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Math Help - Matrix of Linear Transformation

  1. #1
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    Matrix of Linear Transformation

    Determine the matrix representation [T]CB (C is supposed to be the subscript and B the superscript here, sorry) for the linear transformation T and ordered bases B and C.

    T: P3 --> P2 given by T(p(x)) = pí(x),
    B = {1, x, x^2, x^3}; C = {1, x, x^2}

    I really have no idea how to proceed - the textbook is not being very helpful. I would appreciate an explanation rather than a solution, since I have many more of these problems to do. I hope someone can help!
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  2. #2
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    Remember that to find the matrix of linear transformation
    T:\mathcal{A} \to \mathcal{B}

    Let e_1,e_2,....,e_n be a basis for \mathcal{A}
    then
    T(e_1),T(e_2),...,T(e_n) are the columns for the matrix representation of T when expresses in terms of the basis for \mathcal{B}

    This is what I used in the previous problem I helped you with. I hope this gets you started.
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  3. #3
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    So at this point I have e1 = 1, e2 = x and so on for P3. My problem is I have lost track of the geometric meaning of what I am doing, and so I can't see where I am and where I need to go in order to answer the question. If I have these column vectors how to I put them into terms of P2? I know that the kernel will annihilate at least one dimension since it is going into P2 . . .
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    Behold, the power of SARDINES!
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    Quote Originally Posted by dan213 View Post
    So at this point I have e1 = 1, e2 = x and so on for P3. My problem is I have lost track of the geometric meaning of what I am doing, and so I can't see where I am and where I need to go in order to answer the question. If I have these column vectors how to I put them into terms of P2? I know that the kernel will annihilate at least one dimension since it is going into P2 . . .
    Good. So the basis of \mathcal{A}=\{1,x,x^3,x^3\}=\{e_1,e_2,e_3,e_4\} and the basis of \mathcal{B}=\{1,x,x^2\}=\{e_1,e_2,e_3\}

    Now T(e_1)=\frac{d}{dx}(1)=0=0e_1+0e_2+0e_3
    T(e_2)=\frac{d}{dx}(x)=0=1e_1+0e_2+0e_3

    This is the first two columns just keep going....
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  5. #5
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    Thanks, I was able to figure it out last night with your help on the previous thread. I really appreciate it.
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