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Homework Help: Is the set 'V' a vector space?

  1. Sep 16, 2012 #1


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    1. The problem statement, all variables and given/known data

    Let V be the set of all ordered pairs of real numbers, with addition being defined as:

    [itex](x_1 , x_2 ) + (y_1 , y_2 ) = (x_1 + y_1 , x_2 + y_2 )[/itex]

    and scalar multiplication defined as:

    [itex]\alpha \circ (x_1 , x_2 ) = (\alpha x_1 , x_2)[/itex]

    Is V a vector space with these operations? Justify your answer.

    3. The attempt at a solution

    I am thinking yes, because the scalar multiplication rule does not seem to violate any of the 8 axioms for vector spaces, but it seems wrong intuitively.
  2. jcsd
  3. Sep 16, 2012 #2
    What are the axioms defining a vector space with respect to scalar multiplication? Can you confirm each of them? You've probably only ever seen a single example of a vector space (Rn), so your intuition isn't exactly well developed about these things. That's normal, it's important when dealing with algebraic structures to carefully confirm that all necessary conditions are satisfied.
  4. Sep 16, 2012 #3


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    Distributivity of scalar multiplication with respect to vector addition   a(u + v) = au + av
    Distributivity of scalar multiplication with respect to field addition (a + b)v = av + bv
    Compatibility of scalar multiplication with field multiplication a(bv) = (ab)v

    These and the multiplicative identity are the only axioms that would involve the altered scalar multiplicative operator. if u = (u1,u2) and v = (v1,v2)

    "[a(u+v) = au+av]"
    a[(u1,u2) + (v1,v2)] = a(u1+v1, u2+v2) = (au1+av1, u2+v2) = a(u1,u2) + a(v1,v2)
    seems to check out

    "(a + b)(v1,v2) = a(v1,v2) + b(v1,v2)"
    (a+b)(v1,v2) = ((a+b)v1,v2) = (av1+bv1,v2) ≠ (av1, v2) + (bv1,v2) = (av1+bv1, v2+v2)

    I suppose this axiom doesn't really check out, because (av1+bv1, v2+v2)≠(av1+bv1,v2) right?
  5. Sep 16, 2012 #4
    indeed the last one doesn't check out, because you would get a(v1,v2)+b(v1,v2). for the v1's it's ok, but you get the undesired 2 times v2.
  6. Sep 16, 2012 #5


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    Turns out my intuition was correct. ;p
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