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[Linear Algebra] Finding T*; complex conjugate linear transformation

  1. Nov 28, 2011 #1
    [Linear Algebra] Finding T* adjoint of a linear operator

    1. The problem statement, all variables and given/known data

    Consider [itex]P_1{}(R)[/itex], the vector space of real linear polynomials, with inner product

    [itex] < p(x), q(x) > = \int_0^1 \! p(x)q(x) \, \mathrm{d} x [/itex]

    Let [itex]T: P_1{}(R) \rightarrow P_1{}(R)[/itex] be defined by [itex]T(p(x)) = p'(x) + p(x)[/itex]. Find [itex] T^*(p(x)) [/itex] for an arbitrary
    [itex]p(x) = a + bx^2 \in P_1{}(R)[/itex].


    2. Relevant equations

    [itex] < T(p(x)), q(x) > = < p(x), T^*(q(x)) > [/itex]

    3. The attempt at a solution

    Using the standard basis of [itex]P_1{}(R), \alpha = <1, x>[/itex]

    [itex]<T(1),1> = 1[/itex]
    [itex]<T(1), x> = 1/2[/itex]
    [itex]<T(x), 1> = 3/2[/itex]
    [itex]<T(x), x> = 5/6[/itex]

    [itex]
    [T]_\alpha^{\alpha} =
    \left[ {\begin{array}{cc}
    1 & 3/2 \\
    1/2 & 5/6 \\
    \end{array} } \right]
    [/itex]
    [itex]
    [T^*]_\alpha^{\alpha} =
    \left[ {\begin{array}{cc}
    1 & 1/2 \\
    3/2 & 5/6 \\
    \end{array} } \right]
    [/itex]

    Not sure how to find [itex]T^*[/itex] from here...
     
    Last edited: Nov 29, 2011
  2. jcsd
  3. Nov 28, 2011 #2

    micromass

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    What you did only works for an orthonormal basis. So perhaps you should first find an orthonormal basis of your space??
     
  4. Nov 28, 2011 #3
    I just got the new matrices with the orthonormal basis but I'm still at the same sticking point

    The matrices aren't equal so T isn't self-adjoint; that's all I can conclude right now
     
  5. Nov 28, 2011 #4

    micromass

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    If you work in an orthonormal basis, then the adjoint is given by the hermition conjugate. So find the hermitian conjugate and express the matrix as a linear transformation. And there you have your adjoint!!
     
  6. Nov 29, 2011 #5
    [itex]\alpha = <1, \sqrt{3}(2x-1)>[/itex] is an orthonormal basis for [itex]P_1{}(R)[/itex]

    By the definition of the inner product space, have

    [itex] < T(p(x)), q(x) > = \int_0^1 \! p'(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x [/itex]

    So I computed

    [itex]< T(1), 1 > = 1 [/itex]

    [itex]< T(1), \sqrt{3}(2x-1) > = 0[/itex]

    [itex]< T(\sqrt{3}(2x-1)), 1 > = 4\sqrt3[/itex]

    [itex]< T(\sqrt{3}(2x-1)), \sqrt{3}(2x-1) > = 1[/itex]

    [itex]
    [T]_\alpha^{\alpha} =
    \left[ {\begin{array}{cc}
    1 & 4\sqrt{3} \\
    0 & 1 \\
    \end{array} } \right]
    [/itex]

    [itex]
    [T^*]_\alpha^{\alpha} =
    \left[ {\begin{array}{cc}
    1 & 0 \\
    4\sqrt{3} & 1 \\
    \end{array} } \right]
    [/itex]

    Still not sure how to find [itex]T^*[/itex] from this...




    Verifying the matrix...

    [itex] < T(p(x)), q(x) > = \int_0^1 \! p'(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x = < p(x), T^*(q(x))> [/itex]

    [itex] < T(p(x)), q(x) > = \int_0^1 \! p'(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x [/itex]

    [itex] < p(x), T(q(x)) > = \int_0^1 \! p(x)q'(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x = p(1)q(1) - p(0)q(0) - \int_0^1 \! p'(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x[/itex]

    So [itex] T [/itex] is not self-adjoint.
     
  7. Nov 29, 2011 #6

    micromass

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    Doesn't that matrix for T* actually give you T*?? How do you construct a linear mapping from a matrix?
     
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