Show that characteristic polynomial of $Ty = ay$ is power of minimal polynomial of $a$

Question

Let $K/F$ be a finite, separable, algebraic field extension and let $T: K\to K, Ty = ay$. Show that $p = m^n$ where $p$ is $T$'s characteristic polynomial and $m$ is $a$'s minimal polynomial.

$m$ divides $p$ since $p(T) = 0$ and $p(T)(1) = p(a)$, so $a$ is a root of $p$. However I can't show that other irreducible polynomials don't divide $p$.

Answer 1

Let $(y_1,\dots,y_n)$ be a basis of $K$ as an $F(a)$-vector space.

For each $i=1,2,\dots,n,$ the restriction of $T$ to the $F$-subspace $K_i$ spanned by $(y_ia^j)_{0\le j<\deg(m)}$ has $m$ as its characteristic polynomial. Since $K=\oplus_{i=1}^nK_i,$ $p=m^n.$