Timeline for Proof that bases etc. exist in early linear algebra course?
Current License: CC BY-SA 2.5
24 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 11, 2010 at 10:56 | comment | added | Zsbán Ambrus | I don't know what's the right way to teach this, but when I was first-year, the professor proved us that every vector space with a finite generator system has a basis (and we had to be able to prove that on the exam), but she only stated this without proof for infinite dimensions. | |
| Oct 7, 2010 at 19:15 | answer | added | Matthew Daws | timeline score: 2 | |
| Oct 7, 2010 at 18:02 | comment | added | Yemon Choi | Matt: this isn't quite what you want, I think, but have you seen this proof that in char zero any two bases must have the same cardinality? ams.org/mathscinet-getitem?mr=1328020 | |
| Oct 7, 2010 at 8:02 | answer | added | Matthew Daws | timeline score: 1 | |
| Oct 6, 2010 at 23:08 | comment | added | Gordon Royle | Are you really "loathed"? (by your students) :-) | |
| Oct 6, 2010 at 15:14 | comment | added | Matthew Daws | @Simon and Darij: Thanks! Actually, I have today re-written my notes with a form of the exchange lemma which is now very similar to what Wikipedia says. Originally I had a version which had to derive the Wiki version as a corollay. I agree that the version I now have is not "long", and it very quickly gives the other proofs I want. | |
| Oct 6, 2010 at 15:09 | comment | added | Simon Wadsley | Perhaps best of all would be to show inductively that the induction hypothesis is true for $k$ at most the minimum of $n$ and $m$ and then observe if the result is true for $k=n<m$ then we have a contradiction. I feel the idea is getting lost in the details now. But I won't change it. | |
| Oct 6, 2010 at 14:55 | comment | added | Simon Wadsley | Also even if it should be twice as long as it is now it still need not be described as very long. | |
| Oct 6, 2010 at 14:55 | comment | added | darij grinberg | (I remember that I knew some short workaround to this some time ago, but I can't remember it...) | |
| Oct 6, 2010 at 14:53 | comment | added | darij grinberg | The thing is, until you know that your $k$ is $<m$, speaking of $v_{k+1}$ makes no sense. So the first idea that came to me was replacing $0\leq k<n$ in your proof by $0\leq k<m$; but then the use of $w_{k+1}$ became questionable. | |
| Oct 6, 2010 at 14:50 | comment | added | Simon Wadsley | Fair enough. I really just edited it to make it better than it was before. I think it is now easier to improve upon without completely rewriting. The proof shows that if we get to $k=n<m$ then the $v_i$ are not LI. I agree that could be made more explicit. To be honest I was seeing the conclusion that $m\leq n$ as a corollary of the other part. | |
| Oct 6, 2010 at 14:50 | comment | added | darij grinberg | Hmm. I had to stretch it by a factor of 2 to be able to follow it without pen and paper. Is there something trivial that eluded me? | |
| Oct 6, 2010 at 14:24 | history | edited | Alex B. | CC BY-SA 2.5 |
fixed spelling
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| Oct 6, 2010 at 14:13 | comment | added | Simon Wadsley | Does the proof really have to be so long? See the Wiki entry now. It could still be slightly lengthened for maximal clarity but I don't think it is so hard to write out a shortish intelligible proof. | |
| Oct 6, 2010 at 13:33 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
edited title
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| Oct 6, 2010 at 13:32 | answer | added | Hany | timeline score: 3 | |
| Oct 6, 2010 at 13:28 | comment | added | Andrew Stacey | I agree about the state of that wikipedia page. It's awful! Something Ought To Be Done about it. | |
| Oct 6, 2010 at 13:27 | answer | added | Andrew Stacey | timeline score: 14 | |
| Oct 6, 2010 at 11:12 | answer | added | Bob Durrant | timeline score: 2 | |
| Oct 6, 2010 at 10:16 | answer | added | Alex B. | timeline score: 7 | |
| Oct 6, 2010 at 9:59 | answer | added | Robin Chapman | timeline score: 8 | |
| Oct 6, 2010 at 9:59 | comment | added | Matthew Daws | To get an idea of the problems I have, look at the Wiki entry for the Exchange Lemma: en.wikipedia.org/wiki/Exchange_lemma Now, I can follow this proof, but it's basically wrong: there's lots of implicit re-ordering going on, there's some basic lemmas invoked (to do with linear spans) and then an implicit induction is used: but it's not clear that you won't swap out v_1 at a later stage (you don't, because of linear independence). To write this up to be follow-able by my students would make it very long... | |
| Oct 6, 2010 at 9:50 | history | asked | Matthew Daws | CC BY-SA 2.5 |