The minimal polynomial is the determinant of $xI-L_{\alpha}$.

Question

Let $K=F(a)$ a finite field extension of $F$. For $\alpha \in K$, let $L_{\alpha} : K \to K$ be the transformation $L_{\alpha} (x)=\alpha x$. Show that $L_{\alpha} $ is an $F$-linear transformation and show that $det(xI-L_a) =min(a,F)$. For which $\alpha \in K$ do we have that $det(xI-L_{\alpha})= min(\alpha, F)$?

Here, $min(\alpha, F)$ denotes the minimal polynomial of $\alpha$ in $F$, that is, the polynomial with minimal degree with coefficients in $F$ that has $\alpha$ as a root.

It's clear that $L_{\alpha}$ is a linear function. Now, I don't know how to manage the rest of the problem. I know that the basis of $K$ as an $F$-linear space is $\{1,a,...,a^{n-1} \}$, where $n$ is the degree of $min(a,F)$. But then I don't know what to do.

Any help will be very appreciated. Thank you so much!

Answer 1

Let $P=det(xI-L_\alpha)$ Cayley Hamilton implies that $P(L_\alpha)=0$ this implies that $P(L_\alpha)(1)=P(\alpha)=0$. Since $deg P=[F(\alpha):F]$ we deduce that $P$ is the minimal polynomial of $\alpha$.

Answer 2

Let $p(x) = \det(xI - L_\alpha)$. $p(x)$ is a monic polynomial of degree $n$, and by the Cayley-Hamilton theorem we have $p(L_\alpha) = 0$. However, if there were a polynomial $q$ of degree $d$ strictly smaller than $n$ such that $q(\alpha) = 0$, then we would conclude that the elements $\{1, \alpha, \dots, \alpha^{d}\}$ fail to be linearly independent.