Given that the following square matrices are non-invertible:
$\bf A$, $\bf B$, and (A+B)
UPDATE: Assume $\bf (A+B)$ is invertible.
and given that
$\bf (A+I)$, $\bf (B+I)$, and $\bf [(A+I)+(B+I)]$
are invertible, is there a way to calculate $\bf (A^{-1}+B^{-1})^{-1}$?
Other information (which may or may not be necessary):
$\bf A=C^{-1}D$ (C is invertible)
$\bf A^{-1}=D^{-1}C$ (D is not invertible)
$\bf B=E^{-1}F$ (E is invertible)
$\bf B^{-1}=F^{-1}E$ (F is not invertible)
NOTE: All matrices can be assumed to be square matrices.
ALSO NOTE: If $\bf A$ and $\bf B$ are equal, then ${\bf (A^{-1}+B^{-1})^{-1}=(} 2\bf{C)^{-1}D}$
I've been trying to figure this out using The Matrix Cookbook (page 19, section 3.2.5, The Searle Set of Identities), but I seem to be stumped at this point (with the exception of the special case where $\bf A$ and $\bf B$ are equal, but I'd like to figure out a more generalized solution so that $\bf A$ and $\bf B$ can be different). Anyway, here's an example in R:
# C is invertible
C <- matrix(c(1.9,1.4,1.4,1.4,1.9,1.4,1.4,1.4,2.0),3,3)
# D is NOT invertible
D <- matrix(c(.289,.109,0.0,.109,1.46,0.0,0,0,0),3,3)
# E is invertible
E <- matrix(c(4.0,3.0,2.0,3.0,5.0,1.0,2.0,1.0,6.0),3,3)
# F is NOT invertible
F <- matrix(c(0,0,0,0,3,1.5,0,1.5,3),3,3)
A <- solve(C) %*% D
B <- solve(E) %*% F
I <- diag(3)
# How to find solve(solve(A) + solve(B))?
#
# If A and B were equal, then the answer is
# solve(2*C) %*% D
From The Matrix Cookbook, page 19, if $\bf(A+B)$ is invertible, then $\bf(A^{-1}+B^{-1})^{-1}$ can be calculated using $\bf A(A+B)^{-1}B$.