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 $

Question

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.

Answer 1

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.