How smooth are the singular values of a matrix $F$ in terms of entries of $F$? I am hoping for Lipschitz continuity, but was not able to find it.
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$\begingroup$ I think the extreme (i.e smallest and largest) singular-values are $1$-Lipschitz functions of the input matrix (w.r.t operator norm). This can be seen via the variational characterization of these. $\endgroup$– dohmatobCommented Feb 20, 2021 at 15:21
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3 Answers
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The singular values of $F$ are the (square roots of ) eigenvalues of $F F^t,$ and the regularity of the latter have been studied half-to-death. See either T. Kato (perturbation theory of linear operators, ch. 1) or Golub-van Loan (Matrix Computations -- they almost certainly talk about singular values directly, without going through eigenvalues, but at worst talk about eigenvalues).
answered Feb 10, 2012 at 20:37
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2$\begingroup$ Thanks for the answer. Corollary 8.6.2. in Golub-van Loan gives exactly Lipschitz continuity. $\endgroup$ Commented Feb 10, 2012 at 20:48
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Check these references
Dmitri Alekseevsky, Andreas Kriegl, Mark Losik, Peter W. Michor, Choosing roots of polynomials smoothly, Israel J. Math 105 (1998) 203–233, math.CA/9801026,
Andreas Kriegl, Mark Losik, Peter W. Michor, Choosing roots of polynomials smoothly, II, Israel J. Math. 139 (2004) 183–188, math.CA/0208228.
answered Feb 10, 2012 at 22:31
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Since this got bumped to the front page...
Let $\sigma_i$ be the singular values of $A$ and $\tilde{\sigma}_i$ be those of $A+E$. Then, $|\sigma_i - \tilde{\sigma_i}| \leq \|E\|$, by Weyl's inequalities. That gives you 1-Lipschitz continuity.
answered Oct 20, 2021 at 7:12