Timeline for Unitary condition
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2019 at 16:26 | comment | added | LSpice | @SamZbarsky's comment referenced by @IosifPinelis. | |
| Nov 18, 2019 at 13:57 | vote | accept | Toni Mhax | ||
| Nov 17, 2019 at 20:55 | history | edited | Morgan Rogers | CC BY-SA 4.0 |
Added an example.
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| S Nov 17, 2019 at 19:44 | history | suggested | Toni Mhax | CC BY-SA 4.0 |
You meant min i guess. Tx
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| Nov 17, 2019 at 19:06 | comment | added | Iosif Pinelis | Concerning conjecture 2), Sam Zbarsky's comment shows it is false. | |
| Nov 17, 2019 at 18:32 | review | Suggested edits | |||
| S Nov 17, 2019 at 19:44 | |||||
| Nov 17, 2019 at 18:05 | comment | added | Morgan Rogers | The feel of this proof reminded me of the Gershgorin Theorems bounding the spectrum of a matrix in terms of sums of column entries compared with diagonal entries. It's not directly relevant, but may be of interest. | |
| Nov 17, 2019 at 17:57 | history | edited | Morgan Rogers | CC BY-SA 4.0 |
Edited based on comments
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| Nov 17, 2019 at 17:18 | comment | added | Morgan Rogers | Oops yep, good point. I'll edit appropriately. | |
| Nov 17, 2019 at 17:16 | comment | added | Iosif Pinelis | I like this answer a lot. However, your conclusion $n |v_{i,i}|^2\le1$ would only be justified if you had $I(\pi_*)\ge1$, whereas you actually have $I(\pi_*)\le1$. Anyhow, you have proved a slightly weaker conclusion, $n |v_{i,i}|^2\le2$, which will imply the OP's conjecture 1) for $n\ge4$; the cases $n=2,3$ should be easy. Also, your proof applies to the more general setting with a general doubly stochastic matrix $(p_{ij})$ in place of the unistochastic matrix $(|u_{ij}|^2)$; see en.wikipedia.org/wiki/Unistochastic_matrix . | |
| Nov 17, 2019 at 16:19 | history | answered | Morgan Rogers | CC BY-SA 4.0 |