Timeline for Singular values of sequence of growing matrices
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 6, 2016 at 18:37 | vote | accept | Eckhard | ||
| May 6, 2016 at 18:37 | |||||
| Jan 8, 2013 at 17:32 | comment | added | Suvrit | @Eckhard: Sorry for the confusion; indeed, I meant $(1+a)^2$ and $2(1+a^2)$, not $\theta$ and $2\theta$ (stupid carry-over typo on my part!). thanks for catching this error! as soon as i get a chance, i'll edit my answer to reflect this correction. | |
| Jan 7, 2013 at 22:04 | comment | added | Eckhard | @Suvrit: Is it possible that you meant to write in your analysis that $2^LM_L$ is comprised of the two numbers $\theta=(1+\alpha)^2$ and $2(1+\alpha^2)$, as opposed to $\theta$ and $2\theta$? This would lead to the bound $\sigma_1(K_L)\leq \frac{1}{2}\sqrt{3+2\alpha +3 \alpha ^2}$, which is tighter and indeed sharp for $\alpha=\pm 1$. | |
| Jan 6, 2013 at 12:03 | vote | accept | Eckhard | ||
| May 6, 2016 at 7:11 | |||||
| Dec 30, 2012 at 17:09 | history | edited | Suvrit | CC BY-SA 3.0 |
overhauled answer.
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| Dec 30, 2012 at 9:38 | comment | added | Suvrit | Certainly a closer characterization should be possible because the $K_L$ are column stochastic matrices that evolve with $L$ in a fairly regular way; if I get some time, I'll think a bit about this problem; otherwise, numerical experiments will offer a good "guess" that approximates the limit as a function of $\alpha$. | |
| Dec 30, 2012 at 8:59 | history | edited | Suvrit | CC BY-SA 3.0 |
added one more sentence
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| Dec 29, 2012 at 22:32 | comment | added | Eckhard | @Suvrit: Thank you for your answer. Indeed, $\sigma_1(K_2(0))=\sqrt{3/4}$, and after that the sequence $(\sigma_1(K_L(0)))_L$ seems to be decreasing. Do you see a way to characterize the limit more precisely, as alluded to in your last sentence? | |
| Dec 29, 2012 at 17:44 | history | answered | Suvrit | CC BY-SA 3.0 |