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Find a basis

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Let W equal the set of all polynomials in F[x] with degree less than or equal to n-1 such that the sum of the coefficients of the terms is 0. Find a basis of W over F.

    3. The attempt at a solution

    I don't know where to begin to find the basis. Do I use the fact that the sum of the coefficients is 0?
     
    Last edited: Feb 4, 2009
  2. jcsd
  3. Feb 4, 2009 #2

    Office_Shredder

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    Is not a vector space. Do you mean with degree less than or equal to n-1?
     
  4. Feb 4, 2009 #3
    yes I just fixed it.
     
  5. Feb 4, 2009 #4

    Office_Shredder

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    You definitely need to use the fact that the sum of the coefficients is zero. Notice in particular that if f1,...,fm is a basis, then the sum of the coefficients in each fi must be zero also. Try writing down some simple polynomials that satisfy this
     
  6. Feb 4, 2009 #5
    can you just use
    1-x^(n-1)?

    The first coefficient is 1 and the (n-1)th coefficient is -1. But I don't think that is a basis.....I don't see how to find a basis for it and find coefficients at the same time. Can it be alternating positive and negative ones?
     
  7. Feb 5, 2009 #6

    Office_Shredder

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    That's actually a fairly good start.

    Here's something that can help with problems like this: The dimension of the set of all polynomials of degree less than or equal to n-1 is n. Considering the canonical basis{1,x,x2,...,xn-1} you have a single linear equation in the coefficients of linear combinations. So intuitively, the dimension of the new subspace should be n-1. So try finding n-1 polynomials that look like the one in your post
     
  8. Feb 5, 2009 #7
    check that

    The vector space of set of the polynomials is isomorphic to the space X
    a0+a1+a2=...an-1=0

    now find a basis for X
     
  9. Feb 5, 2009 #8
    Can you explain this more? I don't understand the part where you say we have a "linear equation in the coefficients of linear combinations". Do you just mean that W is made up of linear equations?

    And how did you get that the dimension of the new subspace is n-1? Is it because we're saying dimW=n so a subspace of W must be n-1? But why are we trying to find a *new* subspace when we just need a basis for W?
     
  10. Feb 5, 2009 #9

    Office_Shredder

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    Looking back, I definitely didn't phrase it very well.

    Sorry to confuse you. I meant that the condition for the subspace is that your vector has to satisfy a single linear equation. In general, that decreases the dimension of your subspace by 1 for each equation as long as they are independent from each other. So we know we should be looking for a vector subspace of dimension n-1
     
  11. Feb 5, 2009 #10
    By subspace do you mean basis? And why does it decrease the dimension by 1 for each equation?
     
  12. Feb 5, 2009 #11
    Here is what I have so far.

    If the sum of the coefficients (call each coefficient w_i) = 0 then we can have a vector of all the w_i's (n-1 dimension) and an n-1 dimension unit vector. Multiplying these two vectors together we get 0.

    We have w_0= -(summation w_i from 1 to n-1)
    implies -(w_0)= summation w_i from 1 to n-1

    We can write M= w_1 (-1,1,0,..0) + w_2 (-1,0,1,0...0) +.....+ w_n-1 (-1,0,...,1).

    Then the basis for the whole space W (from the question) is made up of the vectors
    (-1,1,0,..0)
    (-1,0,1,0..0)
    .
    .
    .
    .
    .
    (-1,0,...,1)



    Is this right?
     
  13. Feb 6, 2009 #12

    HallsofIvy

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    Yes, that's right. Generally in problems like this you can take whatever equations you have (here is is that [itex]w_1+ w_2+ ...+ w_{n-1}= 0[/itex]) and solve for as many of the numbers as you can in terms of the others. Then take each of those "others" 1 in turn.

    Since here you have only one equation, you can solve for one of them, say [itex]w_1= -(w_2+ ...+ w_{n-1})[/itex]. Now let [itex]w_2= 1[/itex], [itex]w_2= ...= w_{n-1}= 0[/itex] and you get [itex]w_1= -1[/itex] so your first basis vector is (-1,1,0, ..., 0) as you say. Taking [itex]w_3= 1[/itex], [itex]w_2= w_4= ...= w_n= 0]/itex] you get [itex]w_1= -1[/itex] you get (-1, 0, 1, 0, ...,0), again as you have.

    Because this is a set of polynomials, it would be better to write the "vectors" in that way: the basis is the set
    [tex]\{x-1, x^2- 1, x^3- 1, \cdot\cdot\cdot, x^{n-1}-1\}[/tex]
     
  14. Feb 6, 2009 #13
    How do you get from the list of vectors to that basis you just got for polynomials?
    In other words, how do you get from (-1,1,0..0) to (x-1)?

    Is it because x=1
    so x-1=0?
    what happened to the first term that equals -1?
     
    Last edited: Feb 6, 2009
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