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W a subspace of P3

  1. May 28, 2012 #1
    Let [itex]W = \{ax^3 + bx^2 + cx + d : b + c + d = 0\}[/itex] and [itex]P_3[/itex] be the set of all polynomials of degrees 3 or less.

    So say we want to prove the [itex]W[/itex] is a subspace of [itex]P_3[/itex]. We let [itex]p(x) = a_1x^3 + a_2x^2 + a_3x + a_4[/itex] and [itex]g(x) = b_1x^3 + b_2x^2 + b_3x + b_4[/itex]. So, we compute [itex]f(x) + kg(x)[/itex] and the answer should be a polynomial of third degree or less? Is this enough?

    And to find a basis for the subspace, we let [itex]b = -c - d[/itex] and just replace into the polynomial and form a basis. How can we find a vector in [itex]P_3[/itex] but not in the subspace [itex]W[/itex]?
  2. jcsd
  3. May 28, 2012 #2


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    No, that would just show that [itex]P_3[/itex] is a vector space itself. You have said nothing about W! To show that W is a subspace you need to show that if p(x) and q(x) are both members of W, then p+ kq, for any number k, is also a member of W. That is, you must have [itex]a_2+ a_3+ a_4= 0[/itex], [itex]b_2+ b_3+ b_4= 0[/itex], and then show that the corresponding coefficients for p+ kq also sum to 0.

    Yes, any "vector" in W can be written [itex]ax^3+ (-c-d)x^2+ cx+ d= ax^3+ c(x- x^2)+ d(1- x^2)[/itex] so that [itex]\{x^3, x-x^2, 1- x^2\}[/itex] is a basis.

    Choose any numbers, a, b, c, d so that b+ c+ d is NOT equal to 0 and [itex]ax^3+ bx^2+ cx+ d[/itex] will be in [itex]P_3[/itex] but not in W.
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