Show that Polynomials p0 to pn Form Basis of F[t] ≤ n

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Discussion Overview

The discussion revolves around demonstrating that a specific set of polynomials, defined as p0 through pn, forms a basis for the vector space of polynomials F[t] of degree less than or equal to n. The focus includes theoretical aspects of vector spaces, particularly the concepts of spanning and linear independence.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant requests assistance in showing that the polynomials p0 to pn form a basis for F[t] of degree ≤ n.
  • Another participant suggests using the definitions of "basis," "span," and "independent" to approach the problem, emphasizing the need to express any polynomial in the space as a linear combination of the given polynomials.
  • This participant also raises the question of whether it can be shown that the only linear combination of the polynomials that results in the zero vector is the trivial combination where all coefficients are zero.
  • A different participant expresses confidence in handling the spanning part of the proof but indicates uncertainty about proving the linear independence aspect.
  • The initial requester later states they have resolved their confusion regarding the problem.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and requirements for a basis in a vector space, but there is uncertainty regarding the proof of linear independence. The discussion does not reach a consensus on the method for proving that all coefficients must equal zero.

Contextual Notes

Some assumptions about the definitions of span and independence may be implicit in the discussion. The specific steps for proving linear independence remain unresolved.

e.gedge
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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

Thanks!
xo
 
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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]?<br /> <br /> "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<br /> [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?[/itex][/itex][/itex][/itex][/itex][/itex]
 
Uhh... I am 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
 
oh, nevermind, i got it now. Thanks a ton for the help!
 

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