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I had notifications on my phone that I had received responses to my comments to an answer by Norbert on Regarding linear independence on a normed-linear space given a condition. When I went to the question, the answer appeared to have been deleted. Does anyone have sufficient karma to view deleted answers and want to share with me what the response comments were?

Thanks.

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    $\begingroup$ "@ziggurism What do you mean by "..asks to find one tuple (yi)i∈Nn" ? The problem asks to find one ϵ>0 s.t for all tuples (yi)i∈Nn with ∥yi∥≤ϵ the vectors x1+y1,..,xn+yn are linearly independent. This translates to find one ϵ>0 s.t x1+Bϵ,...,xn+Bϵ are linearly independent (informally said as the above are sets). And the ball Bϵ is in the possible infinite dimensional normed space X and thus not compact, as it is if you consider the finite dimensional L." The typesetting was better but in copy paste this does not work and I think you'll be able to read it. $\endgroup$ – quid Oct 6 '16 at 20:50
  • $\begingroup$ @quid thanks but I think I saw that one earlier and replied to it. Were there two that were posted earlier today? $\endgroup$ – ziggurism Oct 6 '16 at 21:16
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    $\begingroup$ It was posted 11h ago. You replied to the earlier (not that one) "The problem is that the yϵi are not supposed to be also in the span of x1,..,xn. Then you cannot represent yϵ1,...,yϵn with the same functionals c1,...,cn for every ϵ in the sequence that you construct. Or you have to justify how you would deal with this in the general case." Your reply "The problem asks to find one tuple yi . While it doesn't require that the tuple be in the span of the xi, it doesn't require that they not be either, so finding any independent tuple solves the problem, even if it is in the span.[...]" $\endgroup$ – quid Oct 6 '16 at 21:20
  • $\begingroup$ Anyway after your last visible comment on the answer there is the comment I gave and another not directed at you. I could not see deleted comments. $\endgroup$ – quid Oct 6 '16 at 21:21
  • $\begingroup$ @quid ok thanks. I guess that's the one. It didn't match the notification I thought I saw, but that must've been the one not directed at me. Thanks again. $\endgroup$ – ziggurism Oct 6 '16 at 21:23
  • $\begingroup$ I have added (specific-answer) tag (see also the tag-info. If I understand your post correctly, you are not asking about what can be done in similar situations in general. (Although answer to that would probably be "usually nothing" or "not much".) If I misunderstood your post, feel free to remove the tag. $\endgroup$ – Martin Sleziak Oct 7 '16 at 8:26
  • $\begingroup$ @MartinSleziak thanks Martin. You're right I was not intending to address generalities around deletion. Although things like this have happened to me before, and I do have opinions about the ease with which questions with active answers and answers with active conversations can be deleted frivolously, and how respectful that is toward the efforts and needs of the community. But I don't imagine such fundamental aspects of the functionality of the site are open to discussion at this stage. $\endgroup$ – ziggurism Oct 7 '16 at 12:55
  • $\begingroup$ Hm... I think almost anything about the site functionality is open to discussion. But there are many things which are very unlikely to be changed. And of course a lot of depends on your definition of the word frivolously. The poster probably had some reasons why they decided to delete answer with score +5. $\endgroup$ – Martin Sleziak Oct 7 '16 at 12:57
  • $\begingroup$ @MartinSleziak my guess is the poster decided there was an issue with the answer so that it was not correct, and deleted it to avoid the embarrassment of having an incorrect answer associated to his account. I think keeping incorrect answers along with explanations of why they are incorrect has value, so I don't approve of such deletions. But Is there a more charitable assumption we can make about just cause for deleting? $\endgroup$ – ziggurism Oct 7 '16 at 13:05
  • $\begingroup$ Well, I think that most user are for deleting incorrect answer. (More than with embarrassment, it has to do with the fact that this might be seen as source of incorrect information.) Somewhat related older discussion: Is sometimes keeping wrong answer reasonable?. Of course, as soon as you reach 10k reputation, it matters less to you whether an answer is deleted or not - you can still see it. $\endgroup$ – Martin Sleziak Oct 7 '16 at 13:19
  • $\begingroup$ @MartinSleziak ok I can understand that argument. Incorrect answers may be misleading or cause confusion for less experienced users, even if they have disclaimers. I don't especially like it, for my own benefit, I, as a more experienced user (if I may be so presumptuous), would prefer to see the arguments reach a logical conclusion and be preserved. But I can appreciate that the needs of the confused student should be paramount on this site. $\endgroup$ – ziggurism Oct 7 '16 at 13:50
  • $\begingroup$ I had never really paid any attention to StackExchange reputation, but maybe I should make an effort to reach that 10k milestone. $\endgroup$ – ziggurism Oct 7 '16 at 14:45
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Here is the whole comment exchange under this answer (currently deleted by the OP.). (At least the part that I can see. Moderators would be able to tell you whether there were additional comments, which were deleted.)


I am afraid that you loose the generality when restricting $y_i^\epsilon\in L$ so that the function $F$ is well defined. Nevertheless, I like your solution. -- Svetoslav

@Svetoslav: couldn't we remove the loss of generality but otherwise keep the proof the same by using the n-fold wedge map, instead of the determinant? -- ziggurism

@ziggurism I am not familiar with this map and its properties. -- Svetoslav

I think it provides a concise elegant proof of OP's question, so you may be interested in learning more. http://en.wikipedia.org/wiki/Wedge_product -- ziggurism

You are losing generality by assuming that $(y_i^\epsilon)_{i\in\mathbb N}\subset L$. In such a way you are answering the OP question at most in the case when $X$ is finite dimensional. -- Svetoslav

@Svetoslav in the very begining I'm saying that it is enough to answer the question for the linear span of vectors x_1, ...,x_n, not the whole X. -- Norbert

The problem is that the $y_i^\epsilon$ are not supposed to be also in the span of $x_1,..,x_n$. Then you cannot represent $y_1^\epsilon,...,y_n^\epsilon$ with the same functionals $c_1,...,c_n$ for every $\epsilon$ in the sequence that you construct. Or you have to justify how you would deal with this in the general case. -- Svetoslav

The problem asks to find one tuple $y_i$. While it doesn't require that the tuple be in the span of the $x_i$, it doesn't require that they not be either, so finding any independent tuple solves the problem, even if it is in the span. There could also be tuples not in the span, this is ok. I think there is no loss of generality. -- ziggurism

@ziggurism What do you mean by "..asks to find one tuple $(y_i)_{i\in\mathbb N_n}$" ? The problem asks to find one $\epsilon>0$ s.t for $\textbf{all}$ tuples $(y_i)_{i\in\mathbb N_n}$ with $\|y_i\|\leq \epsilon$ the vectors $x_1+y_1,..,x_n+y_n$ are linearly independent. This translates to find one $\epsilon>0$ s.t $x_1+B_\epsilon,...,x_n+B_\epsilon$are linearly independent (informally said as the above are sets). And the ball $B_\epsilon$ is in the possible infinite dimensional normed space $X$ and thus not compact, as it is if you consider the finite dimensional $L$. -- Svetoslav

Norbert, your contradiction argument also fits to "there exists $\epsilon>0$ s.t for all tuples $(y_i)_{i\in\mathbb N_n}$ with $\|y_i\|\leq \epsilon,\,i=1,..,n$ ", so you also prove that $x_1+B_\epsilon,...,x_n+B_\epsilon$ are "linearly independent" for some small $\epsilon$. I'm not trolling here, just want to understand what is going on. -- Svetoslav

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  • $\begingroup$ Thank you, Martin. $\endgroup$ – ziggurism Oct 12 '16 at 19:02

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