it is known that if $A=M-N$ is a regular splitting of the matrix $A$ and $A^{-1}\geq 0$, then $$\rho(M^{-1}N)=\frac{\rho(A^{-1}N)}{1+\rho(A^{-1}N)}.$$
And if $AN=NA$, it holds $$\rho(M^{-1}N)=\frac{\rho(A^{-1})\rho(N)}{1+\rho(A^{-1})\rho(N)}.$$ I wonder how to prove this equality. It is known that if $AN=NA$, then $\rho(AN)\leq \rho(A)\rho(N)$. So I guess that because of the special relationship of $A$ and $N$, so $\rho(AN)=\rho(A)\rho(N)$. Am I right? Or is there is anyone who can help to prove the second equatlity?