Nonlinear Operators on Finitely Generated Vector Spaces: Group Property Example

In summary, the conversation discusses the concept of a nonlinear operator on a finitely generated vector space, preferably in ℝn, that also has the group property. A few examples are given and the existence of such a function is shown to depend on the axiom of choice. The properties of linearity are also discussed, with both 1 and 2 being necessary for a function to be linear. The conversation ends with a request for a subcollection of functions that does not satisfy the group property.
  • #1
Zelyucha
25
0
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.
 
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  • #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!
 
  • #3
micromass said:
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!





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]

and

[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).
 
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  • #4
Zelyucha said:
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]

and

[tex](2): \forall λ(scalar), \; λψ(\vec{v})=ψ(λ\vec{v}) \; \forall \vec{v} \in V^n [/tex]
And that is exactly what micromass gave you with n= 1.


Property (1) is what makes ψ Linear
No. Properties 1 and 2 are both necessary to have ψ linear.

and 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).
 
  • #5
Furthermore, what do you mean with the "group property" in this case?
 
  • #6
Zelyucha said:
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).

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.
 

Related to Nonlinear Operators on Finitely Generated Vector Spaces: Group Property Example

1. What is a nonlinear operator?

A nonlinear operator is a function that does not follow the rules of linearity, such as addition and scalar multiplication. Instead, its output is dependent on the input in a nonlinear way.

2. What are finitely generated vector spaces?

Finitely generated vector spaces are vector spaces that can be spanned by a finite set of vectors. This means that any vector in the space can be written as a linear combination of the finite set of basis vectors.

3. How do nonlinear operators act on finitely generated vector spaces?

Nonlinear operators can be defined on finitely generated vector spaces by mapping each basis vector to a unique vector in the space. The resulting function is then extended to the entire space by using the linearity of vector addition and scalar multiplication.

4. What is an example of a nonlinear operator on a finitely generated vector space?

An example of a nonlinear operator on a finitely generated vector space is the logarithm function on the vector space of positive real numbers. This operator maps each positive real number to its logarithm, which is a non-linear transformation.

5. How do nonlinear operators on finitely generated vector spaces behave as a group?

Nonlinear operators on finitely generated vector spaces can behave as a group if they have certain properties, such as being invertible and closed under composition. An example of a nonlinear operator that behaves as a group is the rotation operator on a two-dimensional vector space.

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