Conical combination

 ( Convex Geometry ) 

$\backslash operatorname\{coni\}\; (S)=\backslash left\backslash \{\; \backslash sum\_\{i=1\}^k\; \backslash alpha\_i\; x\_i\; :\; x\_i\; \backslash in\; S,\backslash ,\; \backslash alpha\_i\; \backslash in\; \backslash mathbb\{R\}\_\{\backslash ge\; 0\},\backslash ,\; k\; \backslash in\; \backslash N\; \backslash right\backslash \}.$

By taking $k\; =\; 0$, it follows the zero vector (origin) belongs to all conical hulls (since the summation becomes an empty sum).

The conical hull of a set $S$ is a convex set. In fact, it is the intersection of all containing S plus the origin. If $S$ is a compact set (in particular, when it is a finite set of points), then the condition "plus the origin" is unnecessary.

If we discard the origin, we can divide all coefficients by their sum to see that a conical combination is a convex combination scaled by a positive factor.

Therefore, "conical combinations" and "conical hulls" are in fact "convex conical combinations" and "convex conical hulls" respectively. Moreover, the above remark about dividing the coefficients while discarding the origin implies that the conical combinations and hulls may be considered as convex combinations and in the projective space.

While the convex hull of a compact set is also a compact set, this is not so for the conical hull; first of all, the latter one is unbounded. Moreover, it is not even necessarily a closed set: a counterexample is a sphere passing through the origin, with the conical hull being an open half-space plus the origin. However, if $S$ is a non-empty convex compact set which does not contain the origin, then the convex conical hull of $S$ is a closed set.

• Affine combination
• Convex combination
• Linear combination