Let $K$ be an algebraic number field $K=\mathbb Q(\theta)$ with $\theta$ being a root of a minimal polynomial $f(x)$ of degree $n$ with rational coefficients. Let $\alpha$ be an arbitrary element of $K$. Is it always true that the degree of the minimal polynomial for $\alpha$ is $\le n$?
I suspect that the answer is "yes", and the reasoning I came up with involves companion matrices. Consider a $\mathbb Q$-basis of $K$: $1,\theta,\theta^2,\dots,\theta^{n-1}$. Every element $\alpha\in K$ acts on these basis elements by multiplication, with the result being some $\mathbb Q$-linear combinations of these basis elements. This gives rise to an $n\times n$ matrix $A$ with rational entries. We know from linear algebra that the characteristic polynomial $g$ of $A$ has property that $g(A)=O$ (zero matrix). Thus $g(\alpha)$ acts by multiplication on $K$ as the zero scalar. Probably we can conclude here that $g(\alpha)=0$ and hence the minimal polynomial of $\alpha$ divides $g$ and thus has degree $\le n$.
I am not 100% sure that this is a valid reasoning, as I've just started educating myself on the subject. Maybe someone could validate this argument or provide a slicker reasoning? Is something like that explicitly stated in the literature?
I think your argument is fine.
To conclude that $g(\alpha) = 0$, consider the action of $g(\alpha)$ on the element $1 \in K$. You know that $g(\alpha).1 = 0$, so $g(\alpha) = 0$.
By the way, the standard argument is to say that any $n + 1$ elements in $K$ are linearly dependent over $\mathbb Q$. In particular, the elements $1, \alpha, \dots, \alpha^n$ are linearly dependent over $\mathbb Q$. Hence there exists a polynomial $p(x)\in \mathbb Q[x]$ of degree at most $n$ such that $p(\alpha) = 0$. The minimal polynomial of $\alpha$ divides $p(x)$, so that the minimal polynomial has degree at most $n$.
Edit: In your proof, rather than considering the characteristic polynomial of the matrix $A$, you could have considered the minimal polynomial of the matrix $A$. This too has degree less than or equal to $n$. See if you can convince yourself that the minimal polynomial of the matrix $A$ is the same thing as the minimal polynomial of the element $\alpha$.