ordered vector space
Let $k$ be an ordered field. An ordered vector space over $k$ is a vector space^{} $V$ that is also a poset at the same time, such that the following conditions are satisfied

1.
for any $u,v,w\in V$, if $u\le v$ then $u+w\le v+w$,

2.
if $0\le u\in V$ and any $$, then $0\le \lambda u$.
Here is a property that can be immediately verified: $u\le v$ iff $\lambda u\le \lambda v$ for any $$.
Also, note that $0$ is interpreted as the zero vector of $V$, not the bottom element of the poset $V$. In fact, $V$ is both topless and bottomless: for if $\perp $ is the bottom of $V$, then $\perp \le 0$, or $2\perp \le \perp $, which implies $2\perp =\perp $ or $\perp =0$. This means that $0\le v$ for all $v\in V$. But if $v\ne 0$, then $$ or $$, a contradiction^{}. $V$ is topless follows from the implication^{} that if $\perp $ exists, then $\top =\perp $ is the top.
For example, any finite dimensional vector space over $\mathbb{R}$, and more generally, any (vector) space of realvalued functions on a given set $S$, is an ordered vector space. The natural ordering is defined by $f\le g$ iff $f(x)\le g(x)$ for every $x\in S$.
Properties. Let $V$ be an ordered vector space and $u,v\in V$. Suppose $u\vee v$ exists. Then

1.
$(u+w)\vee (v+w)$ exists and $(u+w)\vee (v+w)=(u\vee v)+w$ for any vector $w$.
Proof.
Let $s=(u\vee v)+w$. Then $u+w\le s$ and $v+w\le s$. For any upper bound $t$ of $u+w$ and $v+w$, we have $u\le tw$ and $v\le tw$. So $u\vee v\le tw$, or $(u\vee v)+w\le t$. So $s$ is the least upper bound of $u+w$ and $v+w$. ∎

2.
$u\wedge v$ exists and $u\wedge v=(u+v)(u\vee v)$.
Proof.
Let $s=(u+v)(u\vee v)$. Since $u\le u\vee v$, $(u\vee v)\le u$, so $s\le v$. Similarly $s\le u$, so $s$ is a lower bound of $u$ and $v$. If $t\le u$ and $t\le v$, then $u\le t$ and $v\le t$, or $v\le (u+v)t$ and $u\le (u+v)t$, or $u\vee v\le (u+v)t$, or $t\le (u+v)(u\vee v)=s$. Hence $s$ the greatest lower bound^{} of $u$ and $v$. ∎

3.
$\lambda u\vee \lambda v$ exists for any scalar $\lambda \in k$, and

(a)
if $\lambda \ge 0$, then $\lambda u\vee \lambda v=\lambda (u\vee v)$

(b)
if $\lambda \le 0$, then $\lambda u\vee \lambda v=\lambda (u\wedge v)$

(c)
if $u\ne v$, then the converse^{} holds for (a) and (b).
Proof.
Assume $\lambda \ne 0$ (clear otherwise). (a). If $\lambda >0$, $u\le u\vee v$ implies $\lambda u\le \lambda (u\vee v)$. Similarly, $\lambda v\le \lambda (u\vee v)$. If $\lambda u\le t$ and $\lambda v\le t$, then $u\le {\lambda}^{1}t$ and $v\le {\lambda}^{1}t$, hence $u\vee v\le {\lambda}^{1}t$, or $\lambda (u\vee v)\le t$. Proof of (b) is similar^{} to (a). (c). Suppose $\lambda u\vee \lambda v=\lambda (u\vee v)$ and $$. Set $\gamma =\lambda $. Then $\lambda u\vee \lambda v=\lambda (u\vee v)=\gamma (u\vee v)=(\gamma (u\vee v))=(\gamma u\vee \gamma v)=((\lambda u)\vee (\lambda v))=((\lambda v\wedge \lambda u))=\lambda v\wedge \lambda u$. This implies $\lambda u=\lambda v$, or $u=v$, a contradiction. ∎

(a)
Remarks.

•
Since an ordered vector space is just an abelian pogroup under $+$, the first two properties above can be easily generalized to a pogroup. For this generalization^{}, see this entry (http://planetmath.org/DistributivityInPoGroups).

•
A vector space $V$ over $\u2102$ is said to be ordered if $W$ is an ordered vector space over $\mathbb{R}$, where $V=W\oplus iW$ ($V$ is the complexification of $W$).

•
For any ordered vector space $V$, the set ${V}^{+}:=\{v\in V\mid 0\le v\}$ is called the positive cone of $V$. ${V}^{+}$ is clearly a convex set. Also, since for any $\lambda >0$, $\lambda {V}^{+}\subseteq {V}^{+}$, so ${V}^{+}$ is a convex cone. In addition^{}, since ${V}^{+}\{0\}$ remains a cone, and ${V}^{+}\cap ({V}^{+})=\{0\}$, ${V}^{+}$ is a proper cone.

•
Given any vector space, a proper cone $P\subseteq V$ defiens a partial ordering on $V$, given by $u\le v$ if $vu\in P$. It is not hard to see that the partial ordering so defined makes $V$ into an ordered vector space.

•
So, there is a onetoone correspondence between proper cones of $V$ and partial orderings on $V$ making $V$ an ordered vector space.
Title  ordered vector space 

Canonical name  OrderedVectorSpace 
Date of creation  20130322 16:37:24 
Last modified on  20130322 16:37:24 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  20 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 46A40 
Classification  msc 06F20 
Synonym  ordered linear space 
Related topic  TopologicalLattice 
Defines  positive cone 