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Proving a transformation is not linear

  1. Jun 9, 2013 #1
    For a certain transformation T, it is known that [itex] T(x+y) = T(x) + T(y) [/itex]

    It is required to determine whether this transformation is linear. Obviously it is not, since it need not satisfy the degree-1 homogeneity property of all linear maps.

    I'm just having trouble cooking up the counterexample. Any ideas? Thanks!

    BiP
     
  2. jcsd
  3. Jun 9, 2013 #2

    micromass

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    Is ##T:\mathbb{R}\rightarrow \mathbb{R}##?
    And are there other things known about ##T## such as continuity?

    If ##T:\mathbb{R}\rightarrow \mathbb{R}## and it is indeed true that the only thing you know is ##T(x + y) = T(x) + T(y)## then there is a counterexample for linearity. This is not an easy counterexample however. It consist in looking at ##\mathbb{R}## as a ##\mathbb{Q}##-vector space. As such, we have a basis ##\{e_i\}_{i\in I}## of ##\mathbb{R}##. So any element ##x\in \mathbb{R}## can be written uniquely as

    [tex]x = \sum_{i\in I}\alpha_i e_i[/tex]

    for some rational numbers ##\alpha_i\in \mathbb{Q}## such that only finitely many ##\alpha_i## are nonzero. Now take ##j\in I## fixed an consider

    [tex]T(x) = \alpha_j[/tex]

    this map satisfies ##T(x+y) = T(x) + T(y)##, but not ##T(\lambda x) = \lambda T(x)## for all ##\lambda \in \mathbb{R}## (it does satisfy it for ##\lambda in \mathbb{Q}## though).
     
  4. Jun 9, 2013 #3
    I see. What about the following transformation [itex] T: ℂ → ℝ [/itex] where [itex] T(x) = Re(x) [/itex]. If the field is the complex field, then the scalar [itex] i [/itex] causes homogeneity to fail. Additivity is trivially shown.

    Is this valid?

    BiP
     
  5. Jun 9, 2013 #4

    micromass

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    If you have a map ##T:V\rightarrow W## then it's assumed that both maps have the same scalar field (otherwise linear makes no sense). So I assume that both ##\mathbb{C}## and ##\mathbb{R}## are ##\mathbb{R}##-vector spaces. But then the map you mention is linear.
     
  6. Jun 9, 2013 #5
    I'm sorry, if I redefined the transformation as:
    ##T:ℂ\rightarrow ℂ## then is my example valid?

    BiP
     
  7. Jun 9, 2013 #6

    micromass

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    Yes, that would be a valid example.
     
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