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Linear transformations, images for continuous functions

  1. Sep 11, 2016 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    Let ##C## be the space of continuous real functions on ##[0,\pi]##. With any function ##f\in C##, associate another function ##g=T(f)## defined by $$g=T(f)\equiv \int_0^\pi \cos(t-\tau) f(\tau) \, d \tau$$
    a) Show ##T## is a linear transformation from ##C## to ##C##.
    b)What is the image of ##T##? Find a basis for it.
    c) List a set of linearly independent vectors that are in the null space of ##T##. What is ##dimN(T)##?

    2. Relevant equations
    Linear algebra stuff. Too much to write I think.

    3. The attempt at a solution
    a)##T(af + bg) = \int\cos(t-\tau)(af + bg) = \int\cos(t-\tau)af + \int\cos(t-\tau)bg = a\int\cos(t-\tau)f + b\int\cos(t-\tau)g = aT(f)+bT(g)##.
    b) I want to say the space of continuous real functions on ##[0,\pi]##. Does this even make sense though?
    c)By inspection, ##a\sin(t-\tau)## but this is all I can see. Are there any others? In this case, I would say ##dimN(T) = 1## so far since I can only think of the one example. Obviously ##0## is as well but this is not linearly independent of what I have listed.
     
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  3. Sep 11, 2016 #2

    Ray Vickson

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    Even if ##f(\tau)## is defined only on ##\tau \in [0,\pi]##, is ##g(t)## defined for ##t## outside the interval ##[0,\pi]##?
     
  4. Sep 11, 2016 #3

    joshmccraney

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    This was not specified, so I'm leaning towards no.
     
  5. Sep 11, 2016 #4

    pasmith

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    [itex]\cos(t - \tau) = \cos t \cos \tau + \sin t \sin \tau[/itex] seems extremely relevant, as it should solve part (b) for you.

    Part (a) required you also to show that [itex]T(f) \in C[/itex] for every [itex]f \in C[/itex]. Part (b) requires you to go further.

    Doing part (b) properly will suggest further examples of vectors in [itex]\ker T[/itex].
     
  6. Sep 11, 2016 #5

    joshmccraney

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    I don't follow you. I can't see how this identity helps us find the image of ##T##. Could you elaborate?



    How to you go about showing this part of a)? I would try but I don't know where to start.

    Thanks for replying!
     
  7. Sep 11, 2016 #6

    pasmith

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    Try expressing [itex]T(f)[/itex] in the form [tex]
    T(f) =\int_0^\pi \left(\sum_{n=1}^N A_n(t)B_n(\tau)\right)f(\tau)\,d\tau = \sum_{n=1}^N A_n(t) \int_0^\pi B_n(\tau) f(\tau)\,d\tau.[/tex]
     
  8. Sep 11, 2016 #7

    Ray Vickson

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    I think that PF rules forbid us from answering that question.
     
  9. Sep 11, 2016 #8

    joshmccraney

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    So something like this $$\int_0^\pi \cos(t-\tau)f(\tau) \, d\tau = \int_0^\pi (\cos t\cos\tau +\sin t \sin \tau)f(\tau) \, d\tau\\ = \cos t\int_0^\pi\cos\tau f(\tau) \, d \tau + \sin t \int_0^\pi\sin \tau f(\tau) \, d\tau$$ But I still don't see how this helps us find a basis. If you feel uncomfortable helping me with this one since I clearly can't do it well, perhaps we could make a simpler example and you could help me through it so I can see the strategy? I don't want to violate the rules on PF, I'm just confused how all of this is related.
     
  10. Sep 12, 2016 #9

    pasmith

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    [itex]\int_0^\pi \cos \tau f(\tau)\,d\tau[/itex] is a definite integral of a real-valued function, so it's just a real number!
     
  11. Sep 12, 2016 #10

    joshmccraney

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    How does this look?
     

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  12. Sep 13, 2016 #11

    pasmith

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    You need to rethink the kernel.

    I sense some confusion. The input of [itex]T[/itex] is a function of one real variable, which we are labelling [itex]\tau[/itex]. The output of [itex]T[/itex] is another function of one real variable, which we are labelling [itex]t[/itex]. Setting [itex]f(\tau) = \sin(t - \tau)[/itex] therefore is nonsense; trying [itex]f(\tau) = \sin (\alpha - \tau)[/itex] for real [itex]\alpha[/itex] makes sense, but it turns out that that is not in the kernel for any [itex]\alpha[/itex].

    Finding functions in the kernel is actually straightforward: You can readily show that if [itex]h(\tau) = a \cos \tau + b \sin \tau[/itex] then [itex]T(h) = \frac{\pi}{2}h[/itex], and having established that the image of [itex]T[/itex] is the space spanned by [itex]\cos[/itex] and [itex]\sin[/itex] it then follows that [tex]
    T^2(f) = \tfrac{\pi}2T(f)[/tex] for all [itex]f \in C[/itex], and hence by linearity of [itex]T[/itex] that [tex]
    T(f) - \tfrac{\pi}{2}f \in \ker T[/tex] for every [itex]f \in C[/itex]. This gives you a way to easily find non-trivial members of the kernel, provided you are careful with your choice of [itex]f[/itex].
     
  13. Sep 14, 2016 #12

    joshmccraney

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    Thanks! I really appreciate your help!
     
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