ordered vector space

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

∈

V

, if

u

≤

v

then

u

+

w

≤

v

+

w

,

2.

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

contradiction

.

V

is topless follows from the

implication

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.

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 bound

of

u

and

v

.
∎

3.

λ

⁢

u

∨

λ

⁢

v

exists for any scalar

λ

∈

k

, and

(a)

if

λ

≥

0

, then

λ

⁢

u

∨

λ

⁢

v

=

λ

⁢

(

u

∨

v

)

(b)

if

λ

≤

0

, then

λ

⁢

u

∨

λ

⁢

v

=

λ

⁢

(

u

∧

v

)

(c)

if

u

≠

v

, then the

converse

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

similar

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

.

•

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 (

http://planetmath.org/DistributivityInPoGroups

).

•

A vector space

V

over

ℂ

is said to be

ordered

if

W

is an ordered vector space over

ℝ

, where

V

=

W

⊕

i

⁢

W

(

V

is the

complexification

of

W

).

•

For any ordered vector space

V

, the set

V

+

:=

{

v

∈

V

∣

0

≤

v

}

is called the

positive cone

of

V

.

V

+

is clearly a

convex set

. Also, since for any

λ

>

0

,

λ

⁢

V

+

⊆

V

+

, so

V

+

is a convex cone. In

addition

, since

V

+

-

{

0

}

remains a cone, and

V

+

∩

(

-

V

+

)

=

{

0

}

,

V

+

is a proper cone.

•

Given any vector space, a proper cone

P

⊆

V

defiens a

partial ordering

on

V

, given by

u

≤

v

if

v

-

u

∈

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.

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

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