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Forming a basis

  1. Jan 15, 2009 #1
    Quick a easy question i need help with, so thanks to anyone who will try it out..

    Show that the polynomials p0= 1 + x + x2+ x3...+ xn, p1= x + x2+ x3... +xn, p2= x2 + x3 +...+ xn, .... pn=xn form a basis of F[t] less than or equal to n

  2. jcsd
  3. Jan 15, 2009 #2


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    For a problem like this it is best to use the definition. A "basis" for a vector space is a set of vectors that (1) span the space and (2) are independent.

    Now look at the definitions of "span" and "independent".
    To span the space means that every vector in the space can be written as a linear combinations in the set. Any vector in this space is a polynomal of degree less than or equal to n: any such vector can be written f(x)= [itex]\alpha_nx^n+ \alpha_{n-1}x^{n-1}+ \cdot\cdot\cdot+ \alpha_1 x+ \alpha_0[/itex]. Can you find numbers [itex]a_0, a_1, \cdot\cdot\cdot, a_n[/itex] so that [itex]a_0(1 + x + x^2+ x^3+ \cdot\cdot\cdot+ x^n)[itex][itex]+ a_1(x+ x^2+ x^3+ \cdot\cdot\cdot+ x^n)+ [/itex][itex]\cdot\cdot\cdot + a_n(x^n)[/itex][itex]= [itex]\alpha_nx^n+ \alpha_{n-1}x^{n-1}+ \cdot\cdot\cdot+ \alpha_1 x+ \alpha_0[/itex]?

    "Independent" means that the only linear combination of the vectors that equals the 0 vector is the "trivial" combination with all coefficients equal to 0. Suppose
    [itex]a_0(1 + x + x^2+ x^3+ \cdot\cdot\cdot+ x^n)[itex][itex]+ a_1(x+ x^2+ x^3+ \cdot\cdot\cdot+ x^n)+ [/itex][itex]\cdot\cdot\cdot + a_n(x^n)[/itex][itex]= 0 for all x. Can you prove that all the "a"s must be 0?
  4. Jan 15, 2009 #3
    Uhh... im pretty sure that i ca do the span part, but i still am unsure as to how to go about proving that all a's must equal 0
  5. Jan 15, 2009 #4
    oh, nevermind, i got it now. Thanks a ton for the help!
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