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Homework Help: Linear Transformations (or lack thereof)

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Let V be the set of complex numbers regarded as a vector space over the real numbers R. Find a linear transformation T: V → V which is not complex linear (i.e. not a linear transformation if V is regarded as a vector space over the complex numbers).

    2. Relevant equations


    3. The attempt at a solution
    I have a solution. What I need is a better explanation of what complex linear means. Or perhaps a rewording of the problem statement so I might have a better understanding of why my solution is a solution.
  2. jcsd
  3. Feb 12, 2014 #2


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    hi nateHI! :smile:
    f is linear if f(λa + µb) = λf(a) + µf(b) for all scalars λ and µ

    so what happens eg if λ = i ? :wink:
  4. Feb 12, 2014 #3


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    "... regarded as a vector space over the real numbers R"

    What does that mean?
  5. Feb 12, 2014 #4
    I think they mean z = u +iv with (u,v) in RxR.
  6. Feb 12, 2014 #5


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    The complex numbers are a 1-dimensional vector space over themselves, so a complex-linear map [itex]\mathbb{C} \to \mathbb{C}[/itex] is a map of the form [itex]z \mapsto az[/itex] for some [itex]a \in \mathbb{C}[/itex], ie. multiplication by a constant.

    The map which takes [itex]z = x + iy[/itex] to its complex conjugate [itex]z^{*} = x - iy[/itex] is not of that form (there is no [itex]a \in \mathbb{C}[/itex] such that [itex]z^{*} = az[/itex] for all [itex]z \in \mathbb{C}[/itex]), and so is not complex linear. But it is real linear (it's just [itex](x,y) \mapsto (x, -y)[/itex]).

    Therefore the map which takes [itex]z[/itex] to its real part [itex]\frac12 (z + z^{*})[/itex] is not complex linear either.
  7. Feb 12, 2014 #6
    No not true, take z = i then z* = -i, take a =-1 and you got it.
  8. Feb 12, 2014 #7


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    Your [itex]a[/itex] must work FOR ALL [itex]z \in \mathbb{C}[/itex]! Finding an [itex]a[/itex] which works for one [itex]z[/itex] is irrelevant!
  9. Feb 12, 2014 #8
    You're right my bad.
  10. Feb 13, 2014 #9
    OK, I get it now. Thanks pasmith and everyone else.
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