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Writing out the span of this polynomial vector space?

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data

    The problems states "All polynomials of the form p(t)= at^2, where a is in R."
    I'm supposed to see if it is a subspace of Pn. I've already done that but the book's answer is that it spans Pn by Theorem 1, because the set is span{t^2}
    2. Relevant equations

    Theorem states "1 If v1....vp are in a vector space V , then Span {v1.... vp.} is a subspace of V ."

    3. The attempt at a solution
    I've already checked the three properties showing that it is a subspace of P2.
    I checked that it was closed under addition and scalar multiplication as well as containing the zero vector. But im not clear how the book comes up with the span{t^2} part? If im correct would the reason be that since a is already a scalar then t^2 is the part that generates the vectors in P2.
     
    Last edited: Oct 13, 2012
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  3. Oct 13, 2012 #2

    tiny-tim

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    hi halo31! :smile:
    yes … the definition of eg span{P,Q,R} is all elements of the form aP + bQ + cR

    when there's only one element in the original set eg span{P}, that's just all elements of the form aP

    here, P = t2 :wink:
     
  4. Oct 13, 2012 #3

    HallsofIvy

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    I recommend that you go back and look at the basic definitions. The set of all polynomials of the form [itex]ax^2[/itex] is a vector space because
    1) the 0 polynomial is of this form with a= 0.
    2) the product of a real number, p, with such a polynomial is of this form: [itex]p(ax^2)= (ap)x^2[/itex] which is of that form with a replaced by ap.
    3) the sum of two such polynomials, [itex]ax^2[/itex], [itex]bx^2[/itex] is of the same form- [itex]ax^2+ bx^2= (a+ b)x^2[/itex].
    4) the "additive inverse" of [itex]ax^2[/itex] is [itex]-ax^2= (-a)x^2[/itex] which is again of that form.

    A basis for a vector space is a set of vectors that both span the space and are independent. Every vector in this space can be written as [itex]ax^2= (a)x^2[/itex] so that each vector is a linear combination (actually a multiple) of [itex]x^2[/itex] so [itex]\{x^2\}[/itex] alone spans the space. And, of course, a set consisting of a single non-zero vector is always independent.
     
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