1

In a paper I've been studying it says:

Let $x$ in the cone $\mathbb R_+^n$ of all vectors in $\mathbb R^n$ with nonnnegative components ($n\in\mathbb N$)

Somebody tell me what does it means, please? $\mathbb R_+^n$ should be $[0,\infty)\times\dots\times [0,\infty)$ ($n$ times), but I don't understand why the cone $\mathbb R_+^n$. Maybe the cone $\mathbb R_+^n$ is different from $[0,\infty)\times\dots\times [0,\infty)$?

Mark
  • 7,841
  • 6
  • 38
  • 72

4 Answers4

4

A cone is a set $C\subseteq\Bbb R^n$ with $x\in C\implies\alpha x\in C$ for all $\alpha\ge0$. The name is motivated by the usualy geometric cones, but $\Bbb R^n_+ := [0,\infty)^n$ satisfies this too.

See this picture of different cones in $\Bbb R^3$, two are familiar, and the right most one is $\Bbb R^n_+$.

enter image description here

M. Winter
  • 29,928
1

A cone $C$ is a set in a vector space with two propertes: $x+y \in C$ whenever $x,y \in C$ and $tx \in C$ whenever $x\in C$ and $t \geq 0$. Since $[0,\infty ) \times [0,\infty ) \times ...[0,\infty ) $ has these properties it is a cone.

Kavi Rama Murthy
  • 311,013
  • Are you sure about $x+y \in C$ ? This seems to belong to the definition of a convex cone instead, not necessarily a cone. – nicomezi Jul 25 '18 at 08:24
  • @Kavi: $ x+y \in C$ is not required ! – Fred Jul 25 '18 at 08:25
  • @nicomezi Some authors add it to the definition because all they gonna do is discuss convex cones. However, I think the preferred definition of only cone should not contain it. – M. Winter Jul 25 '18 at 08:26
1

A subset $A$ of a vector space $X$ on a field $K$ with a notion of positivity is said to be a cone if, for all $x \in A$, $\lambda>0$, we have $\lambda x \in A$.

The infite version of the cone you probably have in mind satisfies this property if the vertex, or apex, is at the origin.

nicomezi
  • 8,254
1

Yes, we have $\mathbb R_+^n= [0,\infty)\times\dots\times [0,\infty)$.

If $C \ne \emptyset$ is a subset of $\mathbb R^n$, then $C$ is called a cone if $x \in C$ and $t \ge 0$ imply that $tx \in C$.

Hence $\mathbb R_+^n$ is a cone in $\mathbb R^n$

Fred
  • 77,394