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Math Help - linear maps

  1. #1
    Super Member Deadstar's Avatar
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    linear maps

    Let T be the linear map on P_2 defined by
    T(p)(t) = (1+t)p^{'}(t)
    Determine the matrix of T with respect to the basis (1, t, t^2) of P_2.
    What are the eigenvalues of T? Is T diagonalisable?
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  2. #2
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    I will help with the basic matrix.
    From T(a + bt + ct^2 ) = (1 + t)(b + 2ct) = \left( {b + (b + 2c)t + 2ct^2 } \right) we get the matrix
    \left[ {\begin{array}{ccc}<br />
   0 & 1 & 0  \\    0 & 1 & 2  \\   0 & 0 & 2  \\ \end{array}} \right]
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Deadstar View Post
    Let T be the linear map on P_2 defined by
    T(p)(t) = (1+t)p^{'}(t)
    Determine the matrix of T with respect to the basis (1, t, t^2) of P_2.
    What are the eigenvalues of T? Is T diagonalisable?
    The columns of the matrix are the images of each of the basis vectors under
    the transformation (in component form with respect to the basis)

    RonL
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  4. #4
    Super Member Deadstar's Avatar
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    cheers guys. One more question, what does the symbol that looks like a "U to the power 'an upside down T'" mean? I dont know how to notate that with latex.
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  5. #5
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    lol, you miss those lectures too? I'm doing the same course.

    The 'U to the power 'an upside down T' thing you speak of is the orthogonal complement of U. The orthogonal compliment of a subspace U of an inner product V is the set of all vectors in V that are orthogonal to every vector in U.

    Know Q3 yet?
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  6. #6
    Super Member Deadstar's Avatar
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    no ive not really looked at it yet, Q.1s sorted. Trying to do the Stats assignment as well, bloody hard...
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  7. #7
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    For T(p)(t) = (1+t)p^{'}(t)

    is T invertible?

    & would T^{-1} = -b - (b+2c)t - (2c)t^2 ?
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by lllll View Post
    For T(p)(t) = (1+t)p^{'}(t)

    is T invertible?

    & would T^{-1} = -b - (b+2c)t - (2c)t^2 ?
    No T is not invertible, the matrix of T has zero determinant.

    Another way to look at it is that if it were invertibable then for every quadratic function f(x) (quadratic function here means a polynomial of degree not more than 2):

    p(t)=\int_0^t \frac{f(\zeta)}{1+\zeta} ~d\zeta + p_0

    would be a unique quadratic, but it is neither unique nor always a quadratic
    (put f(x)=1 to see that it is not always a quadratic)

    RonL
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