Relating the trace and norm in Galois theory with the trace and determinant of a certain matrix 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 $\mbox{Tr}_{F/K}(u) = \sigma_1(a) + \dots + \sigma_r(a)$.
 A: Putting together your embeddings $\sigma_1,\ldots,\sigma_r$ gives an embedding $\Sigma:F\to\Bbb C^r$. The $\sigma_j$ are linearly independent
over $\Bbb C$ (Dededkind's lemma). Therefore the image $\Sigma(F)$
of $F$ under $\Sigma$ spans $\Bbb C^r$ as a complex vector space.
Consider your map $g_u:F\to F$. This induces a map $g'_u:\Sigma(F)\
\to\Sigma(F)$ and then a complex linear map $G_u:\Bbb C^r\to\Bbb C^r$.
The traces of these maps are all the same; likewise the determinants
(it's basically just changing the basis of a vector space). But $G_u(z_1,\ldots,z_r)=(\sigma_1(u)z_1,\ldots,\sigma_r(u)z_r)$
so the matrix of $G_u$ is diagonal with entries $\sigma_1(u),\ldots,\sigma_r(u)$.
A: Here's an alternative argument: by primitive element theorem, we can assume that $F = K(a)$ for some $a \in F$. Let $n = [F : K]$ be the degree of the extension, so that $a$ is a root of an irreducible $f \in K[x]$ of degree $n$.
Then, since $a$ generates $F$ over $K$, different $K$-monomorphisms  $\sigma_i: F \to \mathbb{C}$ correspond to different choices of $n$ different roots of $f \in K[x]$ in $\mathbb{C}$ -- each $\sigma_i$ maps $a$ to a different root of $f$ in $\mathbb{C}$.
As $f$ splits over $\mathbb{C}$, we can write $f(x) = (x-\sigma_1(a))\cdots(x-\sigma_n(a)) \in \mathbb{C}[x]$. Multiplying this out, we get that $\mbox{Tr}_{F/K}(a) = \sigma_1(a) + \ldots + \sigma_n(a)$  is equal to minus coefficient at $x^{n-1}$ of the minimal polynomial of $a$ over $K$.
Now, we will connect this to the matrix above. Let $h$ be characteristic polynomial of $g_a: F \to F$, that is, $h(x) = \det(g_a - x \cdot id) = x^n - b_{n-1} x^{n-1} + \ldots + (-1)^n b_0$. By easy observation, trace of $g_a$ is equal to $b_{n-1}$. We claim that in fact $f = h$, that is, the characteristic polynomial of $g_a$ is indeed the minimal polynomial of $a$ over $K$. 
Indeed, this follows from Cayley-Hamilton theorem, which says that a linear operator satisfies its characteristic equation, that is, $h(g_a) = g_a^n - b_{n-1} g_a^{n-1} + ... + (-1)^n b_0 \cdot id$ is a zero operator. But that means that $h(g_a)(u) = 0$ for any $u \in F$, that is, 
$$
(g_a^n - b_{n-1} g_a^{n-1} + ... + (-1)^n b_0 \cdot id)(u) = g_a^n(u) - b_{n-1} g_a^{n-1}(u) + ... + (-1)^n b_0 \cdot u = 0
$$
which in turns, since $g_a^n(u) = a^n \cdot u$, gives us:
$$
a^n\cdot u - b_{n-1} a^n \cdot u + ... + (-1)^n b_0 \cdot u = 
(a^n - b_{n-1} a^n + ... + (-1)^n b_0) \cdot u = 0
$$
which is only true for all $u \in F$ if $a^n - b_{n-1} a^n + ... + (-1)^n b_0 = 0$, but then it means that $h(a) = 0$. Since $g_a: F \to F$ is a $K$-linear map, and we take characteristic polynomial of it as such, $h \in K[x]$, and $\deg h = \dim_K F = [F : K] = n$, $h$ must be the minimal polynomial of $a$ over $K$, which concludes the proof.
Similar argument also works for the norm, which is left as an exercise for reader.
A: 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 ${f_u}^r$ .
