Field Trace/Norm and Matrix Trace/Norm (Dummit and Foote 14.2.31(c)). I can't quite figure out this final part to 14.2.31 in Dummit and Foote, 3rd edition. I'm given $K/F$ is a finite field extension of degree $n$, and $\alpha\in K$. I've shown that the map $T_\alpha:K\rightarrow K$ given by $k\mapsto \alpha k$  is $F$-linear, and that the minimal polynomial $$m(X)=X^d+a_{d-1}X^{d-1}+\cdots+a_1X+a_0$$ of $\alpha$ over $F$ is the same as the minimal polynomial of the transformation above. 
In previous exercises, I showed that given these conditions, the field norm $$N_{K/F}(\alpha)=\prod_\sigma\sigma(\alpha)=(-1)^n a_0^{n/d}$$ and the field trace $$Tr_{K/F}(\alpha)=\sum_\sigma\sigma(\alpha)=\dfrac{n}{d}a_{d-1}$$
where $\sigma$ ranges over a set of coset representatives for the subgroup $H<Gal(L/F)$ corresponding to $K$ (where $L/K$ is Galois).
I need to show that $Tr(\alpha)$ is the same as the trace of the matrix representation of $T_\alpha$, and that $N(\alpha)$ is the same as its determinant. I'm not sure at all how to proceed, however - I'm having trouble picturing the matrix representation of $T_\alpha$.
 A: The trace is somewhat easier than the determinant, but the idea is sort of the same. I will show you how to do the trace. You can use the fact the field trace behaves well with multiple extensions, that is(I will use the symbol $S$ for the field trace) $$S_{K/F} = S_{F(\alpha)/F} \circ S_{K/F(\alpha)}.$$
This makes things a lot easier. Evaluating the above function in $\alpha$ gives $$S_{K/F}(\alpha) = S_{F(\alpha)/F}([K:F(\alpha)]\alpha) = [K: F(\alpha)]S_{F(\alpha)/F}(\alpha),$$
since $\alpha \in F(\alpha)$ and the number of cosets of the corresponding subgroup is $[K: F(\alpha)]$. But we know $S_{F(\alpha)/F}(\alpha)=-a_{d-1}$, because every coset representative of the subgroup belonging to $F(\alpha)$ sends $\alpha$ to another root of the minimal polynomial of $\alpha$, and there are precisely $\deg(m)$ roots to this polynomial in $L$, if we assume such galois extension $L/F$ exists (which I think is needed for the norm and trace to be well-defined). So if we label these coset representatives by $\tau$, we have $$m(X) = \prod_{\tau} (X - \tau(\alpha)),$$ and by expanding we see why $S_{F(\alpha)/F}(\alpha)=-a_{d-1}$. So $S_{K/F}(\alpha) = -[K: F(\alpha)]a_{d-1}$. It remains to show the trace of $T_\alpha$ is equal to this expression. We do this by choosing a $F$-basis for $K$. Note that since the Galois extension $L/F$ exists, we have that $K/F(\alpha)$ is separable and by the primitive element theorem there is some $\beta$ such that $K = F(\alpha, \beta)$. Now choose the following basis: 
$$\{\alpha^i \beta^j | i \in \{0, .., [F(\alpha): F] -1] \}, j \in \{0, .., [F(\alpha, \beta): F(\alpha)]-1\}\}$$
Now we look at the matrix representation of $T_\alpha$ with respect to this basis. We see that $T_\alpha(\alpha^i \beta^j) = \alpha^{i+1} \beta^j$. If $i< [F(\alpha): F]-1$, the diagonal entry in the column belong to basis element $\alpha^i \beta^j$ is $0$, because in this case $T_\alpha$ outputs another basis element. However, if $i = d-1$ then $$T_\alpha(\alpha^i \beta^j) =\alpha^d \beta^j = -\sum_{k = 0 }^{d-1}a_k\alpha^k\beta^j,$$
now we read off the diagonal entry in the column of $\alpha^i \beta^j$, that is, the coefficient of $\alpha^{d-1} \beta^j$ in the above expression and we vind this is equal to $-a_{d-1}$. There are $[K: F(\alpha)]$ basis elements $\alpha^i \beta^j$ for which $i = d-1$, so if we all sum them we get the desired result: $$\text{Tr}(T_\alpha) = -a_{d-1}[K:F(\alpha)]=S_{K/F}(\alpha).$$
Note: I have only learned the definition of norm and trace of a field extension if the extension is separable. I assumed the existence of a Galois extension $L/F$ that contains all the other fields in the discussion above, like I think you did as well in your question, and then there are no problems. 
