PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (232)
Orphanage
Unclass'd
Unproven (519)
Corrections (22)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: Very high Entry average rating: No information on entry rating
ordered vector space (Definition)

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 $ 0< \lambda \in k$, 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 $ 0< \lambda$.

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 $ \bot$ is the bottom of $ V$, then $ \bot\le 0$, or $ 2\bot\le \bot$, which implies $ 2\bot=\bot$ or $ \bot=0$. This means that $ 0\le v$ for all $ v\in V$. But if $ v\ne 0$, then $ 0<v$ or $ -v<0$, a contradiction. $ V$ is topless follows from the implication that if $ \bot$ exists, then $ \top=-\bot$ is the top.

For example, any finite dimensional vector space over $ \mathbb{R}$, and more generally, any (vector) space of real-valued 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 t-w$ and $ v\le t-w$. So $ u\vee v\le t-w$, or $ (u\vee v)+w\le t$. So $ s$ is the least upper bound of $ u+w$ and $ v+w$. $ \qedsymbol$
  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$. $ \qedsymbol$
  3. $ \lambda u\vee \lambda v$ exists for any scalar $ \lambda\in k$, and
    1. if $ \lambda\ge 0$, then $ \lambda u\vee \lambda v=\lambda(u\vee v)$
    2. if $ \lambda\le 0$, then $ \lambda u\vee \lambda v=\lambda(u\wedge v)$
    3. 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 $ \lambda<0$. Set $ \gamma=-\lambda$. Then $ \lambda u\vee \lambda v=\lambda (u\vee v)=-\gamma (u\vee v)=-(\gamma (u\vee v)... ...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. $ \qedsymbol$

Remarks.

  • Since an ordered vector space is just an abelian po-group under $ +$, the first two properties above can be easily generalized to a po-group. For this generalization, see this entry.
  • A vector space $ V$ over $ \mathbb{C}$ 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^+:=\lbrace v\in V\mid 0\le v\rbrace$ 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^+-\lbrace 0 \rbrace$ remains a cone, and $ V^+\cap (-V^+)=\lbrace 0\rbrace$, $ 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 $ v-u\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 one-to-one correspondence between proper cones of $ V$ and partial orderings on $ V$ making $ V$ an ordered vector space.



"ordered vector space" is owned by CWoo.
(view preamble | get metadata)

View style:

See Also: topological lattice

Other names:  ordered linear space
Also defines:  positive cone

Attachments:
vector lattice (Definition) by CWoo
ordered topological vector space (Definition) by CWoo
Log in to rate this entry.
(view current ratings)

Cross-references: one-to-one correspondence, proper cone, addition, cone, convex set, complexification, po-group, abelian, similar, proof, clear, converse, scalar, greatest lower bound, lower bound, least upper bound, upper bound, ordering, functions, vector, finite dimensional, top, implication, contradiction, implies, bottom, zero vector, iff, property, poset, vector space, ordered field
There are 5 references to this entry.

This is version 17 of ordered vector space, born on 2007-01-26, modified 2007-05-14.
Object id is 8822, canonical name is OrderedVectorSpace.
Accessed 1863 times total.

Classification:
AMS MSC: 06F20 (Order, lattices, ordered algebraic structures :: Ordered structures :: Ordered abelian groups, Riesz groups, ordered linear spaces)
 46A40 (Functional analysis :: Topological linear spaces and related structures :: Ordered topological linear spaces, vector lattices)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add derivation | add example | add (any)