Timeline for Notation for the all-ones vector [closed]
Current License: CC BY-SA 2.5
29 events
| when toggle format | what | by | license | comment | |
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| Nov 12, 2020 at 9:06 | comment | added | seralouk | What if I have a vector of length n with elements 1/n. What's the notation then? | |
| Nov 3, 2019 at 6:39 | comment | added | user76284 | @ThomasDybdahlAhle $n \times \{1\}$ gives you the vector of all 1s directly, if you interpret $n$ as an ordinal. | |
| Jan 24, 2019 at 12:53 | comment | added | Martin Sleziak | On Mathematics: Vector notation of all entries 1. | |
| Oct 6, 2015 at 12:21 | comment | added | Thomas Dybdahl Ahle | For vectors of specific length, you might use that $\{1\}^n$ is the set with one tupple $(1,\dots,1)$ of length $n$. For the vector itself, I'm considering using $\langle1\rangle^n$. | |
| May 18, 2015 at 18:46 | comment | added | LSpice | @EthanAlvaree, I think that physicists are used to $\mathbf i$ and $\mathbf j$ to mean unit vectors in the positive $x$- and $y$-directions. | |
| Apr 24, 2015 at 5:41 | history | closed |
Lucia Emil Jeřábek Joonas Ilmavirta Alex Degtyarev Dima Pasechnik |
Not suitable for this site | |
| Apr 24, 2015 at 3:01 | review | Close votes | |||
| Apr 24, 2015 at 5:41 | |||||
| Apr 15, 2015 at 21:55 | review | Close votes | |||
| Apr 16, 2015 at 1:22 | |||||
| Oct 15, 2014 at 3:16 | comment | added | EthanAlvaree | I have also seen $\mathbf{\iota}$ (boldface iota) denote a column of ones. More often seen, I think, are $\mathbf{i}$ and $\mathbf{j}$. Personally, though, I hope $\mathbf{1}$ catches on. | |
| Dec 28, 2009 at 22:06 | comment | added | Jonas Meyer | Kevin, Bkkbrad already wrote that above. (I'm being pedantic again.) | |
| Dec 28, 2009 at 16:33 | answer | added | Scott Carter | timeline score: 6 | |
| Dec 28, 2009 at 10:41 | comment | added | Kevin H. Lin | There sure are a lot of (pedantic, I would say) comments regarding the choice of basis. Since this seems to be in a graph theory context, perhaps the vector is being used to denote some sort of incidence information. In that case, we do have a preferred basis: that corresponding to the vertices of the graph, perhaps. | |
| Dec 28, 2009 at 0:55 | answer | added | Ilya Nikokoshev | timeline score: 1 | |
| Dec 28, 2009 at 0:00 | answer | added | Douglas S. Stones | timeline score: 2 | |
| Dec 27, 2009 at 23:23 | comment | added | Harry Gindi | Precisely my point. | |
| Dec 27, 2009 at 23:19 | vote | accept | Bkkbrad | ||
| Dec 27, 2009 at 23:17 | comment | added | Mariano Suárez-Álvarez | @Harry: a vector space is not a product of fields unless you fix a basis, and when you do both meanings of 1 denote the same element. | |
| Dec 27, 2009 at 22:42 | answer | added | Mariano Suárez-Álvarez | timeline score: 13 | |
| Dec 27, 2009 at 21:52 | comment | added | Jonas Meyer | Harry's last comment is why I think the arrow on top is a good thing. Nobody writes an arrow over the identity of a ring, I hope. | |
| Dec 27, 2009 at 21:46 | comment | added | Harry Gindi | The "all ones vector" has no meaning without choosing a basis, of course, because there's no canonical choice of a "1" vector. However, if you view your vector space as a direct product of fields, you can use the notation 1, since this element is the unit of the ring. It is terribly misleading, however, to use the notation 1 if you don't care about the ring structure. | |
| Dec 27, 2009 at 21:41 | comment | added | Bkkbrad | @Jonas: Absolutely right; the specific vector space I refer to in my paper has a natural correspondence between basis vectors and indexed vertices of a finite graph. | |
| Dec 27, 2009 at 21:38 | answer | added | Chris Godsil | timeline score: 6 | |
| Dec 27, 2009 at 21:35 | comment | added | Greg Kuperberg | I think the question is fine. | |
| Dec 27, 2009 at 21:34 | answer | added | Greg Kuperberg | timeline score: 30 | |
| Dec 27, 2009 at 21:33 | comment | added | Harry Gindi | This question should be closed for being too localized. | |
| Dec 27, 2009 at 21:31 | answer | added | Anton Petrunin | timeline score: 5 | |
| Dec 27, 2009 at 21:29 | comment | added | Jonas Meyer | "standard basis vector of a given vector space" To nitpick language: A "given" vector space need not have a standard basis. What you probably mean is that a basis for your vector space is fixed, or you are just considering the standard basis of $k^n$. The zero vector is different, because it is all zeros regardless of basis. However, I don't see anything wrong with your notation for all 1's so long as the basis is understood. If it doesn't seem too cumbersome and you want to be careful, you could write $\sum_{k=1}^n e_k$ if your basis is $e_1,\ldots,e_n$. | |
| Dec 27, 2009 at 21:26 | comment | added | Qiaochu Yuan | I've seen it written as 1, but I agree it's a little confusing to read. | |
| Dec 27, 2009 at 21:14 | history | asked | Bkkbrad | CC BY-SA 2.5 |