My current question asks me to prove the following:
Let $K/F$ be a finite field extension. Let $\phi: K \rightarrow K$ be $F$-linear.
Prove the following statement:
If $K = F(\alpha_1, ...,\alpha_n)$, then $\phi$ is determined by its action on $\{\alpha_1,..., \alpha_n\}$.
An earlier part of the problem involves proving that if $p(x) \in F[x]$ has a root $\alpha \in K$, then $\phi(\alpha)$ is also a root of $p(x)$, which is much easier.
I know that from linear algebra, the linear transformation $\phi$ is uniquely determined by its action on the basis, but the basis in this case is a set of products of the $\alpha_i$, not $\{\alpha_1,...,\alpha_n\}$ .
Furthermore, we cannot assert that $\phi$ is multiplicative (i.e. $\phi(ab) = \phi(a)\phi(b)$) since left-multiplication by scalars does not satisfy this property, and yet is a linear transformation.
Can anybody please offer me a solution or at least a hint? I would sincerely appreciate some help.