Is it true that $\lim\limits_{z \to 0^+} \left((A + z I)^{-1} + B\right)^{-1} = A(A + ABA)^+ A$?

Question

Is it true that $\lim\limits_{z\to 0^+} ((A + zI)^{-1} + B)^{-1} = A(A + ABA)^+ A$, where $A, B$ are real symmetric positive semidefinite matrices, and $P^+$ denotes Moore-Penrose pseudo inverse?

Evidently it is true if $A$ is invertible, since $(A^{-1} + B)^{-1} = A(A + ABA)^{-1} A$.

It also holds for diagonal matrices, as can be checked with case work.

My question is what the limit relation is for this quantity; I believe the one given above should hold.

Answer 1

The answer is yes.

This is because

\begin{align*} ((A + zI)^{-1} + B)^{-1} &= (A + zI)(A + zI + (A+zI)B(A+zI))^+ (A + zI)\\ &= (I + (A + zI)B)^{-1} (A + zI) \to (I+AB)^{-1}A, \end{align*} as $z \to 0^+$. But, $$ (I + AB)^{-1} A = A(A + ABA)^+A. $$