# Let $K = F(a)$ be a finite extension of $F$. For $\alpha \in K$, let $L_\alpha$ be the map from $K$ to $K$ defined by $L_\alpha(x)=\alpha x$.

Let $K = F(a)$ be a finite extension of $F$. For $\alpha \in K$, let $L_\alpha$ be the map from $K$ to $K$ defined by $L_\alpha(x)=\alpha x$. Show that $L_\alpha$ is an $F$-linear transformation. Also show that det$(xI — L_\alpha)$ is the minimal polynomial min$(F, a)$ of $a$. For which $\alpha \in K$ is det$(xI — L_\alpha)$= min$(F, \alpha)$?

If $x,y \in K$ and $r\in F$ then $L_\alpha(x+ry)=\alpha (x+ry)=\alpha x+\alpha ry=\alpha x+r\alpha y=L_\alpha (x)+rL_\alpha(y)$ so $L_\alpha$ is an $F$-linear transformation. How can I express $L_\alpha$ as an matrix to be able to compute det$(xI — L_\alpha)$? Thank you very much for your help.

• (1). Use the powers of $\alpha$ as a basis over $F$ and remember the Cayley-Hamilton theorem. (2) The subject line of your post is not well thought out. – KCd Sep 3 '17 at 22:51

Write $P(x)=det(xI-L_{\alpha})$. Cayley Hamilton implies that $P(L_\alpha)=0$. This implies that $P(L_\alpha)(1)=P(\alpha)=0$. The degree of $P=[K:F]$, since $K=F(\alpha)$, we deduce that $P$ is the minimal polynomial of $\alpha$.