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Why is scalar multiplication on vector spaces not commutative?

  1. Oct 15, 2005 #1


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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)
    Last edited: Oct 15, 2005
  2. jcsd
  3. Oct 15, 2005 #2


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    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)
  4. Oct 22, 2005 #3


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    Sorry this is so late. That helped; thank you. :smile:
  5. Oct 25, 2005 #4


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    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: Oct 25, 2005
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