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Math Help - Prove that T is a linear transformation

  1. #1
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    Prove that T is a linear transformation

    Hello,

    T: P2(R) -> P3(R) defined by T(f(x)) = xf(x) + f`(x).

    Let A,B,C,a,b,c be elements of R and x,y elements of P2, where f(x) = (a + bx + cx^2) and f(y) = (A + By + Cy^2)

    Can't I write this as a matrix vector product and prove it to be linear that way? I can also calculate the rank and null this way, right?

    T(f(x)) = x(a + bx + cx^2) + b + 2x
    = b + 2ax + bx^2 + cx^3

    is this right so far? I really want to use the matrix vector product method though, but I'm having a hard time making it into a matrix, any advice?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by zodiacbrave View Post
    Can't I write this as a matrix vector product and prove it to be linear that way?

    Better:

    T[\lambda f(x)+\mu g(x)]=x (\lambda f(x)+\mu g(x))+(\lambda f(x)+\mu g(x))'=

    \ldots =\lambda T[f(x)]+\mu T[g(x)]
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  3. #3
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    Fernando Revilla is correct that the best way to prove something is a linear transformation is to show that it satisfies the definition of a linear transformation.

    Since you specifically ask about writing it as a matrix, remember that a linear transformation from vector space U to vector space V depends on specific choices of (ordered) bases for U and V. Now, apply the linear transformation to each of the basis vector of U in turn, writing the result in terms of the basis for V. The coefficients will be the columns of the matrix.

    Here, U is the function space P2 of quadratic polynomials and V is the function space P3 of cubic polynomials. The standard basis for the first is 1, x, and x^2 and for the second, 1, x, x^2, and x^3.

    Applying the linear transformation to "1" we get x(1)+ 0= x= 0(1)+ 1(x)+ 0(x^2)+ 0(x^3). The first column of the matrix representation is \begin{bmatrix}0 \\ 1\\ 0 \\ 0\end{bmatrix}.

    Applying the linear transformation to "x" we get x(x)+ 1= x^2+ 1= 1(1)+ 0(x)+ 1(x^2)+ 0(x^3). The second column of the matrix representation is \begin{bmatrix}1 \\ 0 \\ 1 \\ 0\end{bmatrix}.

    Applying the linear transformation to " x^2" we get x(x^2)+ 2x= x^3+ 2x= 0(1)+ 2(x)+ 0(x^2)+ 1(x^3). The third column of the matrix representation is \begin{bmatrix}0 \\ 2 \\ 0 \\ 1\end{bmatrix}.

    The matrix representing the linear transformation, using these bases in these orders, is
    \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 2\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}.
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