Vector space definition with respect to field membership

  • Thread starter thelema418
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  • #1
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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?
 

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  • #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]?
 
  • #3
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It is actually the case.
 
  • #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.
 
  • #5
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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.
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?
 
  • #6
pasmith
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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?
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.
 
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  • #7
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Thanks -- that information is definitely not in the textbook or the lecture notes.
 

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