In Carathéodory’s Theorem, “Any $x$ in the convex hull of $S \subset E^n$, can be represented as a convex combination of n+1 elements of $S$.”, why is it that $x$ can be represented with $n+1$ elements and not $n$? Isn’t the number of basis in the Euclidean space $n$?
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1The basis elements might not be in $S$, and even if you write the element as a linear combination of basis elements, it may not be convex. – Sarvesh Ravichandran Iyer Jan 22 '18 at 01:08
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Well a convex hull is different from a linear hull. Think about the 2 dimensional case for a moment. Essentially it says that if $x$ is in the convex hull of $S$, then you can find 3 points such that $x$ is inside the triangle spanned by them.
On the other hand a triangle given as the convex hull of its corners is an obvious counter example as to why 2 points are not enough: the convex hull of any pair of corners is only the line segment connecting them.
So in essence it boils down to the fact that convex hull of $\le n$ points in an n-dimensional space has zero volume.
Hyperplane
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