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.