Inverse of a sum of PSD matrices

Question

I was wondering if anyone knew any techniques to convert the following:

$ (A+B+C+..)^{-1} $

where $A,B,C...$ are positive semi-definite (PSD) matrices into a sum of some other function:

$ f(A)+f(B)+f(C)... $

where $f()$ is an arbitrary functon, if at all possible. I.e.:

$ (A+B+C+..)^{-1} = f(A)+f(B)+f(C)+.... $

If this is not possible, $f()$ can be an estimate or just capture relative differences. I.e., the relative contribution of $A$ in the inverse, $B$ in the inverse, etc.

A similar question was posted here: Inverse of the sum of matrices

and here: Finding the inverse of the sum of two symmetric matrices A+B

but I think this question is a little different

Answer 1

It is not possible, even for $1 \times 1$ matrices (i.e. ordinary numbers). If it were, then we'd have $$ \dfrac{\partial^2}{\partial A \; \partial B} (A + B)^{-1} = 0$$ but that is not the case: in fact $$ \dfrac{\partial^2}{\partial A\; \partial B} (A + B)^{-1} = 2 (A+B)^{-3}$$

On the other hand, you could use the following linear approximation: if $A, B, \ldots$ are close to $A_0, B_0, \ldots$ respectively and $Q = A_0 + B_0 + \ldots$ is invertible, then $$ \eqalign{(A + B + \ldots)^{-1} &\approx Q^{-1} - Q^{-1} (A - A_0 + B - B_0 + \ldots) Q^{-1}\cr = Q^{-1} &- Q^{-1} (A - A_0) Q^{-1} - Q^{-1} (B - B_0) Q^{-1} + \ldots \cr}$$

Answer 2

If $B$, $C$ etc have very low rank, then matrix inversion lemma shows how to start with $A^{-1}$ and sequentially compute the answer. Do a Google search. Not much can be said without further information.