I was given a HW question to:
Show that if $y-x,z-y,x-z\in U$ are all different with $\dim(U)=1$, then one of $x,y,z$ is a convex combination of the other two.
A note: The setting is that all vectors are in $\mathbb{R^n}$
My efforts:
denote $v=y-x$ then $z-y=\alpha v$ for some $\alpha\in\mathbb{R}^{n}$ since every non zero element of $U$ spans it.
note that $$ (1+\alpha)v=v+\alpha v=(y-x)+(z-y)=z-x=-(x-z) $$
thus $$ x-z=-(1+\alpha)v $$
thus $x-z$ can be shown to be in $U$ even without the explicit assumption. Then I tried to discard the use of the redundant data that $x-z\in U$ and just write $v=y-x,z-y=\alpha v$ but I can see how I would get a convex combination from this.
I would appreciate any help on the problem.
Choose a $v$ that spans $U$. Write $y-x = \alpha_1 v$, $z-y = \alpha_2 v$, $z-x = \alpha_3 v$. Without loss of generality, suppose that $0 \leq \alpha_2 \leq \alpha_3$. We then have $$ y = z - \alpha_2 v = z - \frac{\alpha_2}{\alpha_3}(z - x) = \left(1 - \frac{\alpha_2}{\alpha_3} \right)z + \frac{\alpha_2}{\alpha_3}x $$ So, $y$ is a linear combination of $x$ and $z$.