ordered vector space

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

  1. 1.

    for any u,v,wV, if uv then u+wv+w,

  2. 2.

    if 0uV and any 0<λk, then 0λu.

Here is a property that can be immediately verified: uv iff λuλv for any 0<λ.

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 is the bottom of V, then 0, or 2, which implies 2= or =0. This means that 0v for all vV. But if v0, then 0<v or -v<0, a contradictionMathworldPlanetmathPlanetmath. V is topless follows from the implicationMathworldPlanetmath that if exists, then =- is the top.

For example, any finite dimensional vector space over , 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 fg iff f(x)g(x) for every xS.

Properties. Let V be an ordered vector space and u,vV. Suppose uv exists. Then

  1. 1.

    (u+w)(v+w) exists and (u+w)(v+w)=(uv)+w for any vector w.


    Let s=(uv)+w. Then u+ws and v+ws. For any upper bound t of u+w and v+w, we have ut-w and vt-w. So uvt-w, or (uv)+wt. So s is the least upper bound of u+w and v+w. ∎

  2. 2.

    uv exists and uv=(u+v)-(uv).


    Let s=(u+v)-(uv). Since uuv, -(uv)-u, so sv. Similarly su, so s is a lower bound of u and v. If tu and tv, then -u-t and -v-t, or v(u+v)-t and u(u+v)-t, or uv(u+v)-t, or t(u+v)-(uv)=s. Hence s the greatest lower boundMathworldPlanetmath of u and v. ∎

  3. 3.

    λuλv exists for any scalar λk, and

    1. (a)

      if λ0, then λuλv=λ(uv)

    2. (b)

      if λ0, then λuλv=λ(uv)

    3. (c)

      if uv, then the converseMathworldPlanetmath holds for (a) and (b).


    Assume λ0 (clear otherwise). (a). If λ>0, uuv implies λuλ(uv). Similarly, λvλ(uv). If λut and λvt, then uλ-1t and vλ-1t, hence uvλ-1t, or λ(uv)t. Proof of (b) is similarPlanetmathPlanetmath to (a). (c). Suppose λuλv=λ(uv) and λ<0. Set γ=-λ. Then λuλv=λ(uv)=-γ(uv)=-(γ(uv))=-(γuγv)=-((-λu)(-λv))=-(-(λvλu))=λvλu. This implies λu=λv, or u=v, a contradiction. ∎


  • 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 generalizationPlanetmathPlanetmath, see this entry (http://planetmath.org/DistributivityInPoGroups).

  • A vector space V over is said to be ordered if W is an ordered vector space over , where V=WiW (V is the complexification of W).

  • For any ordered vector space V, the set V+:={vV0v} is called the positive cone of V. V+ is clearly a convex set. Also, since for any λ>0, λV+V+, so V+ is a convex cone. In additionPlanetmathPlanetmath, since V+-{0} remains a cone, and V+(-V+)={0}, V+ is a proper cone.

  • Given any vector space, a proper cone PV defiens a partial ordering on V, given by uv if v-uP. 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.

Title ordered vector space
Canonical name OrderedVectorSpace
Date of creation 2013-03-22 16:37:24
Last modified on 2013-03-22 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