I encountered an inequality when reading a paper. Can someone help to show how to prove it?


Let be the spectral radius of matrix $A$ or $\rho(A)=\max\{|\lambda|, \lambda \text{ are eigenvalues of matrix }A\}$. For matrices $S$ and $T$ with positive spectral radii, and two arbitrary real positive numbers $a$ and $b$, such that $\rho(S) < a < b$ Is the following inequality true?
$$b\rho((bI-S)^{-1}T) \leq a\rho((aI-S)^{-1}T)$$
If the above is not true in general, will it be true if $S$ and $T$ are non-negative matrices?



spectral radius monotonicity
Not true in general, as noted by @SergeiIvanov, but true for (element-wise) nonnegative matrices.


Note that if $\rho(S) < b$, then $b(bI-S)^{-1}=(I-\frac{S}{b})^{-1}=\sum\_{i=0}^\infty \frac{S^i}{b^i}$. In particular, thanks to this expansion, if $\rho(S)< a < b$, then $b(bI-S)^{-1}< a(aI-S)^{-1}$ in the componentwise ordering, and thus also $b(bI-S)^{-1}T \leq a(aI-S)^{-1}T$ for any nonnegative $T$. Now, it is a part of the Perron-Frobenius theorem that for any $A,B$ with $0 \leq A \leq B$ then $\rho(A) \leq \rho(B)$, and that's all we need here.


