two products, differentiate, and solve for $dV$ $$\eqalign{ dV\,A &= -V\,dA + dE_v \cr dV\,(I+E_a) &= -V\,dA\,V + dE_v\,V \cr dV &= -V\,dA\,V + (dE_v\,V-dV\,E_a) \cr }$$ You haven't given us any information about your process, so I'll assume that the term in parentheses can be treated as uncorrelated noise.

  • $\begingroup$ Hey Greg, I'm using a point Jacobi method to solve the linear system. Does thay clarify it at all? $\endgroup$ – EMP Oct 13 '18 at 3:43
  • $\begingroup$ I added some additional information $\endgroup$ – EMP Oct 13 '18 at 13:29
  • $\begingroup$ @EMP Your original question was stated in terms of $(A,x,b)$, but your updated information is in terms of $(R,u)$ and various derivatives. I'm having trouble connecting the two. $\endgroup$ – greg Oct 13 '18 at 15:13
  • $\begingroup$ Hi Greg, The original statement was an attempt to be as broad as possible, but some people commented requesting additional information. So I tried to provide the information that indicates I am using newtons method to find the root of the nonlinear function R. the matrix A is Rs derivative, the vector x is the state field update $\Delta u$ and the right hand side vector, b is the negative of the R function evaluated at the current state. $\endgroup$ – EMP Oct 13 '18 at 15:57

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