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Math Help - Spanning sets, linear independence and bases

  1. #1
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    Spanning sets, linear independence and bases

    PROBLEM
    Let B = {1,−1 + x,1 + x^2}
    (a) Show that B is a basis for P2.
    (b) Find the coordinates for p(x) = a + bx + cx^2 with respect to the basis B.
    (c) Hence write down the coordinates, with respect to B, for q(x) = 3 − 4x^2 and r(x) = 4 − x + 2x^2

    ATTEMPT AT SOLUTION
    (a) B as a matrix: \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} which row-reduces to \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
    Each row has a leading entry, so B is linearly independent and thus is a basis for P2. Also, B has 3 polynomials and the dimension of P2 is 3, so B spans P2

    (b) I know this is incorrect, but I got:
    p(x) = a(1,0,0) + b(-1,1,0) + c(1,0,1)
    a=b=c=1 (from (a))
    Therefore p(x) = 1 + x + x^2

    (c) No idea?

    Please help!
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  2. #2
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    Quote Originally Posted by iExcavate View Post
    PROBLEM
    Let B = {1,−1 + x,1 + x^2}
    (a) Show that B is a basis for P2.
    (b) Find the coordinates for p(x) = a + bx + cx^2 with respect to the basis B.
    (c) Hence write down the coordinates, with respect to B, for q(x) = 3 − 4x^2 and r(x) = 4 − x + 2x^2

    ATTEMPT AT SOLUTION
    (a) B as a matrix: \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} which row-reduces to \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
    Each row has a leading entry, so B is linearly independent and thus is a basis for P2. Also, B has 3 polynomials and the dimension of P2 is 3, so B spans P2
    Good! A more fundamental approach, using the definition of "linearly independent"- suppose a(1)+ b(-1+ x)+ c(1+ x^2)= 0 for all x. Taking x= 0, a(1)+ b(-1)+ c(1)= a- b+ c= 0. Taking x= 1, a(1)+ b(0)+ c(2)= a+ 2c= 0. Taking x= -1, a(1)+ b(-2)+ c(2)= a- 2b- 2c= 0. Subtracting the second equation from the first eliminates a: - b- c= 0. Subtracting the third equation from the first b+ 3c= 0. Adding those two equations, 2c= 0 so c= 0 and then b= a= 0.

    Do you have the theorem that "any linearly independent set containing a number of vectors equal to the dimension of the space is a basis" and that the dimension of P2 is 2? If so, then yes, showing that those three vectors are independent show this is a basis. If not, then you should show that for any a, b, c, there exist \alpha, \beta, \gamma such that \alpha(1)+ \beta(-1+ x)+ \gamma(1+ x^2)= a+ bx+ cx^2. That's, fairly easy- just solve for \alpha, \beta, and \gamma.

    (b) I know this is incorrect, but I got:
    p(x) = a(1,0,0) + b(-1,1,0) + c(1,0,1)
    a=b=c=1 (from (a))
    Therefore p(x) = 1 + x + x^2
    No, your "vector" is a+ b+ cx^2 for any a, b, c- you can't just make a= b= c= 1. Your coefficients must be in terms of a, b, and c- just what I said in my last sentence above! Find \alpha, \beta, \gamma such that \alpha(1)+ \beta(-1+ x)+ \gamma(1+ x^2)= a+ bx+ cx^2, for all x.

    One way to do this is to equate like coefficients. Multiplying out the left side, \alpha- \beta+ \beta x+ \gamma+ \gamma x^2= (\alpha- \beta+ \gamma)+ \beta x+ \gamma x^2=  a+ bx+ c for all x so we must have \alpha+ \beta+ \gamma= a, \beta= b and \gamma= c. Putting those into the first equation, \alpha- b+ c= a so \alpha= a+ b- c:
    a+ b + cx^2= (a+ b- c)+ b(-1+ x)+ c(1+ x^2).

    (c) No idea?

    Please help!
    Use the answer to (b) to write out 3- 4x^2= \alpha(1)+ \beta(-1+ x)+ \gamma(1+ x^2 and then 4- x+ 2x^2= \alpha(1)+ \beta(-1+ x)+ \gamma(1+ x^2. The "coordinates" of each are the number \alpha, \beta, and \gamma
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  3. #3
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    Thanks, you deserve a prize in Mathematics, seriously!
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