Connection between Galois trace and matrix trace $\newcommand{\Tr}{\operatorname{Tr}}$
I am having trouble seeing the connection between the two kinds of trace. For a finite extension $K$ of a field $\mathbb F$ 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 $m_{\alpha, F}(x) = \displaystyle\sum_{i=0}^na_ix^i$ 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 $T_\alpha$. 
 A: Take the basis $\{1,\alpha,\alpha^{2},\ldots,\alpha^{n-1}\}$ and then write the matrix of 
$T_{\alpha}$ relate to this basis:
$$T=\begin{bmatrix}0 & 0 & 0 &\ldots& 0 & -a_0\\ 
1 & 0 & 0 & \ldots& 0 & -a_2\\
0 & 1 & 0 &\ldots& 0 & -a_2\\
\vdots & \vdots & \vdots &\ddots&\vdots &\vdots\\
0 & 0 & 0 &\ldots& 1 & -a_{n-1}\\
\end{bmatrix}$$
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 helps 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:
$$a_0+a_1\alpha+\ldots+a_n\alpha^{n}=0$$
Being $a_n=1$ because the miniaml is monic
Then
$$\alpha^{n}=-a_{n-1}\alpha^{n-1}-\ldots-a_1\alpha-a_0$$
And now I hope you can follow the blanks :)
