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Feb 12, 2013 at 18:46 vote accept CommunityBot
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Feb 12, 2013 at 18:45
Feb 12, 2013 at 18:42 vote accept CommunityBot
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Feb 11, 2013 at 19:41 comment added Anton Petrunin I changed my answer.
Feb 11, 2013 at 7:20 history edited user31317 CC BY-SA 3.0
deleted 28 characters in body; edited tags
Feb 11, 2013 at 7:07 comment added user31317 Sure since $a_{1},...,a_{n+1}$ span $R^n$ we get that \lambda_{1}a_{1}+...+\lambda_{n} a_{n} + \lambda_{n+1}a_{n+1} =0 where the scalars are not all zero. Then divide by the one with maximum absolute value, let max the index: a_{max} = \sum_{i\neq max} \frac{\lambda_{i}}{\lambda_{max}}a_{i} with |\frac{\lambda_{i}}{\lambda_{max}}|\leq 1. Hence $a_{max}$ can be written as a convex combination of $nv_{1}-nv_{1},...,nv_{n},-nv_{n}$. Actually I want to solve it (If i can) for $(c\cdot n)v_{1},-(c\cdot)nv_{1},...$
Feb 11, 2013 at 2:18 comment added Anton Petrunin Can you do this for $m=n+1$?
Feb 11, 2013 at 0:32 comment added user31317 Except of Anton the others didn't understand the question i guess. First of all $c$ must be independent of $n$. Also you may assume that $a_{1},...,a_{m}$ span $R^n$ and then you have to find (if there exist) $n$ (as the dimension) vectors $v_{1},...,v_{n}$ s.t that $a_{1},...,a_{m}$ belong to the convex hull of $(c\sqrt{n})v_{1},-(c\sqrt{n})v_{1},...,$. Anton the fact is the number of initial vectors is $m = O(n^2)$, so I want sparse number of them... Thanks again!
Feb 10, 2013 at 17:40 answer added Anton Petrunin timeline score: 4
Feb 10, 2013 at 15:58 answer added Carl Feynman timeline score: 1
Feb 10, 2013 at 14:39 answer added Joseph O'Rourke timeline score: 0
Feb 10, 2013 at 10:30 comment added user31317 Even for the case where we have that $(c⋅n)v_{1},−(c⋅n)v_{1},...,(c⋅n)v_{n},−(c⋅n)v_{n}$ might help. Thanks!
Feb 10, 2013 at 10:19 history asked user31317 CC BY-SA 3.0