# Can 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 $u= \sum_{j=1}^{3}s_jv_j$

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.

• Do you mean chosen from or constructed from? Maybe "chosen from the span of.."? – Paul Nov 30 '18 at 12:47
• @Paul chosen from given vectors. And not from span – Cloud JR Nov 30 '18 at 12:52
• Source : part A 6th question. univ.tifr.res.in/gs2019/Files/GS2012_QP_MTH.pdf – Cloud JR Nov 30 '18 at 12:53
• @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. – DonAntonio Nov 30 '18 at 13:04
• @DonAntonio, well i actually assume u4 is dependent, let me edit it thanks – Cloud JR Nov 30 '18 at 13:45

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.