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Vector space definition with respect to field membership

  1. Sep 20, 2014 #1
    I'm confused by a set of problems my teacher created versus a set of problems in the textbook.

    My textbook states that "A vector space V over a field F consists of a set on which two operations (called addition and scalar multiplication, respectively) are defined so that for each pair of elements, x, y in V there is a unique element x + y in V, and for each element a in F and each element x in V there is a unique element ax in V, such that the following conditions hold..."

    In our problem set, the professor created a problem where V is over ##\mathbb{R}^+##, but the scalars are members of ##\mathbb{R}##. There is a special definition of addition and scalar multiplication, so I can easily prove that the axioms hold.

    My concern is with the use of two different fields. I'm thinking that this is not a vector space because of the field membership.

    The issue I'm having with the textbook problem is that a true or false makes the claim "If f is a polynomial of degree #n# and #c# is a nonzero scalar, then #cf# is a polynomial of degree #n#." If I follow the textbooks definition the way I'm interpreting it, I get the answer True. But if I use two different fields, the answer will be false. For example, let the polynomials be from ##Z_3(x)## but the scalars from ##Z_7## and defined multiplication and addition with mod 3. The #c = 6# is a nonzero scalar, but (6 * 2x) mod 3 = 0.

    In short, I cannot reconcile what the professor is doing in his question with what the definition appears to say to me. Is there something I'm not seeing?
     
  2. jcsd
  3. Sep 21, 2014 #2

    pasmith

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    Is it actually the case that [itex]V = \mathbb{R}^{+}[/itex], and the field of scalars is [itex]\mathbb{R}[/itex]?
     
  4. Sep 21, 2014 #3
    It is actually the case.
     
  5. Sep 21, 2014 #4

    HallsofIvy

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    If you are working in Z3, then multiplying by 6 is exactly the same as multiplying by 0. The theorem you state does NOT apply here because you not multiplying by a non-zero scalar.
     
  6. Sep 21, 2014 #5
    In respect to VS axioms, how are you making that decision when the fields are different? How do you decide what is a scalar zero?

    Example of confusion: In the ##Z_3## situation you are saying that 6 is a zero when it is nonzero ##Z_7##. Yet, in the case of the professor's problem, the zero element of the vectors over ##R^+## is 1, but the zero element of the scalars over ##R## is 0. How do I know (based on the VS axioms) what the zero is? Also, if you end up with non-unique scalar zeros when doing this, is it still a VS?
     
  7. Sep 22, 2014 #6

    pasmith

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    There is a specified field of scalars. Fields have a unique zero.

    In your example of turning [itex]\mathbb{Z}_3[/itex] into a vector space over [itex]\mathbb{Z}_7[/itex], you need to define the scalar multiplication operation [itex]\mathbb{Z}_7 \times \mathbb{Z}_3 \to \mathbb{Z}_3[/itex] in a way which satisfies the vector space axioms. In particular, you need that [itex](-1)v = -v[/itex] for every vector [itex]v[/itex]. But -1 = 6 mod 7, so that [itex](-1)v = 6v = 0 \in \mathbb{Z}_3[/itex]. But in [itex]\mathbb{Z}_3[/itex], [itex]-1 = 2[/itex] and [itex]-2 = 1[/itex]. Thus "multiplication mod 3" cannot be a scalar multiplication rule here.
     
  8. Sep 22, 2014 #7
    Thanks -- that information is definitely not in the textbook or the lecture notes.
     
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