Timeline for Non alternative $k$-linear maps vanishing on $\sum x_i=0$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2017 at 5:40 | vote | accept | Ali Taghavi | ||
| Sep 28, 2017 at 4:43 | comment | added | Ilya Bogdanov | $f(v,v,w)=f(1\cdot v+0\cdot w,v,w)=f(2v+w,v,w)-f(v+w,v,w)=0$. | |
| Sep 27, 2017 at 21:37 | comment | added | Ali Taghavi | My apology if my question is elementary: but may you apply your statement "Since each linear combination is a sum of two linear combinations with non zero coefficients" to the triple $(v,v,w)$? | |
| S Sep 27, 2017 at 20:14 | history | suggested | Luc Guyot | CC BY-SA 3.0 |
Fixed 2 typos
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| Sep 27, 2017 at 20:05 | review | Suggested edits | |||
| S Sep 27, 2017 at 20:14 | |||||
| Sep 27, 2017 at 19:17 | comment | added | Ilya Bogdanov | No, I first explicitly tell that all $\alpha_i$ are nonzero, and second tell that by summing two such expression we can get an arbitrary linear combination, with each coefficient being zero or nonzero on our choice. | |
| Sep 27, 2017 at 17:18 | comment | added | Ali Taghavi | Yes but you implicitly assume that all $\alpha_i$ are non zero. i can imagine a resolution for three vector: $2f(v,v,w)=f(v,v.2w+2v)+f(v,v,-2v)=0$. I guess that the general case can be solved, similarly. Yes? | |
| Sep 27, 2017 at 15:05 | comment | added | Ilya Bogdanov | You may perform the same for the last two vectors instead of the first two. | |
| Sep 27, 2017 at 14:52 | comment | added | Ali Taghavi | Thanks for your answer. How does this argument work for a triple $f(v,v,w)$, for example? | |
| Sep 27, 2017 at 14:09 | history | answered | Ilya Bogdanov | CC BY-SA 3.0 |