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I Questions about linear transformations

  1. Oct 3, 2016 #1
    We learnt that the condition of a linear transformation is
    1. T(v+w) = T(v)+T(w)
    2. T(kv) = kT(v)

    I am wondering if there is any transformation which only fulfil either one and fails another condition. As obviously, 1 implies 2 for rational number k.

    Could anyone give an example of each case? (Fulfilling 1 but 2 and 2 but 1)

    Thanks!
     
  2. jcsd
  3. Oct 3, 2016 #2

    mathman

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  4. Oct 3, 2016 #3

    andrewkirk

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    Consider the real numbers ##\mathbb R## as a vector space over the rationals, and the operator ##T:\mathbb R\to \mathbb R## that is the identity on the rationals and maps to zero on the irrationals. Then T satisfies the second axiom, since it is a linear operator on ##\mathbb Q## considered as a subspace of ##\mathbb R## and, for ##q\in\mathbb Q-\{0\},x\in\mathbb R##, ##qx## is in ##\mathbb Q\cup\{0\}=\ker\ T## iff ##x## is.

    But the addition axiom does not hold, because ##T(1+(\sqrt2-1))=T(\sqrt 2)=0## but ##T(1)+T(\sqrt2-1)=1\neq 0##.
     
  5. Oct 3, 2016 #4

    Stephen Tashi

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    Define ##M((x,y))## by
    if ##x \ne y ## then ##M((x,y)) = (x,y)##
    if ##x = y ## then ##M((x,y)) = (2x, 2y)##
    ##M## satisfies 2, but not 1
     
  6. Oct 4, 2016 #5
    This really gives me new insight into linear transformation. thanks all!
     
  7. Oct 4, 2016 #6
    I am sorry but I don't quite understand. How can we construct a transformation like that?
     
  8. Oct 4, 2016 #7

    mathman

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    Unfortunately Hamel basis exists, but it is not constructable - existence is equivalent to axiom of choice.
     
  9. Oct 4, 2016 #8

    andrewkirk

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    I'm still trying to think of a scenario with a map that satisfies 1 (additivity) but not 2 (scalar mult). Can anybody think of one?

    All the examples I come up with either end up satisfying neither 1 nor 2, or satisfying 2 but not 1.

    I assume there must be one, otherwise some texts would specify 1 as the sole requirement and derive 2 as a consequence of 1.
     
  10. Oct 4, 2016 #9
    Andrew, I am thinking that as we have to apply the transformation to a vector space, and vectors in vector space obeys kv is also in the space. As we can have k be any real number, it seems that it somehow implies axiom2. The transformation only satisfy axiom1 must be of a very weird form.
     
  11. Oct 4, 2016 #10

    Stephen Tashi

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    On the "talk" page for the current Wikipedia article on "Linear transformation", I found:

     
  12. Oct 4, 2016 #11
    That's a clear and direct example!
     
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