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Suppose that you have vectors $a_{1},...,a_{m}$ in $\mathbb{R}^n$. Can you take $n$ among them, let $v_{1},...,v_{n}$ such that $a_{1},...,a_{m}$ belong to the convex hull of $(cn)v_{1},-(cn)v_{1},...,(cn)v_{n},-(cn)v_{n}$ for some constant $c$?

My idea was to use gram-schmidt process where at step $i$ I choose the vector $u_{i}$ with the maximum euclidean norm. The answer would be those $a_{i}$'s that maximize the at each step the euclidean norm. Any ideas? Thanks

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  • $\begingroup$ 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! $\endgroup$
    – user31317
    Commented Feb 10, 2013 at 10:30
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    $\begingroup$ 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! $\endgroup$
    – user31317
    Commented Feb 11, 2013 at 0:32
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    $\begingroup$ Can you do this for $m=n+1$? $\endgroup$
    – Anton Petrunin
    Commented Feb 11, 2013 at 2:18
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    $\begingroup$ 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 $(c\cdot n)v_{1},-(c\cdot)nv_{1},...$ $\endgroup$
    – user31317
    Commented Feb 11, 2013 at 7:07
  • $\begingroup$ I changed my answer. $\endgroup$
    – Anton Petrunin
    Commented Feb 11, 2013 at 19:41

3 Answers 3

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Take $v_1,v_2,\dots,v_n$ which span a parallelepiped of maximal volume. If $$a_i=x_1\cdot v_1+\dots+x_n\cdot v_n$$ then $|x_k|\le 1$, otherwise exchanging $v_k$ to $a_i$ will increase the volume. Hence $a_i$ belongs to the convex hull of $\{\pm n\cdot v_{i}\}$; i.e. $c=1$.

Below is the original answer to the original question.

If you agree to choose $N=\tfrac{n{\cdot}(n+1)}{2}$ points then you can get $c=1$.

Take the ellipsoid of smallest volume which contains all $\{a_i\}$. You may assume that $\{a_i\}$ is in generic position, in this case at most $N$ of the points lie on the boundary of ellipsoid; take them as $\{v_i\}$. If one of $a_i$ does not lie in the convex hull of $\{\pm\sqrt{n}\cdot v_i\}$ then you can decrease the volume of the ellipsoid, by pushing it in one direction and expanding in all the orthogonal directions.

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  • $\begingroup$ Anton: The last version does not work since some $|x_k|$ could be (much) larger than $1$ and some (much) smaller. It is easy to construct examples with $n=2$ and $m=3$. $\endgroup$
    – Misha
    Commented Feb 11, 2013 at 22:33
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    $\begingroup$ @Misha: I do not think you are right; if $|x_i|>1$ once then you can increase the volume. $\endgroup$
    – Anton Petrunin
    Commented Feb 12, 2013 at 5:53
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    $\begingroup$ @Anton: Can you explain why by exchanging $u_{k}$ and $a_{i}$ the volume is increased? $\endgroup$
    – user31317
    Commented Feb 12, 2013 at 10:06
  • $\begingroup$ Observe that $v_{1},...,v_{n}$ are not orthonormal... $\endgroup$
    – user31317
    Commented Feb 12, 2013 at 10:18
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    $\begingroup$ The volume is the absolute value of determinant of matrix $A$ which has as columns the vectors $u_{1},...,u_{n}$. If you change $u_{1}$ with $a_{i} = x_{1}u_{1}+...+x_{n}u_{n}$ (let $|x_{1}|\geq 1$) then the volume of the new parallelepiped will be $|x_{1}||det(A)|$ is that right? I believe you are correct!!! $\endgroup$
    – user31317
    Commented Feb 12, 2013 at 10:29
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This reduces to the problem of finding a set of linearly independent vectors with maximum cardinality. There are in general many such sets, but any of them is a solution if you pick $c$ large enough. Then the convex hull is an $n$-orthoplex (AKA cross-polytope). If you make $c$ big, it will include any set of points in the span of the set, including on particular $a_1,...,a_m$. Some googling reveals that an algorithm for finding such a set is here.

Or are you trying to produce a solution with minimum $c$? This is a much more interesting question, by which I mean that I don't know the answer (:-).

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    $\begingroup$ I think the question is whether a constant $c$ exists for which a solution is possible for all choices of $m$, $n$, $a_1,\ldots,a_m$. $\endgroup$
    – Yoav Kallus
    Commented Feb 10, 2013 at 16:13
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This is just to say that I don't think choosing the vectors with the maximum Euclidean norm will suffice. In the example below, the two longest vectors (red) are collinear, and the hull derived from them—regardless of $c$—will be a line segment, which cannot contain the original (green) points.
           Hull Vectors

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  • $\begingroup$ the idea is to choose pairs $\pm v_i$ of maximal norm, IMHO... $\endgroup$
    – Dima Pasechnik
    Commented Feb 10, 2013 at 14:44
  • $\begingroup$ @Dima: Would not this example still lead to collinear $v_i$? $\endgroup$
    – Joseph O'Rourke
    Commented Feb 10, 2013 at 15:47

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