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Let $u_1,u_2,u_3,u_4$ be vectors in $\mathbb{ R^2}$ and $$ u= \sum_{j=1}^{4} t_ju_j;\text{ }t_j>0 \text{ and } \sum_{j=1}^{4} t_j=1$$

Then three vectors $ v_1,v_2,v_3 \in \mathbb{R}^2$ may be chosen from ${{ u_1,u_2,u_3,u_4}}$ such that $$ u= \sum_{j=1}^{3}s_jv_j ;\text{ } s_j\geq 0 \text{ and } \sum_{j=1}^{3} s_j=1$$

Since $u_1,u_2,u_3,u_4$ are linearly dependent, i can replace $u_4$ (assuming $u_4$ is dependent one) by linear combination of $u_1,u_2,u_3$ so that $ u= \sum_{j=1}^{3}s_jv_j$ but I can't claim $\sum_{j=1}^{3} s_j=1$ .i'm stuck

Please give me a hint! (Using linear algebra ) I didn't studied topology yet, so it's hard for me to understand topological proof! Thanks.

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  • $\begingroup$ Do you mean chosen from or constructed from? Maybe "chosen from the span of.."? $\endgroup$
    – Paul
    Commented Nov 30, 2018 at 12:47
  • $\begingroup$ @Paul chosen from given vectors. And not from span $\endgroup$
    – Cloud JR K
    Commented Nov 30, 2018 at 12:52
  • $\begingroup$ Source : part A 6th question. univ.tifr.res.in/gs2019/Files/GS2012_QP_MTH.pdf $\endgroup$
    – Cloud JR K
    Commented Nov 30, 2018 at 12:53
  • $\begingroup$ @CloudJR you can't do what you say you can. For example, if $\;u_1=u_2=0\;,\;\;u_3=(1,0)\;,\;\;u_4=(0,1)\;$ , there in $\;u=u_4\;$ you can't dispose of $\;u_4\;$...unless some other conditions are given. $\endgroup$
    – DonAntonio
    Commented Nov 30, 2018 at 13:04
  • $\begingroup$ @DonAntonio, well i actually assume u4 is dependent, let me edit it thanks $\endgroup$
    – Cloud JR K
    Commented Nov 30, 2018 at 13:45

1 Answer 1

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By its definition, $u$ lies in the convex hull of the $u_i$. It suffices to show that for any $u$ there are three $u_i$ whose convex hull contains $u$.

  • If there are four points on the convex hull, without loss of generality take them to be $u_1u_2u_3u_4$ in that order. Then $u_1u_2u_3$ and $u_3u_4u_1$ partition the hull, so $u$ lies in at least one of them; the points of the enclosing triangle may be taken as the $v_i$.
  • If there are only three points on the convex hull, a correct choice of $v_i$ is simply those three points.
  • The two- and one-point cases are trivial.
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  • $\begingroup$ I didn't studied convex hull before. Your proof looks geometrically intuitive, but it's hard for me to understand $\endgroup$
    – Cloud JR K
    Commented Nov 30, 2018 at 13:50
  • $\begingroup$ @CloudJR Perhaps you studied "barycentric coordinates" and stuff, which is just next to convex hulls and etc. ? $\endgroup$
    – DonAntonio
    Commented Nov 30, 2018 at 13:52
  • $\begingroup$ But i will try to understand, and i will comment if i understood or having any problem. Please help ! Thanks $\endgroup$
    – Cloud JR K
    Commented Nov 30, 2018 at 13:54
  • $\begingroup$ @DonAntonio nope. I'm not that good at geometry , i am just a second year undergraduate. I think it was a linear algebra problem, but when i tried no linear algebraic tools works... So i ask help here.I'm gonna check it out , in wiki or some books and let you know if i understood. Thanks $\endgroup$
    – Cloud JR K
    Commented Nov 30, 2018 at 13:59
  • $\begingroup$ As far as I remember this is standard 2nd. undergraduate year stuff, and many times it is taught within linear algebra and not until topology... $\endgroup$
    – DonAntonio
    Commented Nov 30, 2018 at 14:08

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