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Linear transformation (minor clarification)

  1. Apr 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Capture.PNG


    3. The attempt at a solution

    I don't think I'm interpreting the question correctly. Maybe someone can point me in the right direction?

    There are 2 conditions: if y =/=0 then f(x,y) = x^2/y and if y=0 then f(x,y) = 0

    Let u =(1,1) and v = (1,1)

    f(v) = f(1,1) = 1^2/1 = 1

    f(u) = f(1,1) = 1

    f(u) + f(v) = 2

    f(u+v) = 2

    testing for the second condition: if y=0 then f(x,y) = 0

    let u = (1,0) v = (1,0)

    f(u) = 0 since if y = 0, f(x,y) = 0

    f(v) = 0

    f(u) + f(v) = 0

    f(u+v) = f(2,0) = 0
     
  2. jcsd
  3. Apr 4, 2014 #2

    jbunniii

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    I assume that the goal is to check whether ##f## is linear. So your approach is fine:
    OK so far.
    No, ##u+v = (1,1) + (1,1) = (2,2)##. What do you get when you evaluate ##f## at this point?

    [edit] Oops, I miscalculated. Yes, you get 2. So this shows that ##f(u+v) = f(u) + f(v)## for that particular choice of ##u## and ##v##. But to conclude linearity, you need to show that it is true for all choices of ##u## and ##v##.

    Hint: try another choice of ##u## and ##v##.
     
  4. Apr 4, 2014 #3
    I presume we are referring to the first condition in the question where x^2/y ?

    I tried a few values but they all preserve addition.
     
  5. Apr 4, 2014 #4

    jbunniii

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    Yes, the ##y=0## case is unlikely to provide a counterexample for additivity.

    You should be able to find a simple counterexample using points of the form ##(a,1)## and ##(b,1)##
     
  6. Apr 4, 2014 #5

    HallsofIvy

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    That "[itex]x^2[/itex]" makes me suspicious! So I would try u= (2, 1) and v= (3, 1).

    What are f(u) and f(v)? What is f(u)+ f(v)?

    u+ v= (5, 2). What is f(u+ v)?
     
  7. Apr 4, 2014 #6
    Halls!

    Any reason as to why you chose those values for u and v? Are they just arbitrary?

    f(u) = f(2,1) = 4 f(v) = f(3,1) = 9
    f(u) + f(v) = 13

    u + v = (5,1)

    f(u+v) = 25
     
  8. Apr 5, 2014 #7

    Fredrik

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    Almost any choice of u and v will work, but the calculation of ##\frac{(u_1)^2}{u_2}+\frac{(v_1)^2}{v_2}## is easier when ##u_2=v_2##, and is especially easy when ##u_2=v_2=1##.
     
  9. Apr 5, 2014 #8

    jbunniii

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    You added ##(2,1)## to ##(3,1)## and got ##(5,1)##?
     
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