Non-singular M-matrix Strictly positive Spectral radius

Question

So I've been looking at M-matrices recently, and by its definition we have $A=sI-B$ where $B$ is strictly positive and we have that the spectral radius ($\rho$) (maximum modulus of the eigenvalues )$s\geq\rho(B)$. It is also said in Wikipedia that for non-singular ones it has $s>\rho(B)$. Does anyone have a proof, i've looked at the nonnegative book by plemmons but it is rather messy

Answer 1

By definition, a scalar $s$ is an eigenvalue of a square matrix $B$ if and only if $Bx=sx$ for some nonzero vector $x$. Since $Bx=sx$ is equivalent to $(sI-B)x=0$, it follows that $s$ is an eigenvalue of $B$ if and only if $sI-B$ is singular.

In your case, if $s>\rho(B)$, then $s$ isn't an eigenvalue of $B$. Therefore $sI-B$ is nonsingular.

If $s=\rho(B)$, then $s$ is the Perron eigenvalue of $B$. Hence $sI-B$ is singular.