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Linear Transformation P2 > R^2

  1. Jun 19, 2008 #1
    1. The problem statement, all variables and given/known data

    If L( p(x) ) = [ integral (p(x)) dx , p(0) ]

    find representation matrix A such that

    L (a + bx) = A[a b]^T

    2. Relevant equations

    3. The attempt at a solution
    I don't quite understand the question.
    I think that:
    if the base from p2 is {1, x} then any vector in p2 is of the form a + bx.
    Then I can find L(a) = (a1 , a2) and L(bx) = (a3 , a4)
    And use it to get A?

    Can someone re phrase the question for me?
  2. jcsd
  3. Jun 19, 2008 #2


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    Given: [tex]L(p(x)) = \left[\begin{array}{c}\int p(x)dx\\p(0)\end{array}\right][/tex]

    Put [tex]L(ax+b) = \left[\begin{array}{c}\int (ax+b)dx\\b\end{array}\right][/tex]

    keeping going... the goal is to come up with a matrix A such that [tex]A\left[\begin{array}{cc}a&b\end{array}\right]^T[/tex] equals the right-hand side of the last equation.
  4. Jun 19, 2008 #3
    [tex]L(ax+b) => \left[\begin{array}{cc}x&x^2 / 2 \\0&1\end{array}\right] [/tex] [tex][\begin{array}{c}\ \alpha\\ \beta \end{array} \right] [/tex]
  5. Jun 20, 2008 #4

    matt grime

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    I think you're missing something important: the map is supposed to go from P^2 to R^2, so you can't map ax+b as you claim as the integral in the first coordinate yields (infinitely many) polynomial expressions. Surely the integral should be over some interval (I'd suggest the integral from 0 to 1 as the most likely).

    Your matrix in the last post can't make sense, since it implies that the polynomial variables are allowed to appear in the positions in vectors in R^2.
  6. Jun 23, 2008 #5
    yup that was it.. I missed the interval, so the xs turned out to be 1's

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