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Are the set of all the polynomials of degree 2 a vector space?

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let P denote the set of all polynomials whose degree is exactly 2. Is P a vector space? Justify your answer.




    2. Relevant equations
    (the numbers next to the a's are substripts
    P is defined as ---->A(0)+A(1)x+A(2)x^2


    3. The attempt at a solution

    I really dont know how to do this problem. i want to say that it isn't a vector space because it violates the property of having an additive inverse. as in, there is no value of x such that

    F(x) + (F(-x)) = F(x)+(-F(x)) = 0 if we keep all of the values for A the same

    A(0) + A(1)x + A(2)x^2 + A(0) + A(1)(-x) + A(2)(-x)^2 == 2A(0) + 2A(2)x^2 != 0

    therefore, there is no additive inverse.
    i probably did it wrong, but i dont know. all i know is that the book says that it isnt a vector space, but it doesnt give the reason.
     
  2. jcsd
  3. Feb 2, 2012 #2

    jbunniii

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    If p(x) is a 2nd degree polynomial, isn't -p(x) also a 2nd degree polynomial? And what happens when you add these two together?
     
  4. Feb 2, 2012 #3

    jbunniii

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    Can you think of two 2nd degree polynomials, p(x) and q(x), such that when you add them together, the resulting polynomial isn't 2nd degree?
     
  5. Feb 2, 2012 #4
    so because P(x) + (-(P(x)) = 0 and therefore, the answer is not a 2nd degree polynomial, then it cant be a vector space because it isnt closed under addition? if so, then i guess i just forgot to check the first property for a set to be a vector space and assumed it to be true.
     
  6. Feb 2, 2012 #5

    jbunniii

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    Yes, any vector space has to contain 0, and 0 isn't a 2nd degree polynomial.

    Another example would be p(x) = x^2 + x + 1, and q(x) = -x^2. Then p(x) + q(x) = x + 1, which is 1st order.
     
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