TITLE: Relating the trace and norm in Galois theory with the trace and determinant of a certain matrix
QUESTION [6 upvotes]: Suppose that $F$ is a finite dimensional extension field of $K$ (both subfields of $\mathbb{C}$). Given $u \in F$, let $g_u : F \to F$ be defined by $v \mapsto uv$. If $B$ is any ($K$-basis) for $F$ and $M = [g_u]_B$, then I would like to show that $\mbox{trace}(M) = \mbox{Tr}_{F/K}(u)$ and moreover $\mbox{det}(M) = \mbox{N}_{F/K}(u)$. I've been trying to prove that this at least holds in the case that $F = K(u)$ (where we would have a basis of the form $\{u^0, u^1, \dots u^n\}$) but haven't succeeded.
I am using the Hungerford notion of trace and norm. In particular, let $\sigma_1, \dots, \sigma_t$ be all the distinct $K$-monomorphisms $F \to \mathbb{C}$. If $a \in F$, then $N_{F/K}(a) = \sigma_1(a)\sigma_2(a) \dots \sigma_r(a)$ and $\mbox{Tr}_{F/K}(u) = \sigma_1(a) + \dots + \sigma_r(a)$.

REPLY [2 votes]: Allow me to reformulate your question. Let $F/K$ be an extension of fields of characteristic zero, of degree $n$. For $u\in F$, define $Tr_{F/K} (u)=$ the sum of the $n$ distinct conjugates of $u$ in an algebraic closure of $K$ (no need of $\mathbf C$), and similarly $N_{F/K} (u)=$ the product of these conjugates. One could also be interested in the other symmetric functions of the conjugates of $u$. Denote by $m_u$ the "multiplication by $u$", considered as a $K$-linear endomorphism of $F$ . Show that $Tr_{F/K} (u)$ and $N_{F/K} (u)$ are resp. equal to the trace $T(m_u)$ and determinant $D(m_u)$ (in the sense of linear algebra) of $m_u$.  
Answering a question of @Cyril L. (see https://math.stackexchange.com/a/2710883/300700), I recalled the following known results:
1° Suppose first that $u$ is a primitive element, i.e. $F=K(u)$. If $f_u=X^n + a_{n-1} X^{n-1} +...+ a_0\in K[X]$ is the minimal polynomial of $u$ over $K$,   then $F\cong K[X]/(f_u)$ admits a basis $(1, u,..., u^{n-1})$ over $F$, and the matrix of the endomorphism $m_u$  wrt. to this basis is the so called "companion matrix" with lines $(0, 0,..., 0, -a_0), (1, 0, ..., 0, -a_1), (0, 1,.., 0,..., -a_2),..., (0, 0,..., 1, -a_{n-1})$. It is then easily checked that the caracteristic polynomial $\chi_u$ of $m_u$ is det ($X.Id_E - m_u$) = $f_u$, and we are done. This is more or less the answer of @xyzzyz. Note that in this particular case, the symmetric functions of the conjugates of $u$ are given by the coefficients of the minimal polynomial $f_u$.
2° In the general case, we must find an adequate basis to compute the characteristic polynomial $\chi_u$ of $m_u$ in $F$. @Lord Shark the Unknown diagonalized the matrix of $m_u$ using Dedekind's lemma. Let us here perform more direct matrix computations (which will give additional informations, see  below). Putting $[F:K(u)]$, of degree $r$, let us show that the characteristic polynomial $\chi_u$ of $m_u$ is equal to the $r$-th power of the minimal polynomial $f_u$ of $u$ (this is stronger than a simple consequence of Cayley-Hamilton). Let $(y_i)$, $1\le i\le q$, be a basis of $K(u)$ over $K$, and $(z_j)$, $1\le j \le r$, be a basis of $F$ over $K(u)$. Then $(y_i z_j)$is a basis of $F$ over $K$, and $[F:K]:=n=qr$. Let $M=(b_{ih})$ be the matrix of $m_u$ in $K(u)$ with respect to the basis $(y_i)$, so that $ay_i = \sum b_{ih}y_h$. Then $a(y_i z_j)=(\sum b_{ih} y_h)z_j = \sum b_{ih}(y_h z_j)$. Ordering lexicographically the basis $(y_i z_j)$ of $F$ over $K$, we see that the matrix $M'$ of $m_u$ in $F$ with respect to this basis is a diagonal table of matrices  $M_1$ of the form $M$ on the diagonal and $0$ elsewhere. The matrix $X.Id - M_1$ is thus a diagonal table of matrices $X. Id_q -M$, hence det ($X. Id_n - M'$) = (det ($X.Id_q - M))^r$. But the LHS is the characteristic polynomial $\chi_u $, and the RHS is ${f_u}^r$ according to the special case 1°. This yields in particular $Det(m_u)={N_{F(u)/k}(u)}^r$ and $Tr(m_u)=r{Tr_{F(u)/k}(u)}$. But by the transitivity of the norm and trace maps in the tower of extensions $F/K(u)/K$ , we know that $N_{F/K}(u)= {N_{F(u)/k}(u)}^r$ and $Tr_{F/K} (u)=r{Tr_{F(u)/k}(u)}$, and we are done. NB: in this general situation, the symmetric functions of the conjugates of $u$ are given by the coefficients of ${f_u}^r$ .