Is a convex cone a convex polyhedron?
Say that I have a convex cone
$C=\{t|Ax = t, x\geq 0\}$.
where $x\in R^n$, and $t\in R^m$, $A\in R^m\times R^n$.
Can I say that this is a convex polyhedron? and why?
EDIT: Just in order to avoid confusion, the definition of convex polyhedron I am thinking of is: the set of solutions to a finite number of linear equalities and inequalites.
$\endgroup$2 Answers
$\begingroup$Yes, or more precisely (as polyhedron may imply three dimensions) a convex polytope. You may readily verify that $C$ is convex, and also that its boundary is piecewise a part of a hyperplane.
$\endgroup$ 1 $\begingroup$No. A cone is not a "polyhedron". A polyhedron, by definition has planar (straight) sides, not curved sides.
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