Let $u_1,u_2,u_3,u_4$ be vectors in $\Bbb{R}^2$ and $$u=\sum_{j=1}^4t_ju_j$$ where $t_i>0$ and $t_1+t_2+t_3+t_4=1$. Then three vectors $v_i ;i=1,2,3$ may be chosen from $u_j;j=1,2,3,4 $ so that $$ u= \sum_{j=1}^3 s_jv_j$$ where $s_j \geq 0$ and $\sum s_j=1$
How to prove this ?
My attempt:
Four vectors in $\Bbb{R}^2$ are linearly dependent, so without loss of generality assume $u_4=a_1u_1+a_2u_2+a_3u_3$. Then $u=(t_1+t_4a_1)u_1 +(t_2+t_4a_2)u_2+(t_3+t_4a_3)u_3$, so choose $s_i=t_i+t_4a_i \;;i=1,2,3$
how to show $s_j \geq 0$ and $\sum s_j =1$ ? Is my approach correct? or anything else?