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?

  • $\begingroup$ @Hans: the statement of your question is incomplete. $\endgroup$ – Chris Godsil May 5 '13 at 17:21
  • $\begingroup$ @Chris Godsil: Thank you for pointing it out. The editor apparently takes the less than symbol "<" followed with no blank as the beginning of an html link and hides all that follows. I have now edited it by inserting spaces after "<". $\endgroup$ – Hans May 5 '13 at 17:29

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.

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  • $\begingroup$ @Federico Poloni: Thank you for the crisp proof. $\endgroup$ – Hans May 6 '13 at 6:35
  • $\begingroup$ @Federico Poloni: I posed another related question regarding spectral radius of non-negative matrices mathoverflow.net/questions/129890/a-spectral-radius-inequality. Please take a look. $\endgroup$ – Hans May 6 '13 at 20:37
  • $\begingroup$ @Federico Poloni: the strict inequality sign in $b\rho((bI-S)^{-1}T)<a\rho((aI-S)^{-1}T)$ is actually incorrect. I have edited the question to reflect that. You proof still goes through with that modification. Would you want to modify it, for prosperity? $\endgroup$ – Hans May 15 '13 at 2:37
  • $\begingroup$ You are right! One needs an additional hypothesis (irreducibility) to get strict inequalities in P-F. $\endgroup$ – Federico Poloni May 15 '13 at 7:18

Update. This answer answers completely different question, see comments. Namely "positive" is substituted by "positive definite", norm is used instead of spectral radius, and quantifiers are different.

Not true in general: take $S=T=-I$. Then the inequality boils down to $\frac{b}{b+1}<\frac{a}{a+1}$ which is always false for $b>a>1$.

For positive symmetric matrices, yes. Fix $a$ and let $b\to+\infty$. The l.h.s equals to $\rho((I-\frac1bS)^{-1}T)$ which goes to $\rho(T)$. And the r.h.s. is greater than $\rho(T)$. Indeed, the matrix $S':=(I-\frac1aS)^{-1}$ satisfies $|S'(v)|>|v|$ for all $v\in\mathbb R^n\setminus 0$ (where $n$ is the size of the matrices). Let $v$ be an eigenvector of $T$ corresponding to the maximal eigenvalue $\lambda=\rho(T)$. Then $|S'T(v)|>|T(v)|=\lambda |v|$, hence $\rho(S'T)>\lambda$ by the minimax principle.

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  • $\begingroup$ @Sergei Ivanov: Nice (partial?) solution. Thank you. But I have two questions. 1. You proved the inequality for positive symmetric matrices with $b\rightarrow\infty$. Is the transition to finite $b>a$ obvious from your present result? 2. Could you please explicate $|S'v|>|v|,\,\forall v\in R^n\\0$? Is this where your symmetry condition on the matrices comes in? $\endgroup$ – Hans May 6 '13 at 6:33
  • $\begingroup$ @Hans: sorry, I meant "positive definite", not "positive elementwise". The inequality follows by diagonalization. The transition from $b\to\infty$ to a large $b$ is basically the definition of limit. $\endgroup$ – Sergei Ivanov May 6 '13 at 10:38
  • $\begingroup$ @Sergei Ivanov: I see about positive definite. Regarding $b\rightarrow\infty$, I understand it's the limit. I point is that the limit case does not prove the monotonicity of the spectral radius on finite parameter $a$. For monotonicity, one need to do more. Do you agree? $\endgroup$ – Hans May 6 '13 at 13:45
  • $\begingroup$ @Hans: yes, sure. I misread the question as "there exist $a$ and $b$ such that...". I am not sure about monotonicity. And actually I now see a flaw in the argument: $S'T$ is not symmetric, so its spectral radius is not equal to the norm. Sorry about this confusion. $\endgroup$ – Sergei Ivanov May 6 '13 at 14:19
  • $\begingroup$ @Sergei Ivanov: Your confusion comes from my inexact statement of the problem "there are two real ..., such that...". Sorry about that. I have now changed it to "... and two arbitrary ... such that...". See the edited version. Can you add explaination, but it'd be better not to change the original statement much, to your answer to reflect your revised opinion of the proof, for posterity? Thank you, Sergei. $\endgroup$ – Hans May 6 '13 at 15:05

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