Timeline for Why is this set of polynomials linearly dependent?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2016 at 23:08 | history | rollback | lisyarus | Rollback to Revision 2 | |
| Dec 30, 2016 at 23:03 | history | edited | asdf | CC BY-SA 3.0 | deleted 356 characters in body |
| May 21, 2016 at 11:31 | answer | added | zawy | timeline score: 1 | |
| Jul 6, 2015 at 14:26 | comment | added | Laurent LA RIZZA | The key in your question is that you looked for one polynomial to be a scalar multiple of another, whereas you should have looked instead for a polynomial to be a linear combination (with scalar multipliers) of the others. | |
| Jul 6, 2015 at 7:52 | comment | added | Colin McLarty | The mathematical fact is that, as you see, polynomials in $x$ do not form a finite dimensional vector space, but polynomials of degree 2 (or less) do form a 3 dimensional vector space with basis $1,x,x^2$. Linear independence of such polynomials can be judged inside that vector space. So you can treat these as (three dimensional) vectors. | |
| Jul 6, 2015 at 0:55 | answer | added | Brian Fitzpatrick | timeline score: 8 | |
| Jul 5, 2015 at 22:52 | vote | accept | asdf | ||
| Jul 5, 2015 at 22:51 | answer | added | Khadija Mbarki | timeline score: 4 | |
| Jul 5, 2015 at 22:46 | answer | added | molarmass | timeline score: 11 | |
| Jul 5, 2015 at 22:45 | answer | added | Joe | timeline score: 2 | |
| Jul 5, 2015 at 22:41 | answer | added | Vadim | timeline score: 3 | |
| Jul 5, 2015 at 22:41 | answer | added | mathreadler | timeline score: 2 | |
| Jul 5, 2015 at 22:40 | answer | added | Matt Samuel | timeline score: 3 | |
| Jul 5, 2015 at 22:38 | answer | added | alkabary | timeline score: 18 | |
| Jul 5, 2015 at 22:36 | history | edited | alkabary | CC BY-SA 3.0 | added 84 characters in body |
| Jul 5, 2015 at 22:30 | history | asked | asdf | CC BY-SA 3.0 |