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I Axiom 10 of Vector Spaces

  1. May 18, 2016 #1
    So I understand how to prove most of the axioms of a vector space except for axiom 10, I just do not understand how any set could fail the Scalar Identity axiom; Could anybody clarify how exactly a set could fail this as from what I know that anything times one results in itself

    1u = u
    1(x,y,z)=(x,y,z)
    1(1,2,3) = (1,2,3)
    1 (1,0,...,1) = (1,0,...,1)

    I don't see how you could ever end up in a situation where you could end up with

    1VV
     
  2. jcsd
  3. May 18, 2016 #2

    Mark44

    Staff: Mentor

    It can happen if you have a vector space with an unusual kind of scalar multiplication. Keep in mind that a vector space consists of a set of vectors over some field (often, the field of real numbers ##\mathbb{R}## or the field of complex numbers ##\mathbb{C}##), together with operations for vector addition and for multiplication by a scalar.

    If scalar multiplication is defined like this
    ##k \cdot <x, y> = <kx, 0>##
    then there is no scalar k for which ##k \cdot <x, y> = <x, y>##.
     
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