Connection between Galois trace and matrix trace

Question

$\newcommand{\Tr}{\operatorname{Tr}}$

I am having trouble seeing the connection between the two kinds of trace. For a finite extension $K$ of a field $F$ of degree $n$, with $\alpha \in K$, we defined the Galois trace as $$ \Tr(\alpha) = \sum_{\sigma \in \mathrm{Gal}(K/F)} \sigma(\alpha) $$ We also showed that $\Tr(\alpha) = -a_{n-1}$, where $m_{\alpha, F}(x) = \sum_{i=0}^na_ix^i$ is the minimal polynomial for $\alpha$ in $F[x]$.

We have also shown that the minimal polynomial for $\alpha$, $m_{\alpha, F}(x)$ and the minimal polynomial for left multiplication by $\alpha$, $T_\alpha(g) = \alpha g$ for $g \in K$, denoted by $m_{T_\alpha}$, are in fact the same polynomial.

My question is: how can I now see that these two traces are equivalent: one is the sum of the Galois conjugates, and one is the sum of the diagonal entries of the $n \times n$ matrix representation of $T_\alpha$.

Answer 1

Take the basis $\{1,\alpha,\alpha^{2},...,\alpha^{n-1}\}$ and then write the matrix of $T_{\alpha}$ relate to this basis: $$T=\begin{bmatrix}0 & 0 & 0 &... & 0 & -a_0\\ 1 & 0 & 0 & ... & 0 & -a_2\\ 0 & 1 & 0 &... & 0 & -a_2\\ \vdots & \vdots & \vdots &\vdots &\vdots &\vdots\\ 0 & 0 & 0 &... & 1 & -a_{n-1}\\ \end{bmatrix}$$

Wich is the rational form of the matrix with minimal polynomial $m_{\alpha,F}(x)$ So the trace of $T_{\alpha}$ is the trace of the matrix $T$, wich is $-a_{n-1}$

The minimal polynomial $m_{\alpha,F}(x)$ is

$$\prod_{\sigma \in \mathrm{Gal}(K/F)} (x- \sigma(\alpha))$$ with coefficient $-a_{n-1}=\displaystyle\sum_{\sigma \in \mathrm{Gal}(K/F)} \sigma(\alpha)$ So there is the conection.

Sorry i don´t have too much time now, i hope this help and can be understood, if not later i can explain some more :)

I explain some more here

$T(\alpha^{n-1})=\alpha^{n}$

So the last column of $T$ is the coordinates of $\alpha^{n}$ in the previous basis. But consider: $$a_0+a_1\alpha+...+a_n\alpha^{n}=0$$ Being $a_n=1$ because the miniaml is monic Then $$\alpha^{n}=-a_{n-1}\alpha^{n-1}-...-a_1\alpha-a_0$$

And now i hope you can follow the blanks :)