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Homework Help: Is this set of polynomials a subspace of P2?

  1. Jun 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Let S denote the collection of all polynomials of the form p(t) = (2a - b)t^2 + 3(c - b)t + (a - c), where a,b,c are real numbers. Determine whether or not S is a subspace of P2.

    3. The attempt at a solution
    Okay, so I know that in order for S to be a subspace, it must satisfy 3 conditions
    1. closure of addition
    2. closure of scalar multiplication
    3. the set is not empty

    I don't know how to show that S is a subspace of P2, my textbook does not give good examples. What should I do first?
  2. jcsd
  3. Jun 24, 2009 #2
    Re: Subspaces

    You must also show that it contains the 0 polynomial and that it is a subset of P2.
    Simply use the definitions. Each of the elements in your subspace is determined by 3 numbers, so two example elements may be P(t)(a, b, c) and P(t)(d, e, f). Show that when you add these two, you get another polynomial of the form given.
  4. Jun 24, 2009 #3


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    Science Advisor

    Re: Subspaces

    Show that the 3 things you list are true!
    1. Take a "vector" in S, u= (2a-b)t^2+ 3(c-b)t+ (a-c). Take another "vector" in S, v= (2a'- b')t^2+ 3(c'-b')t+ (a'-c'). I've used a', b', c' rather than a, b, c because they might be different numbers. What is u+ v? Is it in S? That is, is it of that same form?

    2. Take a "vector" in S, u= (2a-b)t^2+ 3(c-b)t+ (a-c). Take a number, say, p. What is their scalar product? Is in in S? That's easy enough that I don't think it will hurt to show exactly what I mean: The scalar product is pu= p((2a-b)t^2+ 3(c-b)t+ (a-c))= p(2a-b)t^2+ 3p(c-b)t+ p(a-c)= (2pa- pb)t^2+ 3(pc- pb)t+ (pa- pc). That is again in S because it is also of the form "(2a-b)t^2+ 3(c-b)t+ (a-c)" with a replaced by pa, b replaced by pb and c replaced by pc. Therefore S is closed under scalar multiplication.

    3. Choose 3 values for a, b, and c to find a specific "vector" that is in S.
    Last edited by a moderator: Jun 24, 2009
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