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Finding a transformation

  1. Feb 17, 2014 #1
    The two properties every linear transformation T: V -> W has to satisfy is
    T(u + v) = T(u) + T(v), for u,v in V (i)
    T(cu) = cT(u) for u in V and scalar c (ii)

    I'm trying to find a transformation which satisfies (i) but doesn't satisfy (ii) [I've been able to find the opposite for what it's worth].
     
  2. jcsd
  3. Feb 17, 2014 #2

    mathman

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    I suspect it would be very difficult. It is fairly easy to show that (i) => (ii) as long as c is rational.
     
  4. Feb 17, 2014 #3

    jgens

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    This might be overkill, but assuming the axiom of choice, the wild automorphisms of the complex plane provide examples where (1) holds but (2) does not.
     
  5. Feb 17, 2014 #4
    Would you mind briefly mentioning the outline for the proof (i) => (ii) for rational c? Is it simply because the sum of two rational numbers always yields a rational number.

    Yeah that is way above the level I was hoping for. I found an example where (2) holds but (1) doesn't online. The source where I found it did say he wasn't able to find one for the other way around, but he imagines it would be similar to the one for (2) holding and (1) not. Here is the example if it helps.

    T: R^2 -> R
    T[(x,y)] = {x if y = 0
    ..............{0 if y != 0
     
    Last edited: Feb 17, 2014
  6. Feb 17, 2014 #5

    jgens

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    Suppose T satisfies (1). Notice that T(nv) = T(v+...+v) = T(v)+...+T(v) = nT(v) and since T(-v) = -T(v) it follows that (2) holds for all integers. Next consider nT(m/n v) = T(m/n v)+...+T(m/n v) = T(m/n v+...+m/n v) = T(mv) = mT(v) and it now follows that (2) holds for all rational numbers.

    The difficult with finding concrete examples where (1) holds but not (2) is that the arrow needs to be pretty discontinuous. There may well be some simple example illustrating this, but all the examples I can think up that occur "in nature" so to speak are pretty wild.
     
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