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Homework Help: V not vector space with degree 3 polynomials

  1. Oct 31, 2005 #1
    Okay, so i have this problem in my text, and I've almost figured it out (i think) but i need a little help

    "Let V be the set of all polynomials of degree 3. Define addition and scalar multiplication pointwise. Prove that V with respect to these operations of addiont and scalar multiplication is NOT a vector space"

    I have pointwise addition and pointwise scalar multiplication defined... but i'm a little stumped on why V wouldn't be a vector space

    -i know that V cannot have a 0 element (0 is not a degree 3 polynomial) and thus it cannot be a vector space, but that doesn't really explain it via addition/multiplication
    -also, i know that the addition of 2 degree 3 polynomials does not always result in a degree 3 polynomial ie: x^3 + (-x^3)=0 (not a degree 3 polynomial)

    if anyone has any other reasons why V cannot be a vector space according to pointwise addition/scalar multiplication please let me know
     
  2. jcsd
  3. Oct 31, 2005 #2

    EnumaElish

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    Are you sure that 0 is not a 3rd degree polynomial?
     
  4. Oct 31, 2005 #3
    i dont know! :S
     
  5. Oct 31, 2005 #4
    Your reasoning looks OK. You are talking about polynomials of only degree 3 right? Not P3 which would represent all polynomials of degree 3 or less.
     
  6. Nov 1, 2005 #5

    HallsofIvy

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    The (non) existence of the "0" vector certainly is a property of addition!

    Or you look at (x3+ x)+ (x- x3). What does that tell you about closure of addition?

    Or you could look at the scalar multiplication 0(x3).
     
  7. Nov 1, 2005 #6

    EnumaElish

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    I thought that 0 = 0x3 is a 3rd degree polynomial, was I wrong?
     
  8. Nov 1, 2005 #7
    EnumaElish, that would mean that the polynomial x would also be of degree 3, since x = 0x^3 + x. Clearly this is not what we want. Usually, one defines the degree of a polynomial as the largest exponent n of x (or whatever the variable is called) such that the coefficient in front of x^n is not 0. (Sometimes one takes the degree of the zero polyonomial to be -1 or -infinity.)
     
  9. Nov 1, 2005 #8

    EnumaElish

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    Hm, this makes sense, thanks.
     
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