A property of a linear image of the cube Let $[0,1]^3$ denote the unit cube in $\mathbb{R}^3$.
Let $L : \mathbb{R}^3 \to \mathbb{R}^2$ be a surjective linear map,
and let $H := L([0,1]^3)$ (which is generically a hexagon).
Can you provide a proof of this statement ?
If $w_1$, $w_2$, and $w_1 + w_2$ are all in $H$,
then there exist $v_1$, $v_2 \in [0,1]^3$ so that:  


*

*$L(v_1)=w_1$ and  $L(v_2)=w_2$ 

*$v_1 + v_2 \in [0,1]^3$

Property 2. is the hard part.
 A: The answer to your question is NO, and here is a counterexample.
Let $L(x,y,z)=(x-z,y-z)$, $w_1=(1,1)$, $w_2=(-\frac{1}{2},0)$, $v_1=(x_1,y_1,z_1)$, $v_2=(x_2,y_2,z_2)$.
Clearly $L(v_1)=w_1$ iff $v_1$ is of the form $v_1=F_1(z_1)=(1+z_1,1+z_1,z_1)$, and this $F_1(z_1)$ is in $[0,1]^3$ iff $z_1=0$. In particular, $w_1\in H$.
Clearly $L(v_2)=w_2$ iff $v_2$ is of the form $v_2=F_2(z_2)=(z_2-\frac{1}{2},z_2,z_2)$, and this $F_2(z_2)$ is in $[0,1]^3$ iff $z_2\in[\frac{1}{2},1]$. In particular, $w_2\in H$.
Moreover, the vector $F_3(z_1,z_2)=F_1(z_1)+F_2(z_2)=(\frac{1}{2}+(z_1+z_2),1+z_1+z_2,z_1+z_2)$ is in $[0,1]^3$ iff $z_1+z_2=0$. In particular, $w_1+w_2\in H$ (take $z_1=z_2=0$).
Finally, if we want all three of $F_1,F_2,F_3$ to live in $H$, we get the impossible system $z_1=0,z_2\in[\frac{1}{2},1],z_1+z_2=0$. This finishes the proof.
A: Incorrect answer, affine but not linear
I think I can provide a counter example.
Suppose the mapping is such that $(x,y,1)$ maps to $(x,y)$, and $(x,y,0)$ maps to $(x-1000,y-1000)$.  Then $w_1 = (0.1,0.2)$, $w_2 = (0.2,0.1)$ and $w_1+w_2 = (0.3,0.3)$.
This gives us $v_1 = (0.1+n_1, 0.2+n_1, 1-{n_1\over1000})$ and $v_2 = (0.2+n_2, 0.1+n_2, 1- {n_2\over1000})$
where $n_1,n_2 \in [0,0.8]$
This places the third coordinate of $v_1+v_2$ at $\ge 1.9994$, which is outside the cube.
