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?


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, $\chi(a) = \det 0 = 0$.

  • $\begingroup$ I got it, Thank you very much! $\endgroup$ – nicksohn Jan 2 '16 at 18:40
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    $\begingroup$ 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

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