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
  • 2
    $\begingroup$ If $A$ is not invertible, then what do you mean by $A^{-1}$? $\endgroup$ – Mankind Aug 2 '15 at 18:17
  • $\begingroup$ They are singular -- $\bf A^{-1}$ and $\bf B^{-1}$ are carrying divide by zero errors. However, that issue cancels out in the end product of $\bf (A^{-1}+B^{-1})^{-1}$, for instance, if the inverses were being calculated by hand. $\endgroup$ – Eric Aug 2 '15 at 18:27
  • $\begingroup$ This is technically an ill-posed problem. "Canceling out" divide by zero when dealing with $A^{-1}$ and $B^{-1}$ is a non-trivial endeavor, and may depend heavily on how you define your way of doing it. $\endgroup$ – user2566092 Aug 2 '15 at 18:32
  • $\begingroup$ I think there is something very wrong with having "divide by zero errors" in a matrix. For example, try to calculate the inverse of a 2x2 zero matrix, say $A^{-1} = [a,b;c,d]$ where $a,b,c,d$ can be divide by zero errors, i.e. terms that are not necessarily $0$ when multiplied by $0$. Then $A^{-1}A = I$. One of the implications of this is that $a\cdot 0 + b\cdot 0 = 0$, yet at the same time it equals $1$. $\endgroup$ – user49719 Aug 2 '15 at 18:37
  • $\begingroup$ Gotcha. OK, I'm convinced! It seems to work if the matrices are identical (using ${\bf (A^{-1}+A^{-1})^{-1}=(} 2\bf{C)^{-1}D}$), but not in any other case. $\endgroup$ – Eric Aug 2 '15 at 18:59

From The Matrix Cookbook, page 19, if $\bf(A+B)$ is invertible, then $\bf(A^{-1}+B^{-1})^{-1}$ can be calculated using $\bf A(A+B)^{-1}B$.


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