Why is scalar multiplication on vector spaces not commutative?

In summary: Commutativity is about the order of operations: you can't perform two scalar multiplications in the same order and have them give the same result. For example, \left( 2x \right) = \left( 2x \right) + xThis doesn't work for SM because the first equation says that x gets multiplied by 2 before being added to the second equation, which is not what you want.
  • #1
honestrosewater
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(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)
 
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  • #2
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)
 
  • #3
Sorry this is so late. That helped; thank you. :smile:
 
  • #4
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!
 
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1. Why is scalar multiplication not commutative on vector spaces?

Scalar multiplication is defined as the product of a scalar (a real or complex number) and a vector. In vector spaces, the scalar is always multiplied on the left side of the vector. This means that the order of multiplication cannot be changed, making scalar multiplication non-commutative.

2. How does non-commutativity affect vector operations?

Non-commutativity means that the order of vector operations matters. For example, if we have two vectors A and B, and we want to multiply them by a scalar c, we will get different results depending on the order of operations. Specifically, c(A+B) will not be equal to cA + cB.

3. What are the implications of non-commutativity in vector spaces?

Non-commutativity affects many aspects of vector space operations. For example, it means that the dot product and cross product of two vectors will not be the same when the order of the vectors is switched. This can also affect calculations in fields such as physics and engineering.

4. Are there any exceptions to the non-commutativity of scalar multiplication in vector spaces?

No, scalar multiplication is always non-commutative in vector spaces. This is because the definition of scalar multiplication is based on the order of the scalar and vector, and this definition is consistent across all vector spaces.

5. How can we work with non-commutativity in vector spaces?

To work with non-commutativity, it is important to always be aware of the order of vector operations. This means paying close attention to the placement of scalars in equations and being careful when switching the order of vectors in operations such as dot and cross products. Additionally, understanding the properties and rules of vector spaces can help in navigating non-commutativity.

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