Definition of a Cone: Does it Include Zero Vector?

[Bashyboy]

On this wikipedia page https://en.wikipedia.org/wiki/Cone_(linear_algebra) , "a subset $C$ of a real vector space $V$ is a cone if and only if $\lambda x$ belongs to $C$ for any $x$ in $C$ and any positive scalar $\lambda$ of $V$."

The book in this link https://books.google.com/books?id=P...does the cone contain the zero vector&f=false defines the cone as always containing the zero vector.

I am slightly perplexed, especially since a few other sources I have come across define it as having zero, or simply define $\lambda \ge 0$, which implies that it contains the zero vector. So, my first question is, does the cone contain the zero vector? How is it typically defined?

My next question is, why does $V$ have to be a real vector space? Can't we have cones in $\mathbb{C}^n$ or $M_n(\mathbb{C})$? In the wiki article, I see that they say the concept of a cone can be extended to those vector spaces whose scalar fields is a superset of the ones they mention. So, because "$\mathbb{R} \subseteq \mathbb{C}$," the very same definition of a cone can be extended to $\mathbb{C}^n$, without any modification?

 

[S.G. Janssens]

So, my first question is, does the cone contain the zero vector? How is it typically defined?

It is typically defined as in the Wikipedia entry. Sometimes $\mathbf{0}$ is left out, but this should be explicitly stipulated.

My next question is, why does $V$ have to be a real vector space? Can't we have cones in $\mathbb{C}^n$ or $M_n(\mathbb{C})$? In the wiki article, I see that they say the concept of a cone can be extended to those vector spaces whose scalar fields is a superset of the ones they mention. So, because "$\mathbb{R} \subseteq \mathbb{C}$," the very same definition of a cone can be extended to $\mathbb{C}^n$, without any modification?

Cones are related intimately to the theory of ordered vector spaces. In this Wikipedia article, these spaces are always taken over $\mathbb{R}$ but you can in fact replace $\mathbb{R}$ by any ordered field. That explains why you will not (or: not often) encounter complex ordered vector spaces, because $\mathbb{C}$ is not an ordered field.

 

[Bashyboy]

Okay, I see. So, the set of all positive semidefinite matrices in $M_n(\mathbb{C})$ as a convex cone hasn't been studied much?

 

[S.G. Janssens]

Okay, I see. So, the set of all positive semidefinite matrices in $M_n(\mathbb{C})$ as a convex cone hasn't been studied much?

Probably not, would be my first guess.

You could try to see where you get when you regard $M_n(\mathbb{C})$ as a vector space over $\mathbb{R}$ instead of $\mathbb{C}$.