Define $\rho(A)$ to be the spectral radius of a square matrix $A$. Let $S$ and $T$ be two non-negative square matrices and $h$ a real number such that $\rho(S+T) < h$. Show that $\rho((hI-S)^{-1}T) < 1$.
A hint is $hI-S$ are invertible, and $hI-(S+T)=(hI-S)(I-(hI-S)^{-1}T)$. Since $hI-(S+T)$ is invertible, $I-(hI-S)^{-1}T$ is too. But I do not see how the last statement leads to the result. Can someone show the way?
Since $(hI-S)^{-1} = \frac{1}{h} \sum_{k=0}^{\infty}\left(\frac{S}{h}\right)^k$, $(hI-S)^{-1}T$ is non-negative. From the Perron-Frobenius theorem spectral radius is equal to the greatest (positive) eigenvalue. It is then enough to prove that for $\lambda \geq 1$ the matrix $\lambda I - (hI-S)^{-1}T$ is invertible. But this is equal to $(hI-S)^{-1}(\lambda h I - \lambda S - T)$ and, once again from Perron-Frobenius theorem, $\rho(\lambda S + T) \leqslant \rho(\lambda (S+T)) < \lambda h$, because $\lambda S + T \leqslant \lambda (S+T)$ entrywise.
The non-negative definite case: observe that $$hI - S -T = (hI-S)^{\frac{1}{2}}(I - (hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}})(hI-S)^{\frac{1}{2}}$$ so $$I - (hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}} = (hI-S)^{-\frac{1}{2}}(hI - S - T) (hI-S)^{-\frac{1}{2}}.$$ Since $h > \rho(S+T)$, $(hI - S - T)$ is positive definite, hence the right-hand side is positive definite, and that implies $$I > (hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}}$$ so that $$\rho((hI-S)^{-1}T)=\rho((hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}})<1.$$
