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Math Help - Polynomial

  1. #1
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    Polynomial

    f: P1(R) -> P1(R)

    f(1+4x) = 1-2x
    f(-2-9x) = 2+4x

    Find the transformation matrix for f in the monomial basis.

    |1 -2| |1 2|
    |4 -9| |-2 4|

    I tried to multiply these two matrices but that doesn't give me the correct facit.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Polynomial

    Quote Originally Posted by MathIsOhSoHard View Post
    f: P1(R) -> P1(R) f(1+4x) = 1-2x f(-2-9x) = 2+4x Find the transformation matrix for f in the monomial basis.
    Write \begin{Bmatrix} f(1)+4f(x)=1-2x\\-2f(1)-9f(x)=2+4x\end{matrix} . Express, f(1)=a+bx,\;f(x)=c+dx then, the solution is \begin{bmatrix}{a}&{c}\\{b}&{d}\end{bmatrix} .
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    Re: Polynomial

    Quote Originally Posted by FernandoRevilla View Post
    Write \begin{Bmatrix} f(1)+4f(x)=1-2x\\-2f(1)-9f(x)=2+4x\end{matrix} . Express, f(1)=a+bx,\;f(x)=c+dx then, the solution is \begin{bmatrix}{a}&{c}\\{b}&{d}\end{bmatrix} .
    I'm not sure that I understand this entirely.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Polynomial

    Quote Originally Posted by MathIsOhSoHard View Post
    I'm not sure that I understand this entirely.
    Have you studied the theory? If T:V\to W is a linear map, B_V=\{v_1,\ldots,v_m\} and B_W=\{w_1,\ldots,w_n\} are basis of V and W respectively then, [T]_V^W=[\;[T(v_1]_{B_W},\ldots,\;[T(v_n)]_{B_W}] .
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  5. #5
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    Re: Polynomial

    for P_1(\mathbb{R}) the monomial basis is \{1,x\}. so whatever matrix represents f in this basis, we have:

    \begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}1\\4\end{bmatrix} = \begin{bmatrix}1\\-2\end{bmatrix}

    and

    \begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}-2\\-9\end{bmatrix} = \begin{bmatrix}2\\4\end{bmatrix}

    now the column (a,c) is f(1,0) = f(1+0x) = f(1). so let's find this first. remember, matrix multiplication is linear. this means that 9f(1,4) + 4f(-2,-9)
    = f(9,36) + f(-8,-36) = f(1,0).

    but f(1,4) = (1,-2), and f(-2,-9) = (2,4). so 9f(1,4) + 4f(-2,-9) = 9(1,-2) + 4(2,4) = (9,-18) + (8,16) = (17,-2). so a = 17, c = -2.

    now let's find the second column (b,d), which is f(0,1) = f(0+1x) = f(x).

    -2f(1,4) - f(-2,-9) = f(-2,-8) + f(2,9) = f(0,1). so f(0,1) = -2(1,-2) - (2,4) = (-2,4) + (-2,-4) = (-4,0). so b = -4, d = 0.

    finally, let's check:

    \begin{bmatrix}17&-4\\-2&0\end{bmatrix} \begin{bmatrix}1\\4\end{bmatrix} = \begin{bmatrix}1\\-2\end{bmatrix}

    and

    \begin{bmatrix}17&-4\\-2&0\end{bmatrix} \begin{bmatrix}-2\\-9\end{bmatrix} = \begin{bmatrix}2\\4\end{bmatrix}

    so that is indeed the correct matrix.
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