# Show that there exist real numbers $s_i≥0$ for $i=1,2,3$ such that $u=s_1v_1+s_2v_2+s_3v_3$ with $s_1+s_2+s_3=1$.

Let $u_1$,$u_2$,$u_3$,$u_4$ are four vectors in $R^2$. Let a vector $u$ be defined as $u=t_1u_1+t_2u_2+t_3u_3+t_4u_4$, where each $t_i>0$ and $t_1+t_2+t_3+t_4=1$. Let us choose any three vectors $v_1$,$v_2$,$v_3$ from the set $\{u_1,u_2,u_3,u_4\}$. Then show that there exist real numbers $s_i≥0$ for $i=1,2,3$ such that $u=s_1v_1+s_2v_2+s_3v_3$ with $s_1+s_2+s_3=1$.

I see that the set $\{u,u_1,u_2,u_3,u_4\}$ is linearly dependent. But I have to select such $s_i$ to show the set of vectors $\{u,v_1,v_2,v_3\}$ is linearly dependent. Please help me to solve this.

$u=t_1u_1+t_2u_2+t_3u_3+t_4u_4$, where each $t_i>0$ and $t_1+t_2+t_3+t_4=1$, then $u$ can be any point inside (not on the edges) the quadrilateral with vertices $u_1,u_2,u_3,u_4$.
$s_i≥0$ for $i=1,2,3$ with $s_1+s_2+s_3=1$, then $s_1v_1+s_2v_2+s_3v_3$ is a point inside the triangle with vertices $v_1,v_2,v_3$.
But $u$ can very well not lie in that triangle.
For example take $u_1=(0,0),u_2=(0,1),u_3=(1,0),u_4=(1,1)$. Let $v_1=u_1,v_2=u_2,v_3=u_3$ and $u=(3/4,3/4)$.