Timeline for Bounding the non-multiplicativity of isometric projection
Current License: CC BY-SA 3.0
17 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 19, 2016 at 12:58 | comment | added | fedja | Indeed, extra ones might come from the outside :-) I just was too much concentrated on not losing any. You are certainly welcome. As I said, the right bound is $C(\log n)\delta$ rather than $5n\delta$ but I doubt it makes big difference for you, so I stayed with simple tools. | |
| Sep 19, 2016 at 12:47 | comment | added | Asaf Shachar | Thanks! Your proof is really great, and I have learned a lot in the process. One minor remark: It is not true that $A_1$ has exactly the same number of eigenvalues in $K$ as $A_t$ for $t <1$. If we do not assume $A_1$ has no eigenvalues on $\partial K$, then the number of eigenvalues in $K$ might increase. Indeed, take $A_t=\begin{pmatrix} 2-t& 0 \\ 0 & t\end{pmatrix}$ and $K$ to be the standard unit disk. Indeed, in our application (phrased as lemma 3 in my answer) we could show $\partial K$ is free from eigenvlaues of $A_t$ only for $t<1$, so this distinction was relevant. | |
| Sep 18, 2016 at 20:07 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
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| Sep 18, 2016 at 14:39 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
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| Sep 17, 2016 at 22:53 | comment | added | fedja | You know more: namely that $|\lambda-\lambda_j(A_0)|\ge\|Q\|$ for every $j$. So, the minimal singular value of $A_0-\lambda I$ is large. | |
| Sep 17, 2016 at 20:34 | comment | added | Asaf Shachar | Thanks! I am almost there... :) Can you please say why the normality of $A$ implies $|(A_0-\lambda I)x|\ge \|Q\||x|$ for all $x \neq 0$? I only know that $|\lambda - \lambda_i(A_0)| = \|Q\|$ for some eigenvalue $\lambda_i(A_0)$ of $A_0$ (since $\lambda$ is on the boundary of the union of the disks with radius $\|Q\|$ around these eigenvalues). I know normality tells you that the singular values are the absolute values of the eigenvalues, and in particular the operator norm equals the maximal eigenvalue, but I do not see how it helps here. Thanks again. | |
| Sep 17, 2016 at 19:27 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
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| Sep 17, 2016 at 15:56 | comment | added | fedja | I edited mine :-). See if it is clear now | |
| Sep 17, 2016 at 13:08 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
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| Sep 17, 2016 at 13:03 | comment | added | Asaf Shachar | @fedjabut ... but the gap cannot be bridged using all the Gershgorin disks of $U$. However, I still do not understand how you show there is an eigenvalue of $U$ which is close to one of $A$, nor the precise version of the Gershgorin circle theorem you are using (I did not quite understand the relevance of your previous comment about using different norms, doesn't it become relevant only when using "blocks version" of the theorem?) I would be very happy if you could write more explicitly your full argument. Feel free to edit my answer if you want to. | |
| Sep 17, 2016 at 13:03 | comment | added | Asaf Shachar | Thanks, I see that now; Pieces are starting to fall into place... your proof seems very nice indeed. I am now beyond the obstacle of the eigenvalue which is far from the non-negative reals. I still do not understand how exactly did you use the Gershgorin disks. I think you are supposed to use the fact $|U-A|_{op} \le \delta$ to claim there is an eigenvalue of $U$ which is close to an eigenvalue of $A$ (which we know must be real positive). Then, you get that on the one hand $U$ has an eigenvalue close to the $x$-axis, and one eigenvalue that is quite far from it... | |
| Sep 17, 2016 at 12:28 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
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| Sep 17, 2016 at 11:16 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
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| Sep 16, 2016 at 17:28 | comment | added | fedja | I replied in the previous comment thread. Also, you do not need large $b$: the point $-1$ is also far from the positive semi-axis. All that I'm saying is that on the unit circle the distance to $1$ can only be twice bigger than the distance to non-negative reals. | |
| S Sep 16, 2016 at 16:15 | history | answered | Asaf Shachar | CC BY-SA 3.0 | |
| S Sep 16, 2016 at 16:15 | history | made wiki | Post Made Community Wiki by Asaf Shachar |