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Linear independent vector space

  1. Oct 24, 2009 #1
    I have a quick question about vector spaces.

    Consider the vector space of all polynomials of degree < 1. If the leading coefficient (the number that multiplies [tex]x^{N-1}[/tex]) is 1, does the set still constitute a vector space?

    I am thinking that it doesn't because the coefficient multiplying [tex]x^{N-1}[/tex] is the same as the coefficient multiplying [tex]x^{0} = 1[/tex], and then it would not be linearly independent, or something like that, but I am not totally sure about this. Any clarification would be greatly appreciated.
     
  2. jcsd
  3. Oct 24, 2009 #2

    HallsofIvy

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    No, it is not a matter of being "linearly independent"- that is a property of a basis for a subspace, not the subspace itself. Polynomials in a subspace of polynomial can have the same coefficient for different powers. There is nothing wrong with that.

    A subspace must have two properties:
    a) It is closed under vector addition.
    b) It is closed under scalar multiplication.

    It should be easy to see that neither of those is satified by a set of polynomials with leading coefficient 1. If you add two such polynomials, you get a polynomial with leading coefficient 2, not 1. If you multiply such a polynomial by the number "a", you get a polynomial with leading coefficient "a", not 1.

    By the way, did you really mean "the vector space of all polynomials of degree < 1"? That is the set of all constant functions and further requiring that "the leading coefficient is 1" reduces it to a single "vector"!
     
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