ordered vector space (original) (raw)

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. for any u,v,w∈V, if u≤v then u+w≤v+w,
    1. if 0≤u∈V and any 0<λ∈k, then 0≤λ⁢u.

Here is a property that can be immediately verified: u≤v iff λ⁢u≤λ⁢v 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 0≤v for all v∈V. But if v≠0, 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 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≤g iff f⁢(x)≤g⁢(x) for every x∈S.

Properties. Let V be an ordered vector space and u,v∈V. Suppose u∨v exists. Then

    1. (u+w)∨(v+w) exists and (u+w)∨(v+w)=(u∨v)+w for any vector w.
Proof.

Let s=(u∨v)+w. Then u+w≤s and v+w≤s. For any upper bound t of u+w and v+w, we have u≤t-w and v≤t-w. So u∨v≤t-w, or (u∨v)+w≤t. So s is the least upper bound of u+w and v+w. ∎ 2. 2.
u∧v exists and u∧v=(u+v)-(u∨v).

Proof.

Let s=(u+v)-(u∨v). Since u≤u∨v, -(u∨v)≤-u, so s≤v. Similarly s≤u, so s is a lower bound of u and v. If t≤u and t≤v, then -u≤-t and -v≤-t, or v≤(u+v)-t and u≤(u+v)-t, or u∨v≤(u+v)-t, or t≤(u+v)-(u∨v)=s. Hence s the greatest lower boundMathworldPlanetmath of u and v. ∎ 3. 3.
λ⁢u∨λ⁢v exists for any scalar λ∈k, and

  1. (a)
    if λ≥0, then λ⁢u∨λ⁢v=λ⁢(u∨v)
  2. (b)
    if λ≤0, then λ⁢u∨λ⁢v=λ⁢(u∧v)
  3. (c)
    if u≠v, then the converseMathworldPlanetmath holds for (a) and (b).
Proof.

Assume λ≠0 (clear otherwise). (a). If λ>0, u≤u∨v implies λ⁢u≤λ⁢(u∨v). Similarly, λ⁢v≤λ⁢(u∨v). If λ⁢u≤t and λ⁢v≤t, then u≤λ-1⁢t and v≤λ-1⁢t, hence u∨v≤λ-1⁢t, or λ⁢(u∨v)≤t. Proof of (b) is similarPlanetmathPlanetmath to (a). (c). Suppose λ⁢u∨λ⁢v=λ⁢(u∨v) and λ<0. Set γ=-λ. Then λ⁢u∨λ⁢v=λ⁢(u∨v)=-γ⁢(u∨v)=-(γ⁢(u∨v))=-(γ⁢u∨γ⁢v)=-((-λ⁢u)∨(-λ⁢v))=-(-(λ⁢v∧λ⁢u))=λ⁢v∧λ⁢u. This implies λ⁢u=λ⁢v, or u=v, a contradiction. ∎

Remarks.