Timeline for Distance of vectors versus distance of their difference vectors
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 26, 2017 at 16:40 | comment | added | Moritz Firsching | @RodrigodeAzevedo I see, now I understand your reasoning. | |
| Apr 26, 2017 at 16:33 | comment | added | Rodrigo de Azevedo | @MoritzFirsching I did not quite prove the statement. I proved that the statement holds in the extremely special case where the same permutation puts both vectors in non-decreasing order. That is not the case in the counterexample you gave. I am still working on the general case, trying to find an use for $\rm x^\top y \geq 0$. | |
| Apr 26, 2017 at 16:27 | comment | added | Moritz Firsching | Wasn't $\mathrm x^{\top} \mathrm y \geq 0$ a condition that was necessary, since otherwise a counterexample exists? (see my comment above, after which the OP added the condition). It is weird, that it would not be used, when trying to prove the statement. | |
| Apr 26, 2017 at 15:46 | history | edited | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor improvements
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| Apr 25, 2017 at 9:33 | history | edited | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor improvement
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| Apr 25, 2017 at 9:25 | history | edited | Rodrigo de Azevedo | CC BY-SA 3.0 |
added 7 characters in body
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| Apr 25, 2017 at 9:18 | history | answered | Rodrigo de Azevedo | CC BY-SA 3.0 |