1

I know there is a common answer that the weights used to make a linear combination should be non-negative and sum up to one. But I think this condition is just a sufficient condition, not a sufficient and necessary condition. For example, there are four points in a two-dimensional plane, $x_1=(1,1) , x_2=(1,2), x_3=(2,1)$, and $x_4=(2,2)$. Is the point $(1/4)*x_1+(1/6)*x_2+(1/4)*x_3+(1/6)*x_4$ in the convex hull of $\{x_1, x_2, x_3, x_4\}$? Yes, it is, but its weights do not satisfy the condition. Is my opinion correct? Thanks!

Ricky
  • 3,148
Xu Yang
  • 13
  • Yes, that looks correct. – Tzimmo Nov 03 '23 at 08:36
  • 2
    You are correct. However the condition is "necessary" in the sense that a point belongs to the convex hull if and only if it has that nonnegative and adding up to 1 representation (regardless of the fact that it may have other representations, like in your question). Moreover the barycentric condition will be invariant under isometries and yours will not, etc., so it is a geometric notion. That's what makes it interesting. – Michal Adamaszek Nov 03 '23 at 08:37

1 Answers1

1

The point is in the convex hull if and only if there is at least one way to express it as a linear combination that satisfies the condition, not every combination has to satisfy the condition. Let $S$ be the convex hull of $\{0,1\}$ in $\mathbb{R}$, then $0.5$ is in $S$. It doesn't matter if $0.5$ can be expressed as $0.5\cdot 1+0\cdot 0$ and $0.5+0\ne 1$, since it can also be expressed as $0.5\cdot 1+0.5\cdot0$.

You will notice that since $x_1,x_2,x_3,x_4$ are linearly dependent, $$\frac14x_1+\frac16x_2+\frac14x_3+\frac16x_4=(\frac54,\frac76)$$ can be expressed in more than one way. For example, $$(\frac54,\frac76)=\frac{7}{12}x_1+\frac16x_2+\frac14x_3+0x_4$$ and these weights satisfy the condition.

Ricky
  • 3,148