How would I have to go about proving that the minimal polynomial $p_\beta$ of a field extension $F\subseteq G$ coincides (modulo the sign) with the characteristic polynomial of the linear mapping $f:F(\beta)\rightarrow F(\beta), \ v \mapsto \beta v$ between the $F$-spaces $F(\beta)$ ?
My attempt was trying to express $f$ in a basis; and since $B:=(1,\beta,\ldots,\beta^{d-1})$, where $d$ is the degree of $p_\beta$, is a basis, this seemed natural; if we assume that $p_\beta=X^d-a_{d-1}X^{d-1}-\cdots-a_1 X-a_0, $ for some $a_i\in K$, and compute the characteristic polynomials $c_\beta $ of $f$ with the help of the above basis, we get $ (-1)^{2d}(-X)^d + (-1)^{2d}a_{d-1}(-X)^{d-1}+(-1)^{2d-1}\beta^{d-1}a_{d-2}(-X)^{d-2}+\cdots+(-1)^{d+2}\beta^2\cdots\beta^{d-1}a_1(-X)+(-1)^{d+1}\beta^1 \beta^2\cdots\beta^{d-1}a_0.$ Now this doesn't really look like the above $p_\beta$, so either I think I made a mistake (but I double checked) or somehow the $\beta$'s would have to vanish - but I don't know how/why they would do that. Any ideas please ?