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Math Help - differentiation operator basis

  1. #1
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    differentiation operator basis

    Let V be the vector space

    f(x)=\sum^2_{k=1}(a_Kcoskx+b_ksinkx)

    Find the matrix [D]_B of the differentiation operator D on V in the basis \{ 1,sinx, cosx,sin2x, cos2x \}.



    So I don't understand how they got that matrix for D! To find the matrix for D, what I did was to compute the images of the standard unit vectors under the transformation. These are

    T(e_1)=(1,0,0,0,0)=(0, cos0, -sin0, 2cos0, 2sin0)

    And so on for

    D(e_2)=(0,1,0,0,0)
    D(e_1)=(0,0,1,0,0)
    D(e_1)=(0,0,0,1,0)
    D(e_1)=(0,0,0,0,1)

    And then form the matrix D_B=[D(e_1), D(e_2), D(e_3), D(e_4), D(e_5)]. But the resulting matrxix I got was very different from the answer. So what do I need to do?
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  2. #2
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    Quote Originally Posted by demode View Post
    Let V be the vector space

    f(x)=\sum^2_{k=1}(a_Kcoskx+b_ksinkx)

    Find the matrix [D]_B of the differentiation operator D on V in the basis \{ 1,sinx, cosx,sin2x, cos2x \}.



    So I don't understand how they got that matrix for D! To find the matrix for D, what I did was to compute the images of the standard unit vectors under the transformation. These are

    T(e_1)=(1,0,0,0,0)=(0, cos0, -sin0, 2cos0, 2sin0)

    And so on for

    D(e_2)=(0,1,0,0,0)
    D(e_1)=(0,0,1,0,0)
    D(e_1)=(0,0,0,1,0)
    D(e_1)=(0,0,0,0,1)

    And then form the matrix D_B=[D(e_1), D(e_2), D(e_3), D(e_4), D(e_5)]. But the resulting matrxix I got was very different from the answer. So what do I need to do?

    They must be using the definition that 99% of all linear algebraists use: the matrix for D is the transpose of the coefficients matrix of the system: for example, you got D(\cos x)=-\sin x, and since \cos x is the 3rd vector in the given matrix and \sin x the 2nd one , that means that the 3rd column of the matrix will have -1 in the 2nd position, which it indeed has.
    BTW, the given matrix is the one for D wrt the given basis: the standard basis' vectors you use there have nothing to do with this problem since they don't belong to the space you're working with.

    Tonio
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