how to draw the space of such linear combinations?

Question

We have the linear combination

$$ {2 \choose 1 } x_1 + {1 \choose 2} x_2 + {1 \choose -2} x_3 + {1 \choose 1} x_4 + {-1 \choose 0 } x_5 + {0 \choose -1 }x_6 $$

As $x_i \geq 0 $ is given, according to the definition, the linear combination above generates a cone. But, how can we draw it? Isnt the linear combination above just the entire $\mathbb{R}^2$ plane?

Also, I need to decide whether the vector ${6 \choose 4}$ lies in the cone, but since the linear combination above represesnt the entire plane, then ${6 \choose 4}$ must lie in.

Am I missing something here?

Answer 1

Just to illustrate Yves' answer: in the below figure the original points are drawn in blue. The blue polygon is their convex hull. As Yves rightly pointed out, the origin (black) is inside this convex hull. To draw the cone, you now need to inflate this blue convex hull. This is illustrated by the red polygon. You will notice that you will ultimately end up with the whole plane when you keep on inflating the convex hull. Hence, the cone you are looking for is indeed the whole plane and since ${6 \choose 4 }$ is obviously in the whole plane, it is also in the cone. Hope this helps.

Answer 2

You are right. As the origin lies inside the convex hull of the vectors, any point can be reached.

Answer 3

A possible way is as follows:

• Let's call the vectors $v_1={2 \choose 1 }, v_2 ={1 \choose 2}, v_3={1 \choose -2}, v_4 = {1 \choose 1}, v_5={-1 \choose 0 }, v_6={0 \choose -1 }$
• Excluding the trivial case $x_1 = \ldots = x_6 = 0$ note that
• $S = \sum_{i=1}^6x_i > 0 \Rightarrow \sum_{i=1}^6\frac{x_i}{S}v_i$ is a convex combination of the given vectors.
• So, the convex hull of the given vectors $\{v_1, \ldots , v_6\}$ must lie in the cone.
• If this covex hull contains an open ball around ${ 0\choose 0 }$ - which is the case - then the "cone" is the whole plane, because any positive multiple of the elements of this convex hull must lie in the cone, as well.

Answer 4

If $\max(u,v)\lt0$, then $$ \begin{pmatrix}u\\v\end{pmatrix}=(-u)\begin{pmatrix}-1\\0\end{pmatrix}+(-v)\begin{pmatrix}0\\-1\end{pmatrix} $$ If $\max(u,v)\ge0$, then $$ \begin{pmatrix}u\\v\end{pmatrix}=\max(u,v)\begin{pmatrix}1\\1\end{pmatrix}+\max(v-u,0)\begin{pmatrix}-1\\0\end{pmatrix}+\max(u-v,0)\begin{pmatrix}0\\-1\end{pmatrix} $$