proof that a continuous additive homomorphism $\mathbb{R}^n\to\mathbb{R}^m$ is $\mathbb{R}$-linear How can we prove that a continuous additive homomorphism $ \Phi \colon \mathbb{R}^{n}\to \mathbb{R}^{m} $ is $\mathbb{R}$-linear. i.e. satisfies $ \Phi (rv)=r \Phi (v)$ for $ r\in \mathbb{R} $ and $ v \in \mathbb{R}^{n} $?
 A: HINT. Use induction and additivity to show that it is $\mathbb{N}$-linear; then use that to show that it is $\mathbb{Z}$-linear, and from there to show that it is $\mathbb{Q}$-linear.
Once you know it is $\mathbb{Q}$-linear, use continuity to deduce it is $\mathbb{R}$-linear.
A: Let $V=\mathbb R^n$ and let $f:V\to V$ be a continuous homomorphism.
Let $\{e_1,\dots,e_n\}$ be a basis of $V$ over $\mathbb R$ and let $g:V\to V$ be the unique $\mathbb R$-linear map such that $g(e_i)=f(e_i)$ for all $i\in\{1,\dots,n\}$, which is of course continuous.
The difference $h=f-g$ is also continuous, so its kernel $K=\ker h$ is therefore a closed subgroup of $V$ which contains all the $e_i$. If $\frac pq\in\mathbb Q$ with $p$, $q$ integral, then $q\cdot h(\frac pqe_i)=h(pe_i)=ph(e_i)=0$ for all $i$, so in fact $h(\frac pqe_i)=0$ because $V$ has no torsion: we thus see that $H$ contains the subgroup $\mathbb Q e_1+\mathbb Q e_2+\cdots+\mathbb Q e_n$. Since $H$ is closed, this implies that $H=V$.
It follows from this that in fact $f=g$.
