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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.

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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

1 Answer 1

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

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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

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