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Polynomial Span related problem Linear Algebra

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the vector space F(R) = {f | f : R → R}, with the standard operations. Recall that the zero of F(R) is the function that has the value 0 for all
    x ∈ R:
    Let U = {f ∈ F(R) | f(1) = f(−1)} be the subspace of functions which have
    the same value at x = −1 and x = 1.
    Define functions g; h; j and k ∈ F[R] by
    g(x) = 2x^3− x − 2x^2+ 1;
    h(x) = x^3+ x^2− x + 1;
    k(x) = −x^3+ 5x^2+ x + 1 and
    j(x) = x^3− x; ∀x ∈ R:
    a) Show that g and h belong to U.
    b) Show that k ∈ span{g; h}.
    c) Show that j /∈ span{g; h}.
    d) Show that span{g; h} /= span{g; h; j}.

    2. Relevant equations



    3. The attempt at a solution


    My Attempt:

    I have no idea how to do a) I tried plugging -1 and +1 in g and h however it doesn't meet the requirement of f(1)=f(-1).

    I am sure I know how to do party b,c and d. However, I am doing something wrong and have no clue what it is...

    So what I did for b) was:

    k = a(g) + b(h)

    after that you get something like (....)x^3 + (....)x^2 + (....)x + (a+b)

    So I set the coefficient (....) equal to the coefficients of k (-1,5,1,1)
    however I am unable to solve the linear system.
    But it says it is in the span so I don't know what happened maybe mtah error, but I redid it a lot of times.
     
  2. jcsd
  3. Sep 28, 2012 #2

    HallsofIvy

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    I suspect you have copied this wrong. It should be g(x)= 2x^3- x^2- 2x+ 1.

     
  4. Sep 28, 2012 #3
    no I didn't copy anything wrong that was exactly how it was.
     
  5. Sep 28, 2012 #4
    it is right it is -x* not -2x
     
  6. Sep 28, 2012 #5

    HallsofIvy

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    Then it is an error in your text. The statements are true if g(x)= )= 2x^3- x^2- 2x+ 1, not 2x^3- x- 2x^2+ 1. (It is also peculiar that the polynomial is not in standard form. That is why I thought about 2x^3- x^2- 2x+ 1.)
     
  7. Oct 1, 2012 #6
    HallsofIvy, you are right. g(x) = 2x^3 - x^2 -2x + 1. I did part a by calculating g(1) = 0 and g(-1) = 0 therefore g(1) = g(-1) so g belongs to U. Then I did h(1) = 2 and h(-1) = 2 so h(1) = h(-1) therefore h also belongs to U. I am stuck at the span part and not really sure what I am supposed to do. It would be great if someone could help me out :)
     
  8. Oct 2, 2012 #7
    anyone wanna help me with part b please? I understand that it is asking whether k is a linear combination of g + h. So basically k = ag + bh of a and b are all real numbers. I just don't understand what number to use for a and b because none of them get be k :( Any help would be appreciated.

    Thanks!
     
  9. Oct 2, 2012 #8
    It is 2x^3-x^2-2x+1 the teacher made a mistake.

    To do this you do what you said set ag+bh=k

    You don't have to sub in for x

    so ag+bh=k would be: a(2x^3-x^2-2x+1)+b(x^3+x^2-x+1)=-x^3+5x^2+x+1

    so after you expand and factor out the a,b and c you get

    (2+b)x^3-(a-b)x^2-(2+b)x+a+b = -x^3+5x^2+x+1

    so you set the coefficients equal to each other you get:

    2a+b=-1
    -a+b=5
    -2a-b=1
    a+b=1

    Solve this linear system and you has the answer, it is solvable so it is part of the span.
     
  10. Oct 2, 2012 #9
    Oh okay, thanks for your help! See you around uOttawa :D
     
  11. Oct 2, 2012 #10
    how do you do part d i dont understand it
     
  12. Oct 3, 2012 #11
    after you verify c) you can just say that since j is not in {g,h} therefore {g,h} is not in {g,h,j}.

    It is basically the same process if you don't do that^. :)
     
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