Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Does anybody know the inequality of singular value for differences of matrices, i.e.

$\sigma_{max}\left(\begin{array}{c} A-B\end{array}\right)\leq??$

in term of $\sigma_{max}\left(\begin{array}{c} A\end{array}\right)$, $\sigma_{min}\left(\begin{array}{c} A\end{array}\right)$, $\sigma_{max}\left(\begin{array}{c} B\end{array}\right)$, or $\sigma_{min}\left(\begin{array}{c} B\end{array}\right)$

A and B are not Hermitian though.

share|cite|improve this question
Is $\bar \sigma$ the spectral radius? – Fabian Jan 8 '13 at 19:25
It is a norm, hence any norm inequality applies? – copper.hat Jan 8 '13 at 19:29
It is the maximum singular value, I presume. – copper.hat Jan 8 '13 at 19:29
I have edited the question, yes, it's singular value – aning Jan 8 '13 at 19:35

Recall that $\sigma_\max(M)=\max_{\|x\|=1}\|Mx\|$. By triangular inequality, clearly we have $\sigma_\max(A-B)\le\sigma_\max(A)+\sigma_\max(B)$.

share|cite|improve this answer
Thanks, but I need the tighter one, We know that $\sigma_{max}\left(\begin{array}{c} A\end{array}-B\right)\geq\max\left\{ \sigma_{max}\left(\begin{array}{c} A\end{array}\right),\sigma_{max}\left(\begin{array}{c} B\end{array}\right)\right\} -\min\left\{ \sigma_{min}\left(\begin{array}{c} A\end{array}\right),\sigma_{min}\left(\begin{array}{c} B\end{array}\right)\right\} $ Perhaps there's inequality like that for the upper bound? – aning Jan 8 '13 at 19:42
@RahayuPrihatin Admittedly, this one is not very useful, but it is tight. For instance, $\sigma_\max(I-(-I))=2=\sigma_\max(I)+\sigma_\max(-I)$. – user1551 Jan 8 '13 at 20:11

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.