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Linear Transformation - Complex -> Complex

  1. Mar 22, 2009 #1
    The problem:

    T(x + yi) = x

    C -> C (Complex Numbers)

    Show that the above is:

    Linear
    Isomorphic


    This is what i have for showing it's linear:

    T(x+yi + a + bi) = x + a + i(y + b) => T(x+a) => T(x) + T(a)
    T(k(x+yi)) = k(x) + k(yi) = T(kx) = kT(x)

    I assume that i(y+b) = 0. Is this the proper assumption?

    Therefore it is not isomorphic because of the fact that i(y+b) = 0?

    The answer in the back is states it's not isomorphic because 5i = 0 exists.
    So I'm sort of confused as to what they are trying to say.
    Aare they saying that by choosing a point on the plane, (x,y) that the component of 5i = 0 (I'm assuming 5 is random) exists for this particular transformation, and since thats impossible the transformation is not isomorphic? Or am I totally on the wrong track.
     
  2. jcsd
  3. Mar 22, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Completely wrong. T(x+yi + a + bi) is NOT equal to x + a + i(y + b), it is equal to x+ a. And I can make no sense of "=> T(x)+ T(a)".

    NO, T(kx+ yi)= T(kx+ kyi)= kx, not kx+ k(yi). And T(kx)= kT(x) is what you want to prove.


    No, it is not. Where did you get that idea?

    Probably the problem is that your book doesn't say anything of the sort! Surely you know that "5i" is NOT equal to 0. What your book says is that T(5i)= 0. That is true because T(5i)= T(0+ 5i)= 0. Also T(0)= 0, T(3i)= T(0+ 3i)= 0, etc. Since a an isomorphism must be one-to-one, T is not an isomorphism.

    That makes no sense at all. I have no idea what you mean by "the component of 5i".

     
  4. Mar 22, 2009 #3
    Wow, now I feel like a moron.
    ha ha ha
    Grr...

    I got confused when I saw that the transformation went from x + yi to x. :/

    Sorry, you're right, the book did say T(5i) = 0 exists, therefore it's not an isomorphism.

    I think I get it now
     
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