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Matrix of a linear mapping

  1. Jul 25, 2012 #1
    Hi, I have the following problem that is solved, but I get lost at one step and cannot find how to do it in the notes. I would really appreciate it if someone could tell me where my teacher gets the result from.

    The problem says:

    "Find the matrix of linear mapping [itex]T:P_3 → P_3[/itex] defined by

    [itex](Tp)(t)=p(t)+p'(t)+p(0)[/itex]

    with respect to the basis {[itex]1,t,t^2,t^3[/itex]} of [itex]P_3[/itex]. Deduce that, given [itex]q \in P_3[/itex], there exists [itex]p \in P_3[/itex] such that
    [itex]q(t)=p(t)+p'(t)+p(0)[/itex]."

    And I get lost here... It says:

    "We have

    [itex]T(1)=2[/itex]
    [itex]T(t)=1+t[/itex]
    [itex]T(t^2)=2t+t^2[/itex]
    [itex]T(t^3)=3t^2+t^3[/itex]"

    So I don't know why it says [itex]T(1)=2[/itex].... I think [itex]T(t)=1+t[/itex] because it is the derivative of t plus t, and [itex]T(t^2)[/itex] is the derivative of [itex]t^2[/itex] plus [itex]t^2[/itex]... But why T(1)=2?

    Thanks a lot!
     
  2. jcsd
  3. Jul 25, 2012 #2
    I think this would be clearer written slightly differently.

    [tex]T[p(t)] = p(0) + \frac{dp}{dt} + p(t)[/tex]

    When you say [itex]T(1)[/itex], you substitute [itex]p(t) = 1[/itex] into the above. Clearly, then, [itex]p(0) = 1[/itex] because [itex]p(t) = 1[/itex] for any [itex]t[/itex]. That takes care of the first and last terms. The derivative is zero because [itex]p[/itex] is a constant.

    In short, you get

    [tex]T(1) = 1 + 0 + 1 = 2[/tex]
     
  4. Jul 25, 2012 #3

    Fredrik

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    T takes polynomials to polynomials, so when he writes T(1), 1 denotes a polynomial. The only polynomial that it makes sense to denote by 1 is the function that takes every real number to 1. It might be less confusing to denote it by a symbol like I instead. Then for all real numbers t, we have ##(T(I))(t)=I(t)+I'(t)+I(0)=1+0+1##, as Muphrid has already said. So T(I) is the polynomial that takes every real number t to 2. In this context, it seems to be standard to denote this polynomial by 2.

    I think this notation is worse, because now it looks like T is acting on the real number p(t) instead of on the polynomial p.
     
    Last edited: Jul 25, 2012
  5. Jul 25, 2012 #4

    Fredrik

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    I don't understand this argument. To find T(t), you need to do something similar to what I did above.

    Here t denotes the identity map on the set of real numbers, i.e. the function that takes every real number to itself. If we use this notation, then for all real numbers s, we have
    ##(T(t))(s)=t(s)+t'(s)+t(0)=s+1+0##. So T(t) is the polynomial that takes s to 1+s. In this notation, that polynomial is denoted by 1+t.

    This problem shows how confusing it can be to use notations like t2 both for a number (the square of the number t) and a function (the function that takes every real number to its square). I would prefer to use a different notation for the basis vectors, for example ##\{e_1,e_2,e_3,e_4\}## instead of ##\{1,t,t^2,t^3\}##, where the ##e_i## are defined by
    ##e_1(s)=1## for all s.
    ##e_2(s)=s## for all s.
    ...and so on.

    Now what the problem writes as T(1) and T(t) can be written as ##Te_0## and ##Te_1## respectively, and for all ##t\in\mathbb R##,
    ##Te_1(t)=e_1(t)+e_1'(0)+e_1(0)=1+0+1=2=2(e_1(t))=(2e_1)(t),##
    ##Te_2(t)=e_2(t)+e_2'(0)+e_2(0)=t+1+0=e_2(t)+e_1(t)=e_1(t)+e_2(t)=(e_1+e_2)(t)##.

    Since this holds for all t, we have ##Te_1=2e_1## and ##Te_2=e_1+e_2##.

    Can you do ##T(t^2)## and ##T(t^3)## now? You can stick to the t^something notation if it doesn't confuse you, but then you will have to write weird things like ##{t^2}'(s)=2s##.
     
    Last edited: Jul 25, 2012
  6. Jul 26, 2012 #5

    HallsofIvy

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    The derivative of the polynomial p(t)= 1 (for all t) is 0 so p(t)+ p'(t)+ p(0)= 1+ 0+ 1= 2.

    More precisely p(t)+ p'(t)+ p(0)= t+ 1+ 0= t+ 1

    Again, [itex]p(t)+ p'(t)+ p(0)= t^2+ 2t+ 0= t^2+ 2t[/itex]
     
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