How to determine if a linear combination of $n$ variables is inside the convex hull of that $n$ variables

Question

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!

Answer 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.