If $T:\ \mathbb R^n\longrightarrow\mathbb R^n$ is additive then $T$ is linear

Question

Let the bijection $T:\, \mathbb R^n\longrightarrow\mathbb R^n,\ n\geq2$ satisfy \begin{align} T(u+v)=T(u)+T(v),\ \forall u,v\in\mathbb R^n\tag1 \end{align} and \begin{align} T\big(\langle u\rangle\big)=\big<T(u) \big>,\ \forall u\in\mathbb R^n,\tag2 \end{align} this means \begin{align} \forall u\in\mathbb R^n,\ \forall\lambda\in\mathbb R,\ \exists\,\mu=\mu(\lambda)\in\mathbb R:\ T(\lambda u)=\mu(\lambda,u)T(u),\tag3 \end{align} Moreover, we assume that \begin{align} T(\lambda u)\,=\lambda T(u),\ \forall\lambda\in\mathbb Q. \tag4 \end{align}

Then is $T$ $\mathbb R-$linear ?

My attempt :

Let us define the maps as follow \begin{align} \mathbb R\,\overset{f}{\longrightarrow}\,\langle u\rangle\,\overset{T}{\longrightarrow}\,\big< T(u)\big>\,\overset{g}{\longrightarrow}\,\mathbb R \end{align} given by \begin{align} \lambda\,&\longmapsto\,f(\lambda)=\lambda u \\ \lambda u\,&\longmapsto\,T(\lambda u)=\mu T(u) \\ \mu T(u)\,&\longmapsto\,g\big(\mu T(u)\big)=\mu. \end{align} From the additivity of $T$,we can show that
\begin{align} \mu(\lambda+\varepsilon,u)&=\,\mu(\lambda,u)+\mu(\varepsilon,u) \end{align} In other hand, we have \begin{align} \mu(\lambda\varepsilon,u)T(u) \,=\,T(\lambda\varepsilon u)\,&=\,\mu(\lambda,\varepsilon u)T(\varepsilon u) \\ &=\,\mu(\lambda,\varepsilon u)\mu(\varepsilon,u)T(u) \end{align} Thus \begin{align} \mu(\lambda\varepsilon,u)\,=\,\mu(\lambda,\varepsilon u)\mu(\varepsilon,u) \end{align} I want to show that the function $\mu(\cdot,u)$ is multiplication preserving, and in order to get that, the first thing I think to show is \begin{align} \mu(\lambda,\varepsilon u)\,=\,\mu(\lambda,u).\tag5 \end{align} Then, we can use this fact to show that $\mu(\epsilon^2,u)=\mu(\epsilon,u)^2$, this will implies $\mu(\lambda,u)>0,\ \forall \lambda >0$. Hence $$\mu(\lambda,u)-\mu(\varepsilon,u)\,=\,\mu(\lambda-\varepsilon,u)\,>\,0,\ \forall \lambda >\varepsilon. $$ Thus $\mu$ is strictly increasing.

Let $\lambda\in\mathbb R$, suppose $\lambda <\mu(\lambda,u)$. By density of $\mathbb R$, there exists $q\in\mathbb Q$ such that $\lambda<q<\mu(\lambda,u)$. Since $\lambda<q$, we have $\mu(\lambda,u)<\mu(q,u)=q$ from (4), a contradiction. By similar deduce, we claim that $\lambda=\mu(\lambda,u)$.

So all remains is to show (5), but it seems not trivial.

My another try is to consider the function \begin{align} f:\ \mathbb R&\longrightarrow\mathbb R \\ \lambda&\longmapsto f(\lambda)=T(\lambda u)\cdot v \end{align} for a fixed $v\in\mathbb R^n$. From additivity of $T$, we have $f(\lambda+\varepsilon)=f(\lambda)+f(\varepsilon),\ \forall\lambda,\varepsilon$. Since $f(\lambda)$ is a inner product with respect to $u$, I think we can someway show that $f(\lambda)$ is continuous, which will lead $f$ is linear. But this is not simple.

May anyone provide me some interesting hint ? Thanks

Answer 1

A $\mathbb Q$-linear bijection which sends $\mathbb R u$ (into and) onto $\mathbb R T(u)$ for $u \in \mathbb R^n$ in particular sends $\mathbb R u + v$ onto $\mathbb R T(u) + T(v)$ for $u, v \in \mathbb R^n$, i.e., affine lines are sent to affine lines. Such a function is therefore semi-linear assuming $n>1$, i.e., linear up to twisting by an automorphism of $\mathbb R$. But $\operatorname{Aut}(\mathbb R)$ is trivial, so $T$ is simply $\mathbb R$-linear.

Answer 2

Not really an answer but too long for a comment.

• First of all, assumption (4) it's not necessary since it follows from additivity i.e. $T(u+v)=T(u)+T(v)$ implies $T(q\cdot v)=q\cdot T(v)$ for all $q\in\mathbb{Q}$.
• Another remark, if $T$ is not injective then additivity alone doesn't suffice to prove $\mathbb{R}$-homogeneity. Indeed consider $T:\mathbb{R}\to\mathbb{R}$ additive but not $\mathbb{R}$-homogeneous and $\phi:\mathbb{R}^2\to\mathbb{R}^2$ defined as $\phi(x,y)=(0,T(y))$.

Now, let us suppose that for $T$ isomorphism (of additive groups $\mathbb{R}^n$) it holds that:

$u,v\in\mathbb{R}^n$ linearly independent implies $T(u),T(v)$ linearly independent.

Consider $T(\lambda(u+v))=T(\lambda u)+T(\lambda v)$ hence $$ \mu(\lambda,u+v)T(u+v)=\mu(\lambda,u)T(u)+\mu(\lambda,v)T(v)$$ and from this $$\mu(\lambda,u+v)T(u)+\mu(\lambda,u+v)T(v)=\mu(\lambda,u)T(u)+\mu(\lambda,v)T(v)$$ If $T(u),T(v)$ are linearly independent it follows that $\mu(\lambda,u+v)=\mu(\lambda,u)$ and $\mu(\lambda,u+v)=\mu(\lambda,v)$ i.e. $\mu(\lambda,u)=\mu(\lambda,v)$ for every $u,v$ l.i.

Since, for $\epsilon\neq0$, $u,v$ l.i. implies $u,\epsilon v$ l.i. it follows that $\mu(\lambda,u)=\mu(\lambda,v)=\mu(\lambda,\epsilon v)$.

The map $\mu(\cdot,\cdot)$ is a function of the first argument only.

Remark It remains to prove that for $T:\mathbb{R}^n\to\mathbb{R}^n$ injective and additive: $u,v$ l.i implies $T(u),T(v)$ l.i.