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

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!

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 $$\alpha$$.
• Thank you! This shows that $det(xI-L_a) =min(a,F)$.Now I'm having trouble with the general case: "For which $\alpha \in K$ do we have that $det(xI-L_{\alpha})= min(\alpha, F)$? " It would be great if you could help me with this. Thanks again! – user392559 Dec 21 '18 at 21:28
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.