Timeline for How bad could $\|A^k\|$ be when $\rho(A) < 1-\delta$ [closed]
Current License: CC BY-SA 3.0
23 events
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| Nov 2, 2015 at 8:57 | history | closed |
Yemon Choi Ryan Budney Terry Tao Qiaochu Yuan abx |
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| Nov 2, 2015 at 6:59 | answer | added | Federico Poloni | timeline score: 5 | |
| Nov 2, 2015 at 3:16 | comment | added | Alex Wenxin Xu | @YemonChoi, I see. Sorry I am very new to this forum.. I guess all I should expect is that the question in my mind is indeed an open question and I didn't miss any common knowledge. Thanks! | |
| Nov 2, 2015 at 3:09 | comment | added | Yemon Choi | @AlexWenxinXu That is because MathOverflow is not meant for open-ended questions, and not meant as some kind of discussion forum for evolving conversations | |
| Nov 2, 2015 at 3:07 | comment | added | Alex Wenxin Xu | @YemonChoi, alright .. You don't seem to like open-ended question.. | |
| Nov 2, 2015 at 1:52 | comment | added | Yemon Choi | I'm voting to close this question because it is a moving target and does not seem to show signs of enough thought before asking | |
| Nov 2, 2015 at 0:26 | history | edited | Alex Wenxin Xu | CC BY-SA 3.0 |
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| Nov 2, 2015 at 0:15 | comment | added | Alex Wenxin Xu | @YemonChoi, Thanks for the comments. Please see my reply to Christian as well. I mainly just wanted to control $\|A^k\|$ in a non-asymptotic way. In other words, I'd like to have some version of non-asymptotic Gelfand's formula. | |
| Nov 2, 2015 at 0:10 | comment | added | Alex Wenxin Xu | @ChristianRemling, Thanks for all your comments. I guess it's not clear what I want to prove and that's the beauty of its right? I just wanted to have something to bound about $\|A^k\|$ in terms of $\rho(A)^k$ weakly. For example, if $\rho(A) < 1/2$, can you say $\|A^k\| \le ||A||^{100} (2/3)^k$. I am looking for the correct theorem to be proved here -- so what's the right bound is just my question probably. Thanks! | |
| Nov 1, 2015 at 22:02 | comment | added | Igor Rivin | @FanZheng The OP wanted just ONE $k.$ $k=1$ is fine in that case. | |
| Nov 1, 2015 at 21:15 | comment | added | Fan Zheng | @IgorRivin You probably overshot this: if A=[[0 x][0 0]] then $A^2=0$. | |
| Nov 1, 2015 at 18:34 | comment | added | Stefan Kohl♦ | @ChristianRemling: Ah, I see -- there was some history ... . I remove my comment. | |
| Nov 1, 2015 at 18:27 | comment | added | Christian Remling | @StefanKohl: I was referring to this question and its predecessor: mathoverflow.net/questions/222205/… | |
| Nov 1, 2015 at 18:01 | comment | added | J.J. Green | This may be of interest (Section 3.3 in particular). | |
| Nov 1, 2015 at 17:25 | comment | added | Christian Remling | Sebastian's example also refutes your latest version since $\|A^k\|\sim k \|A\|\rho(A)^k$. As Yemon pointed out, it really can't work very well if your question + edits is a livestream of your thought process. Please try to think it through before you go public. | |
| Nov 1, 2015 at 15:18 | comment | added | Yemon Choi | Alex, I think that you need to figure out what you actually want to prove, rather than continually adding extra conditions every time someone points out a counterexample | |
| Nov 1, 2015 at 13:50 | comment | added | Alex Wenxin Xu | @IgorRivin Please see my edit of the question and my comments to Sebastian. I would really like to get something constructive here.. | |
| Nov 1, 2015 at 13:48 | history | edited | Alex Wenxin Xu | CC BY-SA 3.0 |
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| Nov 1, 2015 at 13:45 | comment | added | Alex Wenxin Xu | @SebastianGoette Thanks! yeah I kind of thought such counter-examples. but this cases have a simple fix -- the frobenius norm of the matrix is bounded. So my point here is not to find a counter example but I'd like to get some constructive theory that really rule out the bad instances. | |
| Nov 1, 2015 at 13:45 | review | Close votes | |||
| Nov 2, 2015 at 8:58 | |||||
| Nov 1, 2015 at 13:26 | comment | added | Igor Rivin | To amplify on the previous comment by @SebastianGoette: replace the $1/2$ in his example by $0.$ | |
| Nov 1, 2015 at 13:23 | comment | added | Sebastian Goette | Consider $\binom{1/2\;x}{0\;1/2}$. It has spectral radius $1/2$ and operator norm $\ge|x|$. Now put $k=1$ and $x$ as large as you like ... | |
| Nov 1, 2015 at 11:33 | history | asked | Alex Wenxin Xu | CC BY-SA 3.0 |