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, there are two real positive numbers $a$ and $b$, such that $\rho(S)<a<b$. Is it true that $b\rho((bI-S)^{-1}T)<a\rho((aI-S)^{-1}T)$? If the above is not true in general, will it be true if $S$ and $T$ are non-negative matrices?
I have now modified the desired spectral inequality to $b\rho((bI-S)^{-1}T)\leq a\rho((aI-S)^{-1}T)$ in response to user1551's counterexample.
Without imposing further conditions, the statement is false in general. For a counterexample, let $$S=\pmatrix{1&0\\ 0&0},\ T=\pmatrix{0&0\\ 0&1},\ b>a>\rho(S)=1.$$ Then $b(bI-S)^{-1}T = a(aI-S)^{-1}T = T$.