Nonlinear Operators on Finitely Generated Vector Spaces: Group Property Example

[Zelyucha]

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.

 

[micromass]

Finding a nonlinear operator is of course very easy. Take f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x^2.

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

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

x=\sum_{i\in I} \alpha_i e_i

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

f:\mathbb{R}\rightarrow \mathbb{R}: \sum_{i\in I} \alpha_i e_i\rightarrow \alpha_j

This function satisfies f(x+y)=f(x)+f(y) (because it is \mathbb{Q}-linear), but it is not \mathbb{R} linear!

 

[Zelyucha]

Finding a nonlinear operator is of course very easy. Take f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x^2.

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

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

x=\sum_{i\in I} \alpha_i e_i

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

f:\mathbb{R}\rightarrow \mathbb{R}: \sum_{i\in I} \alpha_i e_i\rightarrow \alpha_j

This function satisfies f(x+y)=f(x)+f(y) (because it is \mathbb{Q}-linear), but it is not \mathbb{R} 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:

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

and

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


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

\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

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).

 

[HallsofIvy]

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:

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

and

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

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

\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

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).

 

[micromass]

Furthermore, what do you mean with the "group property" in this case?

 

[pwsnafu]

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.