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Can you find a basis without deg. 2 polynomials?

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Can you find a basis {p1, p2, p3, p4} for the vector space ℝ[x]<4 s.t. there does NOT exist any polynomials pi of degree 2? Justify fully.


    2. Relevant equations



    3. The attempt at a solution
    We know a basis must be linearly independant and must span ℝ[x]<4. So intuitively if there are no polynomials of degree 2 we cannot span V. It just isn't possible.

    I'm just struggling to justify it rigorously.

    Many thanks - I appreciate the time and help.
     
  2. jcsd
  3. Oct 1, 2013 #2
    Yes. Try [itex]\{1,x,x^3+x^2,x^3\}[/itex].
     
  4. Oct 1, 2013 #3
    Nice. Can't believe I didn't see that. It is clearly linearly independant. Is it enough to say that since [itex]\{1,x,x^3+x^2,x^3\}[/itex] is a linear combination of pi that it will span the vector space?

    Thanks
     
  5. Oct 1, 2013 #4

    D H

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    Consider {1,x,ax3+bx2,cx3+dx2}. That's obviously a linear combination of {1,x,x2,x3}. What happens if ad=bc?
     
  6. Oct 1, 2013 #5

    arildno

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    Another simple choice for a basis is, for 4 different constants a,b,c,d to look at

    (x-b)(x-c)(x-d), (x-a)(x-c)(x-d), (x-a)(x-b)(x-d), (x-a)(x-b)(x-c)

    These polynomials are closely related to those we typically call Lagrange interpolation polynomials.
     
  7. Oct 1, 2013 #6
    If ad=bc, then we have linear independance, thus a basis.
     
  8. Oct 1, 2013 #7

    D H

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    Try that again. Consider a=b=c=d=1 as a simple example of ad=bc. Do you really think that {1,x,x3+x2,x3+x2} is a basis?
     
  9. Oct 1, 2013 #8

    arildno

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    To transform one valid basis to another valid basis, think of each function in the base as a vector component. Say we've got n basis functions.
    Now, we may transform that n-vector into another n-vector by having it multiplied with an n*n matrix with constant coefficients.

    What do you think will be a condition under which the new vector of function components will necessarily be a linear independent set of functions?
     
    Last edited: Oct 1, 2013
  10. Oct 1, 2013 #9
    ok - taking a=b=c=d=1. our basis would be:{1,x,x3+x2}

    I am not sure where this is going? the question asked for a basis consisting of 4 vectors.

    Would an alternative method to show that we have a basis be the following:

    Take {1,x,x3+x2,x3} which is linear independant.

    Now we have e1, e2, e3={x3+x2}-{x3}, e4 and so we have the standard basis. It follows that since the span contains the standard basis, it contains all of ℝ[x]<4.
     
  11. Oct 1, 2013 #10

    arildno

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    Incorrect.
    DH's set is {1,x,x3+x2, x3+x2}
    That set is a linearly dependent set.
     
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