Why is scalar multiplication on vector spaces not commutative?

Click For Summary

Discussion Overview

The discussion revolves around the properties of scalar multiplication in vector spaces, particularly focusing on the lack of commutativity and the implications of this characteristic. Participants explore theoretical aspects, definitions, and the relationship between scalar multiplication and different algebraic structures, such as modules and rings.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the definition of scalar multiplication, noting its lack of commutativity and inverses, and wonders if the multiplicative identity applies in the context of vector spaces.
  • Another participant introduces the concept of modules, explaining the distinction between left-modules and right-modules, and how scalar multiplication is defined differently in each case.
  • A participant mentions that when the ring of scalars is commutative, left-modules and right-modules can be isomorphic, allowing for a conversion between the two.
  • There is a challenge to the idea of commutativity in scalar multiplication, emphasizing that the scalar and vector cannot be interchanged in the multiplication process, despite the notation suggesting otherwise.

Areas of Agreement / Disagreement

Participants express differing views on the nature of scalar multiplication and its commutativity. Some propose that scalar multiplication can be defined in a commutative manner under certain conditions, while others argue against the possibility of interchanging scalars and vectors in the multiplication process. The discussion remains unresolved regarding the implications of these definitions.

Contextual Notes

Participants highlight the dependence on the definitions of scalar multiplication and the structures involved, such as fields and rings. The discussion reflects the complexity of these concepts and the varying interpretations that arise from different algebraic contexts.

honestrosewater
Gold Member
Messages
2,133
Reaction score
6
(Or if you prefer: Why are things defined this way?) I noticed that, in my book's definition, scalar multiplication (SM) on vector spaces lacks two familiar things: commutativity and inverses.

The multiplicative inverse concept doesn't seem to apply to SM. Can it? I can't imagine how it could because, for one thing, the multiplicative identity is a scalar and the product of SM is a vector. (Right? I can't find a definition that actually says that 1 is the identity, but that's what I take 1v = v for all v in V to mean. And is 1 meant to just be the multiplicative identity of the set over which the vector space is defined, whatever it happens to be?)

I guess that SM isn't required to be commutative because you want to be able to define vector spaces over different kinds of sets? But isn't SM commutative on vector spaces over fields? That is, for example, if Fn is a field, a is in F, and (x1, ..., xn) is in Fn, then I could define SM' as

a(x1, ..., xn) = (ax1, ..., axn) = (x1, ..., xn)a

and as long as everything else holds, Fn with SM' would be a vector space? Is commutativity for SM interpreted in another way? For example

ab(v) = ba(v)
 
Last edited:
Physics news on Phys.org
The general concept is that of a module, and there are two kinds: left-modules and right-modules, and you have a ring of scalars. (Rings need not be commutative, nor must they have inverses. The ring of all nxn matrices is a good example)

A left-module is one where scalar multiplication is written on the left, and a right-module is one where scalar multiplication is written on the right.

If m is an element of a right-module, and a, b are elements of the ring of scalars, then (ma)b = m(ab)

Generally, a right and left modules are different things. If we have a right module, where (ma)b = m(ab), and we try to write it as left-multiplication, then we'd have b(am) = (ab)m. :frown:

But when the ring of scalars is commutative, the concepts of left-modules and right-modules are isomorphic, since we can define ma := am to convert a left-module into a right-module.

By convention, we write vector spaces as left-modules, so only av is defined when a is a scalar, and v is a vector, but we could define va if we wanted, when over a field. However, when over a division ring (we have inverses, but not commutativity), we still call it a vector space, but there's a difference between a left and a right vector space. (If you want an example, take a vector space over the quaternions)
 
Sorry this is so late. That helped; thank you. :smile:
 
How in the world could scalar multiplication be commutative?
Scalar multiplication is the product of a scalar and a vector- you can't interchange them. Of course, if just want to say that it doesn't matter how you write the product [itex]\lambda v= v \lambda[/itex] where [itex]\lambda[/itex] is a scalar and v is a vector, then that's trivially true but that is not what "commutative" means!
 
Last edited by a moderator:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 0 ·
Replies
0
Views
9K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K