Timeline for are intersections of kernels also kernels? [closed]
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2012 at 15:06 | comment | added | JHM | "When you know something for real, ask again"--okay, I like that advice. | |
| Sep 25, 2012 at 4:00 | comment | added | Qiaochu Yuan | But it also doesn't seem to be the question the OP meant to ask, which is really a question about the exterior algebra of a single vector space rather than linear maps between two vector spaces; in that setting there are more natural maps (I think they are essentially generated by wedge products). | |
| Sep 24, 2012 at 14:41 | comment | added | Steven Landsburg | I take Qiaochu's definitive answer as evidence that this was in fact a perfectly reasonable question. | |
| Sep 24, 2012 at 6:34 | comment | added | Yemon Choi | While I sympathise a little with the OP's frustration, I find the claims "This question (phrased in terms of elementary linear algebra) is non-standard, not to be found in any textbook...In my opinion, this question has more substance and more significance than half of the daily MO posts. At bottom, almost every problem is a problem in linear algebra. And when we find an honest unanswered linear algebra question, we should appreciate its relevance" somewhat overdone. Note that this is "English-speak", so those of you who know how to decode may get some impression of what I actually think... | |
| Sep 24, 2012 at 5:13 | history | closed |
Andreas Blass Goldstern Fernando Muro Will Jagy Andrés E. Caicedo |
off topic | |
| Sep 23, 2012 at 21:59 | comment | added | Will Jagy | @Goldstern, I wouldn't worry about it. From what I can see there are two correct answers now. If the question is closed, no additional answers may be added, but you will be able to edit your own answer and, from your reputation points, the question. Both answers give consideration to what "canonical" should mean in this problem. That ought to suffice. | |
| Sep 23, 2012 at 21:54 | comment | added | Goldstern | I voted to close, which is inconsistent with my attempt at an answer. But I do not see how to retract my vote. | |
| Sep 23, 2012 at 21:53 | answer | added | Goldstern | timeline score: 1 | |
| Sep 23, 2012 at 21:52 | answer | added | Qiaochu Yuan | timeline score: 5 | |
| Sep 23, 2012 at 21:42 | comment | added | Will Jagy | Mr. Martel, I recommend you find the books by Greub on linear algebra and multilinear algebra, I believe those are separate volumes. Greub maintained a steady interest in coordinate free presentation, at the price of length of presentation. I have little interest in this myself. However, I am definitely against questions where someone, asking for help, then repeatedly says No, that one's not right. As you have such a clear idea of what you want, work on this for a few weeks, order some books on interlibrary loan. When you know something for real, ask again. | |
| Sep 23, 2012 at 21:25 | comment | added | JHM | "Not research level" is different than "elementary". This question (phrased in terms of elementary linear algebra) is non-standard, not to be found in any textbook, and even more importantly: still unanswered! The point of the question, ie. its entire substance, is not to take a different target space. In my opinion, this question has more substance and more significance than half of the daily MO posts. At bottom, almost every problem is a problem in linear algebra. And when we find an honest unanswered linear algebra question, we should appreciate its relevance. | |
| Sep 23, 2012 at 21:17 | comment | added | JHM | Among those who have voted to close, any further comments on why or how to improve the question? | |
| Sep 23, 2012 at 21:12 | history | edited | JHM | CC BY-SA 3.0 |
omitted first sentence
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| Sep 23, 2012 at 20:36 | comment | added | JHM | @Ivanov: I erred in not emphasized my main motivation. In the case of linear maps $\mathbb{R}^{2n} \to \wedge^{n+1}\mathbb{R}^{2n}$ there is no issue of `small' dimension, and a reasonable answer may be possible. | |
| Sep 23, 2012 at 20:28 | history | edited | JHM | CC BY-SA 3.0 |
clarifed motivation
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| Sep 23, 2012 at 20:15 | comment | added | Sergei Ivanov | If you assume that the target space $V'$ is always the same, then $R$ may fail to exist, because $\dim V'$ may be too small. If you allow a different target space, just take a direct product $S\times T:V\toV'\times V'$. | |
| Sep 23, 2012 at 20:13 | comment | added | JHM | @yang: sure, IF. | |
| Sep 23, 2012 at 20:09 | comment | added | JHM | @Stanislav: no good since any such $R$ is not canonical, ie. you still have to extend. Actually, one maybe cannot expect to do better than this: take the pullback of $S,T$ to obtain a subspace $P$ of $V \times V$ and a linear map $R_o: P \to V'$. Take $R$ to be the restriction of $R_o$ along the diagonal $\Delta \subset V \times V$. Then the kernel of $R$ is exactly the intersection $ker S \cap ker T$. This is canonical, but totally worthless. | |
| Sep 23, 2012 at 20:08 | comment | added | Deane Yang | I don't know of any canonical way to do this. If you allow the range to be changed, then $(S, T): V \rightarrow V' \oplus V'$ works. | |
| Sep 23, 2012 at 20:00 | comment | added | Stanislav | If $e_1, \ldots, e_n$ is the basis of the intersection of the kernels of $S$ and $T$, which is known, one can find $R$ as a solution of the linear system $R e_i = 0$, $i = 1, \ldots, n$. | |
| Sep 23, 2012 at 19:57 | comment | added | JHM | expanded and clarified the question. | |
| Sep 23, 2012 at 19:55 | history | edited | JHM | CC BY-SA 3.0 |
expanded and clarified.
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| Sep 23, 2012 at 19:39 | comment | added | Andreas Blass | Not research level, so I'm voting to close. | |
| Sep 23, 2012 at 19:37 | history | asked | JHM | CC BY-SA 3.0 |