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Nonlinear Operators

  1. Mar 6, 2013 #1
    Can someone give an example of a nonlinear operator on a finitely generated vector space(preferably ℝn)? I'd be particularly interested to see an example of such that has the group property as well.
  2. jcsd
  3. Mar 6, 2013 #2
    Finding a nonlinear operator is of course very easy. Take [itex]f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x^2[/itex].

    But you seem to be interested in a function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] such that [itex]f(x+y)=f(x)+f(y)[/itex] for all [itex]x,y\in \mathbb{R}[/itex] but that is still not [itex]\mathbb{R}[/itex]-linear.

    The existence of such a function depends on the axiom of choice. One construction is as follows. We know that [itex]\mathbb{R}[/itex] is a [itex]\mathbb{Q}[/itex]-vector space. Let [itex]\{e_i\}_{i\in I}[/itex] a basis of [itex]\mathbb{R}[/itex] as [itex]\mathbb{Q}[/itex]-vector space. We can write every [itex]x\in \mathbb{R}[/itex] uniquely as

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

    where each [itex]\alpha_i[/itex] is rational and only finitely many of them are nonzero. Now take an arbitrary but fixed [itex]j\in I[/itex]. Take the function

    [tex]f:\mathbb{R}\rightarrow \mathbb{R}: \sum_{i\in I} \alpha_i e_i\rightarrow \alpha_j[/tex]

    This function satisfies [itex]f(x+y)=f(x)+f(y)[/itex] (because it is [itex]\mathbb{Q}[/itex]-linear), but it is not [itex]\mathbb{R}[/itex] linear!
  4. Mar 6, 2013 #3

    That's not quite what I meant. When I spoke of a non-linear operator, I was talking about a mapping η : Vn→Vn (where Vn is a vector space of positive integer dimension n) such that η is a non-linear transformation. The term "transformation" is used in vector algebra to mean a function mapping of a vector space into itself . So if ψ is a mapping from Vn→Vn that maps n-tuples to n-tuples, then for any vectors u,v in Vn, ψ is linear if:

    [tex](1): ψ(\vec{u})+ ψ(\vec{v}) = ψ(\vec{u}+\vec{v}) \; \forall \; \vec{u},\vec{v} \in V^n[/tex]


    [tex](2): \forall λ(scalar), \; λψ(\vec{v})=ψ(λ\vec{v}) \; \forall \vec{v} \in V^n [/tex]

    Correct me if I'm wrong but I believe that statement (1) is what is called the superposition principle. Much like for any real number r, the function f(x)=rx obeys this principle for any (x,y) in ℝ: f(x+y) = r(x+y) = rx + ry = f(x) + f(y). As you probably know, any non-singular invertible square matrix An qualifies as a linear operator in ℝn and the collection of all such n-square invertible matrices is a group called the general linear group denoted by GL(n,ℝ).

    So for any finite dimensional vector space V, let C(Vn,Vn) be the collection of all functions mapping V → V(for dimension n). So what I'm looking for is a subcollection D in C(Vn,Vn) such that

    [tex]\forall ζ \in D \subseteq C(V^n,V^n), \; ζ(\vec{u}+\vec{v}) \neq ζ(\vec{u})+ζ(\vec{v}) \; whenever \; \vec{u} \neq \vec{v} \; \; \forall (\vec{u},\vec{v}) \in V^n [/tex]

    And in particular, a collection D in C(ℝn,ℝn) where D satisfies the group property under multiplication(and perhaps addition too but that's optional).
    Last edited: Mar 6, 2013
  5. Mar 6, 2013 #4


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    And that is exactly what micromass gave you with n= 1.

    No. Properties 1 and 2 are both necessary to have ψ linear.

  6. Mar 6, 2013 #5
    Furthermore, what do you mean with the "group property" in this case?
  7. Mar 6, 2013 #6


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    So let n = 1. Take all constant functions f(x) = c with c > 0.
    Then f(x+y) = c but f(x)+f(y) = 2c.
    And you have a group under point-wise multiplication.
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