MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

The separation between two square matrices $A$ and $B$, often used as a measure of the sensitivity of invariant subspace problems, is defined as $$ \operatorname{sep}(A,B)=\min_{X\neq 0}\frac{\left\Vert AX-XB\right\Vert_F}{\left\Vert X\right\Vert_F} $$ (see e.g. Golub and Van Loan, Section 7.2.4).

I would like to express the separation between two M-matrices in terms of their Perron vector and values $Au=\lambda u$, $Bv=\mu v$. All I can do is estimating $\operatorname{sep}(A,B)\geq \lambda+\mu$. Is there any better bound, maybe from above? The question looks like an "eigenvalues vs. singular values" bound, so I am not sure that the answer is positive.

Alternatively, do you know any "handier" way to deal with $\operatorname{sep}(A,B)$, other than using its definition and its alternative formulation as the smallest singular value of a Kronecker sum $\sigma_{min}(B \otimes I + I \otimes A^T)$? I always find it unwieldy to use this definition, it is not easy to squeeze something simple to compute/estimate out of it. For instance, if $A$, $A'$, $B$ are M-matrices and $A'\geq A$, does $\operatorname{sep}(A',B)\geq \operatorname{sep}(A,B)$ hold?

share|cite|improve this question
Nitpick: $\mu$ is supposed to be the dominant eigenvalue of one of those M-matrices, 'no? – J. M. Aug 25 '10 at 16:04
You're right, I've corrected it. – Federico Poloni Aug 25 '10 at 17:02
What do you mean by "$\sigma_{min}$", the smallest singular value? Your last question is not true, even for the scalars, in this case one may take $X=1$ or $X=-1$. – Sunni Sep 17 '10 at 3:47
Yes, with $\sigma_{min}$ I mean the smallest singular value. And for the last question, I intended $A$ and $B$ to be M-matrices. I'll edit the question, thanks for pointing out. – Federico Poloni Sep 17 '10 at 17:07

Your Answer


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

Browse other questions tagged or ask your own question.