Let $\mathbb{Z}$ be the integers and $\widehat{\mathbb{Z}}$ its profinite completion. Suppose I have a group homomorphism
$$f: \mathbb{Z} \to GL_n(\mathbb{Q}_p).$$ I want to show that $f$ is continuous with respect to the profinite topology (i.e. the topology with fundamental open nbhds finite index subgroups) if the eigenvalues of $f(1),$ in an algebraic closure, are units.
I have a messy proof of this fact, using the operator norm of $GL_n(\mathbb{Q_p}).$ Is there a short conceptual proof of this fact?
Suppose that $K/\mathbf{Q}_p$ is a finite extension. Then $f$ is continuous if and only if the composite: $$g: \mathbf{Z} \rightarrow \mathrm{GL}_n(\mathbf{Q}_p) \rightarrow \mathrm{GL}_n(K)$$ is continuous. Since conjugation by an invertible matrix is also continuous, we may assume that the image $M$ of the generator of $\mathbf{Z}$ under $g$ is in Jordan canonical form. In particular, if the eigenvalues are units, then $M \in \Gamma:=\mathrm{GL}_n(\mathcal{O}_K)$ where $\mathcal{O}_K$ is the ring of integers of $K$. But $\Gamma$ is a pro-finite group (the inverse limit of the finite groups $\mathrm{GL}_n(\mathcal{O}_K/\pi^i_K)$), so any map from $\mathbf{Z}$ to this group automatically extends to a continuous map from $\widehat{\mathbf{Z}}$.