Timeline for Which norms on vectors can be consistently decomposed?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2018 at 15:36 | comment | added | Mateus Araújo | On a second thought, the proof did require me to consider $\|1^{(n)}\|$ for all $n$, so it was incorrect. I rewrote it so that it now applies for all dimensions larger or equal to $3$. Thanks for the heads-up. | |
| Feb 26, 2018 at 15:33 | history | edited | Mateus Araújo | CC BY-SA 3.0 |
added 244 characters in body
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| Feb 25, 2018 at 22:20 | comment | added | Mateus Araújo | "e.g." means "for example". But no, you cannot insert $1^{(4)}$ if you're talking about norms in $\mathbb R^3$. Why would you? | |
| Feb 25, 2018 at 20:56 | comment | added | Lutz Mattner | Then we must have $n=3$, and you can't insert $1^{(4)}$. | |
| Feb 25, 2018 at 19:47 | comment | added | Mateus Araújo | Sure I can. I'm using the convention that e.g. $\|(a,b)\| := \|(a,b,0)\|$. Just pad the vectors with zeroes. | |
| Feb 25, 2018 at 17:58 | comment | added | Lutz Mattner | But then you cannot, as in your answer, insert $1^{(n)}$ for different $n$. | |
| Feb 25, 2018 at 17:43 | comment | added | Mateus Araújo | Indeed, it doesn't make any sense, so that's why I'm not talking about an arbitrary normed space. I thought the question already made clear that I'm talking about a fixed norm in $\mathbb R^n$. | |
| Feb 24, 2018 at 12:52 | comment | added | Lutz Mattner | Where are your norms defined? In an arbitrary normed space, permutation-invariance of the norm, or disjointness of the supports of vectors, just don't make sense. And perhaps you are considering some family of norms on different spaces without saying so? | |
| Feb 24, 2018 at 9:24 | history | edited | Mateus Araújo | CC BY-SA 3.0 |
deleted unnecessary assumption
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| Feb 23, 2018 at 16:39 | vote | accept | Mateus Araújo | ||
| Feb 23, 2018 at 15:48 | history | answered | Mateus Araújo | CC BY-SA 3.0 |