1
$\begingroup$

Let $F$ be a field and $K/F$ be a finite extension.

For any $x \in K$, there is a minimal polynomial for $x$.

On the other hand, the multiplication by $x$ induces a $F$-linear map $K \to K$. This map has matrix representation, say $A$.

I want to show that $x$ is in fact root of a characteristic polynomial of $A$.

Is this true? And if so, how can I approach this?

$\endgroup$
3
$\begingroup$

Let $a \in K$ and $L_a : K \to K$ be the multiplication map $L_a(x) = ax$. The characteristic polynomial induced by $a$ is $\chi(x) = \det (x I - L_a) \in K[x]$. Notice that

$(a I - L_a)(x) = ax - ax = 0$,

so that $a I - L_a$ is the zero endomorphism of $K$. Hence, $\chi(a) = \det 0 = 0$.

$\endgroup$
  • $\begingroup$ I got it, Thank you very much! $\endgroup$ – nicksohn Jan 2 '16 at 18:40
  • 1
    $\begingroup$ The relation between these polynomials is the following: if $p_a(x) \in F[x]$ is the minimal polynomial for $a \in K$, then $\chi_a(x) = p_a(x)^{[K : F(a)]}$. $\endgroup$ – Eduardo Longa Jan 2 '16 at 18:41
  • $\begingroup$ @EduardoLonga, where can we find a proof that $\chi_a(x)=p_a(x)^{[K:F(a)]}$? $\endgroup$ – rmdmc89 Mar 10 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.