Vector subspace of $\mathbb{C}$ spanned by $n$ complex numbers over the field of rational numbers.

Question

Let $\alpha_{1} \dots \alpha_{n}$ be complex numbers and V = $\{ \sum_{i=1}^{n} a_{i}\alpha_{i} : a_{i} \in \mathbb{Q} \}$ be the vector subspace of $\mathbb{C}$ spanned by them over the field of rational numbers $\mathbb{Q}$. Let $\beta$ be a complex number such that $\beta$ be a complex number such that $\beta V \subset V$. Show that $\beta$ is the root of a degree $n$ polynomial with rational coefficients.

I think I understand the gist of this question, but not sure how to formally prove it. If we treat multiplication by $\beta$ as the linear transformation $T_{\beta}: V \to V$, where $V$ is has dimension $n$, then we can use Cayley-Hamilton theorem to show that $P(T_{\beta}) = 0$. My question is how do we know that $P$ has rational coefficients, and how can we show that $P(\beta) = 0$?

Answer 1

This does't need anything as non-trivial as Cayley-Hamilton. Say $\alpha\in V$, $\alpha\ne0$. Since $V$ has finite dimension (as a vector space over $\Bbb Q$) the vectors $\beta\alpha,\beta^2\alpha,\dots\in V$ cannot all be ($\Bbb Q$-)independent. So some linear combination (with rational coefficients) vanishes.

Answer 2

The fact that $P$ has rational coefficients follows from the fact that the coefficients of the characteristic polynomial are in $\mathbb Q$.

For showing $P(\beta)=0$, hint: You may want to assume $V \neq \{0\}$.