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Math Help - Transpose of a linear transformation

  1. #1
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    Transpose of a linear transformation

    If F is a field and f is the linear functional on F^2 s.t. f(x_1,x_2) = ax_1+bx_2
    For the following linear operators T, let g = T^{t}f and find g(x_1,x_2)

    lets say T(x_1,x_2) = (x_1,0)

    So do I find f first, get f(x_1,x_2) = ax_1+bx_2

    do i now treat " ax_1" and " bx_2" as a vector with two rows and multiply by the transpose of the matrix representation of the linear transformation?

    in this case, i get ax_1

    I read this part of the book, it really isn't clear on what to do. Thank you for your help.
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  2. #2
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    Re: Transpose of a linear transformation

    the definition of g doesn't make sense unless g = f \circ T^t, because the domain of T^t is F^2, and the co-domain of f is F.

    we're missing some crucial information here. please state the entire problem.
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  3. #3
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    Re: Transpose of a linear transformation

    Sorry. To clarify, it is composition. g = f \; o \; T^{t}
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  4. #4
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    Re: Transpose of a linear transformation

    but you wrote g = T^tf which isn't the same thing. you can't compose two functions unless the co-domain of the first one you do, is the domain of the 2nd one you do.

    to make matters even more confusing, some authors means g(f(x)) by gf, and some mean, f(g(x)).

    IF g = f \circ T^t, there isn't any problem:

    g(x_1,x_2) = f(T^t(x_1,x_2)) =

    =f(\begin{bmatrix}t_{11}&t_{21}\\t_{12}&t_{22}\end  {bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix})

    =f(t_{11}x_1+t_{21}x_2,t_{12}x_1+t_{22}x_2)

    =a(t_{11}x_1+t_{21}x_2) + b(t_{12}x_1+t_{22}x_2)

    where T = \begin{bmatrix}t_{11}&t_{12}\\t_{21}&t_{22} \end{bmatrix} is the matrix for T in the standard basis.

    if we use the example you gave, T(x_1,x_2) = (x_1,0), T has the matrix:

    \begin{bmatrix}1&0\\0&0\end{bmatrix} and its transpose is: \begin{bmatrix}1&0\\0&0\end{bmatrix}, which is the same as T, in which case:

    g(x_1,x_2) = f(x_1,0) = ax_1.
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