MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
such that ?? in the first paragraph? – Betrand May 6 '13 at 20:54
@Betrand: fixed the taking-"<"-as-html-link problem. – Hans May 6 '13 at 21:06
Maybe this question is not appropriate for this site, but neverheless let me give a hint: $(hI-S)^{-1}T$ and $(hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}}$ have the same eigenvalues. – Mateusz Wasilewski May 6 '13 at 21:13
There, of course, should have been "nevertheless" in my previous comment. – Mateusz Wasilewski May 6 '13 at 21:14
Yes, that "hint" thing sounded like homework. I have one question: are $S$ and $T$ non-negative definite (this is the case that I referred to) or element-wise non-negative? – Mateusz Wasilewski May 6 '13 at 21:31
up vote 2 down vote accepted

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 $$\rho((hI-S)^{-1}T)=\rho((hI-S)^{-\frac{1}{2}}T(hI-S)^{-\frac{1}{2}})<1.$$

share|cite|improve this answer
@Mateusz Wasilewski: Fabulous. Thank you, Mateusz! – Hans May 6 '13 at 22:15
You are welcome. – Mateusz Wasilewski May 6 '13 at 22:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.