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Homework Help: Linear Algebra Problem

  1. Oct 11, 2006 #1
    Let F be the vector space of all functions mapping [tex] R [/tex] into [tex] R [/tex], and let [tex]T:F\rightarrow F[/tex] be a linear transformation such that [tex]T(e^{2x})=x^{2}[/tex], [tex] T(e^{3x})=sinx[/tex], and [tex]T(1)=cos5x[/tex]. Find the following, if it is determined by this data.

    [tex]T(e^{5x})[/tex]
    [tex]T(3e^{4x})[/tex]
    [tex]T(3+5e^{3x})[/tex]
    [tex]T(\frac{e^{4x}+2e^{5x}}{e^{2x}})[/tex]
     
    Last edited: Oct 11, 2006
  2. jcsd
  3. Oct 11, 2006 #2
    I have no idea what to do here. Does this involve the kernel of T?
     
  4. Oct 11, 2006 #3

    radou

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    Homework Helper

    Try using the properties of a linear transformation: http://mathworld.wolfram.com/LinearTransformation.html" [Broken].
     
    Last edited by a moderator: May 2, 2017
  5. Oct 11, 2006 #4
    So am I supposed to factor out the [tex]e^{x}[/tex] term like this

    [tex]T(e^{3x})=sinx[/tex]
    [tex]e^{x}T(e^{2x})=sinx[/tex]
    [tex]T(e^{2x})=\frac{sinx}{e^{x}}[/tex]

    and since [tex]T(e^{2x})=x^{2}[/tex]

    it must be true that [tex]\frac{sinx}{e^{x}}=x^{2}[/tex]

    Am I on the right track?
     
  6. Oct 11, 2006 #5
    Oh wait, that can't be right, because [tex]e^{x}[/tex]
    is not a constant. Duh. :rolleyes:
     
  7. Oct 11, 2006 #6
    So the only one that I actually can figure out is
    [tex]T(3+5e^{3x})[/tex]
    since it is preserved by scaler multiplication and vector addition. Right?

    [tex]T(3+5e^{3x})=T(3)+T(5e^{3x})[/tex]

    [tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3T(1)+5T(e^{3x})[/tex]

    [tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3cos5x+5sinx[/tex]

    Is that right?
     
    Last edited: Oct 11, 2006
  8. Oct 11, 2006 #7

    radou

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    Yes, looks okay. Hint regarding the last one: [tex]T(\frac{e^{4x}+2e^{5x}}{e^{2x}}) = T(\frac{e^{4x}}{e^{2x}}+\frac{2e^{5x}}{e^{2x}})[/tex].
     
    Last edited: Oct 11, 2006
  9. Oct 11, 2006 #8
    [tex]T(\frac{e^{4x}+2e^{5x}}{e^{2x}}) = T(\frac{e^{4x}}{e^{2x}}+\frac{2e^{5x}}{e^{2x}})[/tex]
    Is this what you mean?
     
  10. Oct 11, 2006 #9
    Thank you so much bro. I REALLY apreciate the assistance. I owe you one. Gotto go to my L.A. class now. THanks.
     
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