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Vector Spaces and polynomial functions

  1. Feb 7, 2004 #1
    I'm really confused about a question I came across in my textbook.
    It basically says this:

    Consider the set of polynomial functions of degree 2. Prove that this set is not closed under addition or scalar multiplication (and therefore not a vectorspace).
    I'm confused because I think it is closed under addition and scalar mult.
    example:

    f(x) = ax^2 + bx + c
    g(x) = dx^2 + ex + f
    (f+g)(x) = (a+d)x^2 + (b+e)x + (c+f)

    (sf)(x) = (sa)x^2 + (sb)x + sc

    both results should be in the set of polynomial functions of degree 2.
    Why would the question say it is not closed under addition and scalar mult. ??
    Am I missing something very basic here, or could it be a trick question or something?
    Thanks!
     
  2. jcsd
  3. Feb 7, 2004 #2
    The set of polynomials of degree less than or equal to 2 is a vector space. If you only have polynomials of degree 2, this is no longer true.
     
  4. Feb 8, 2004 #3
    I'm not sure I understand why this is.
    So then the set of functions is only: ax^2
    Is this what you mean? If it is, I'm still not sure why the set wouldn't be closed under addition and scalar mult. :

    f(x) = ax^2
    g(x) = bx^2
    (f+g)(x) = (a+b)x^2
    (sf)(x) = (sa)x^2

    {where s is a scalar; a and b are coefficients)

    I'm not sure what I'm missing in this problem. :(

    And thanks for the quick reply master_coda

    -Confused
     
  5. Feb 8, 2004 #4
    x^2 + (-x^2) is not a degree 2 polynomial.
     
  6. Feb 8, 2004 #5
    no but x^2 + (-x^2) = 0
    0 is in the vector set.
     
  7. Feb 8, 2004 #6
    then what is the additive zero element of your set?
     
  8. Feb 8, 2004 #7
    I'm not sure what you mean by additive zero element.

    And just so I'm clear, is 2x^2, for example, a degree 2 polynomial that would be in the set that the above question is asking?
    i.e. degree 2 polynomials can still have coefficients, right?
     
  9. Feb 8, 2004 #8
    every vector space must contain an element denoted O or 0 such that for all vectors u, u+O=O+u=u.
     
  10. Feb 8, 2004 #9
    Oh, yeah, I get what you mean now.
    Wouldn't the additive zero element be 0^2?
     
  11. Feb 8, 2004 #10
    what's the degree of 0^2=0?
     
  12. Feb 8, 2004 #11
    I'm sorry. But I don't see the point your trying to get across here.

    f(x) = ax^2
    g(x) = bx^2
    (f+g)(x) = (a+b)x^2
    (sf)(x) = (sa)x^2

    {where s is a scalar; a and b are coefficients)

    Why doesn't this show that its closed under addition and scalar mult?

    when you said:

    "x^2 + (-x^2) is not a degree 2 polynomial."

    doesn't that support the following axiom:
    "For every element u of V, there exists an element called the negative of u, denoted -u, such that u + (-u) = 0."
    (This is one of the axioms from my text that needs to be satisfied in order for something to be a vector space)

    Which further supports why I think this is a vector space.
    Thanks for the hints so far, but don't seem to be grasping why this isn't a vector space. :(
     
  13. Feb 8, 2004 #12
    If I'm understanding this correctly...

    First, you must realise that (for example) x + 1 is not a member of the set of second degree polynomials. It is however an element of the set of polynomials of degree <= 2. This is crucial.

    Now, add two polynomials of degree 2. Can you say with certainty that this sum will also be a polynomial of degree 2 (unless you place certain restrictions on the coefficients)? No, consider (x^2 + 3x + 1) + (-x^2 + x + 3) = 4x + 4, which is a polyonomial of degree 1. Hence the set of second degree polynomials is not closed under addition.

    However, you /can/ say that the sum of two second degree polynomials will be /a/ polynomial, and that its degree will be at most 2, hence the set of polyonomials of degree <= 2 is closed under addition (and is a vector space).
     
    Last edited: Feb 8, 2004
  14. Feb 8, 2004 #13

    matt grime

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    And to REALLY get home something that you need ton understand: 0 is a constant polynomial, it has degree 0, as do all the polys f(x)=constant. How did you arrive at 0^2? Or if you like write down a polynomial in x with a non zero x^2 coeff that when added to every other polynomial leaves it unchanged.
     
  15. Feb 8, 2004 #14
    In order for you to use that axiom, you need to assume the set of degree 2 polynomials is a vector space. Since the set of degree 2 polynomials has no zero element, it is clearly not a vector space.

    Thus you cannot draw any valid conclusions by assuming that it is a vector space, since that would be a false assumption.
     
  16. Feb 8, 2004 #15

    Hurkyl

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    Because (a+b)x^2 isn't always a degree 2 polynomial.
     
  17. Feb 8, 2004 #16
    Thanks very much to everyone who posted to my questions. Everybody's reply helped be understand and solve the problem.
    Thanks again, I really appreciate it!
     
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