Matrix $A$ of multiplication with some $a$ in a field extension $K/F$ for some basis of $K$. Show that $a$ is a root of the characteristic polynomial.

Question

Let $A$ be the matrix of the linear map that is multiplication with some $a$. Show that $a$ is a root of the characteristic polynomial of $A$.

I've thought about this problem for some time, and feel like I'm missing something easy. I've worked out examples like $\mathbb{C} = \mathbb{R}(i)$, where for $a = p + qi$ the matrix $A$ (over the basis $\{1,i\}$) takes the form $$ \left (\begin{array}{cc} p & -q \\ q & p \\ \end{array} \right), $$ which has solutions $p + qi$ and $p - qi$. It works out in other field extensions like $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$, but I cant find the general solution.

Answer 1

If multiplication by $a$ is an $F$-linear map on $K$ then $a$ is an eigenvalue of this linear map with eigenvector $1\in K$. Hence $a$ is a root of the characteristic polynomial of this linear map.

Answer 2

It's better not to think about matrices in this context, but about linear maps. Let $\chi$ be the characteristic polynomial of the map $f:x\mapsto\alpha x$. Then we know that $\chi(f)$ is the zero map. At the same time, $\chi(f)(x)=\chi(\alpha)\cdot x$. And if $\chi(\alpha)\cdot x=0$ for all $x\in K$, then $\chi(\alpha)=0$.

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