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Determination of Linear transformation

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine if the following T is linear tranformation, and give the domain and range of T:

    T(x,y) = (x + y2, [tex]\sqrt[3]{xy}[/tex] )

    2. Relevant equations

    T ( u + v) = T(u) + T(v)

    T(ru) = rT(u)



    3. The attempt at a solution
    1)
    let u = (x1, x2);
    T(ru ) = T(rx1, rx2)
    T(ru )= r(x + y2) , r([tex]\sqrt[3]{xy}[/tex] )
    T(ru ) = r(x + y2 , [tex]\sqrt[3]{xy}[/tex] )

    so it suffices the first condition, right?

    2)
    let u = (x1, y1) and let v = (y1, y2);
    T ( u + v ) = T ( x1 + y1, x2 + y2)
    T ( u + v ) = Here I am confused with the term ( x + y2)
    T ( u + v )


    Any help please !
     
  2. jcsd
  3. Nov 16, 2008 #2

    Redbelly98

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    If u=(x,y), what is ru?
     
  4. Nov 16, 2008 #3
    (rx, ry)
     
  5. Nov 16, 2008 #4

    Redbelly98

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    Yes.

    So T(ru) = T(rx, ry) = ???
     
  6. Nov 16, 2008 #5
    yeah i have done that
     
  7. Nov 16, 2008 #6

    Redbelly98

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    Okay, but you did it wrong.

    What do you get when you substitute rx for x, and ry for y, into
    T(x,y) = (x + y2, [tex]\sqrt[3]{xy}[/tex] )

    EDIT: FYI this is the part that I'm trying to correct:
     
    Last edited: Nov 16, 2008
  8. Nov 16, 2008 #7
    should it be

    let u = (x1, x2);
    T(ru ) = T(rx1, rx2)
    T(ru )= ( (rx1 + (rx2)2), [tex]\sqrt[3]{(rx)(ry)}[/tex] )
     
  9. Nov 16, 2008 #8

    Redbelly98

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    Yes. Although x1 is x, and x2 is y, of course.

    Next (as you know), you compare that expression with rT(u).
     
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