# Thread: Characteristic polynomial of left-multiplication transformation

1. ## Characteristic polynomial of left-multiplication transformation

Let $K$ be a field, $[K:\mathbb{Q}]=n$, and take $\alpha \in K$ such that $\alpha$ has degree $d$ over $\mathbb{Q}$. Let $f(x)$ be the monic irreducible polynomial over $\mathbb{Q}$ having $\alpha$ as a root.

Let $T_\alpha : K \rightarrow K$ be given by $k \mapsto \alpha k$. Show that the characteristic polynomial of $T_\alpha$ is $f(x)^{n/d}$.

2. Originally Posted by Bruno J.
Let $K$ be a field, $[K:\mathbb{Q}]=n$, and take $\alpha \in K$ such that $\alpha$ has degree $d$ over $\mathbb{Q}$. Let $f(x)$ be the monic irreducible polynomial over $\mathbb{Q}$ having $\alpha$ as a root.

Let $T_\alpha : K \rightarrow K$ be given by $k \mapsto \alpha k$. Show that the characteristic polynomial of $T_\alpha$ is $f(x)^{n/d}$.
from the definition of $T_{\alpha},$ it's easily seen that for any $g(x) \in \mathbb{Q}[x]$ and $k \in K$ we have $g(T_{\alpha})k=g(\alpha)k.$ thus $g(T_{\alpha})=0$ if and only if $g(\alpha)=0.$ in particular $T_{\alpha}$ and $\alpha$ have the same minimal

polynomials. so $f(x)$ is the minimal polynomial of $T_{\alpha}.$ now by, Cayley-Hamilton, the irreducible factors of the characteritsic polynomial and the minimal polynomial of a linear transformation are

the same. thus the caracteristic polynomial of $T_{\alpha}$ is $(f(x))^r,$ for some integer $r \geq 1.$ comparing the degrees gives us $r=\frac{n}{d}.$