Calculating the Matrix of a Transformation Using Bases of Field Extensions I'm trying to understand this topic in my Abstract Algebra class:
Suppose that we have a finite field extension $L/F$ and let us choose $a \in L$.
We'll define the transformation $T_{a} : L \to L$ by $T_{a}(x) = ax$
and we wish to prove that $P(x) = m_{a}^{[L:F(a)]}(x)$ where P is the characteristic polynomial of $T_{a}$.
Here are my steps so far:
I define a basis for $L$ over $F(a)$ as $\{l_{j}\}_{j=1}^{k}$ for $[L:F(a)] = k$
Similarly $\{a_{i}\}_{i=1}^{m}$ defines a basis for $F(a)$ over $F$.
It can be shown that the set $B = \{l_{j}a_{i}\}_{i,j=1}$ is a basis for $L$ over $F$.
We now look for the matrix of $T_{a}$ with respect to $B$:
Write $a = \sum_{j=1}^{k}\sum_{i=1}^{m}f_{j,i}a_{i}l_{j}$ and $f_{j,i} \in F$
Im having a hard time understanding how to write the matrix since every element of the basis, say $l_{1}a{1}$, will be multiplied by $\sum_{j=1}^{k}\sum_{i=1}^{m}f_{j,i}a_{i}l_{j}$ under the transformation. How can I account for the coefficients from $F$ when 1 element in every column is squared in this representation?
 A: By ordering your basis of $L$ over $F$ correctly, you can make the matrix of $T_a$ a block diagonal matrix.
Recall that $k = [L:F(a)]$ and $\deg(m_a)=m.$
Let $V_i \subseteq L$ be defined by:
\begin{align*}
V_i &= \textrm{span}_F\{a_1l_i, a_2l_i,\dots,a_ml_i\}\\
&= l_i \textrm{ span}_F\{a_1,a_2,\dots,a_m\}.
\end{align*}
Then clearly $L = \oplus_{i=1}^k V_i$. I claim that each $V_i$ is $T_a$-invariant.
Let $x\in V_i$. Then $x= l_i(b_1a_1 + \cdots + b_ma_m)$. Notice that $aa_j \in F(a) = \textrm{ span}_F\{a_1,\dots,a_m\}$ for all $j = 1,\dots,m$.
Now,
\begin{align*}
T_a(x)&= ax\\
&= al_i(b_1a_1 + \cdots + b_ma_m)\\
&= l_i(b_1aa_1 + \cdots + b_maa_m) \in l_i \textrm{ span}\{a_1,\dots, a_m\} = V_i.
\end{align*}
So $V_i$ is $T_a$-invariant.
We have shown that the matrix of $T_a$ is block diagonal if you order the basis $a_1l_1, a_2l_1, \dots,a_ml_1, a_1l_2,a_2l_2,\dots$.
The minimal polynomial of a block diagonal matrix is the least common multiple of the minimal polynomials of its blocks.
Thus we have reduced the problem to computing the minimal polynomial of $T_a|_{V_i}$. I haven't worked through this, but I would think that it turns out to be $m_a(x)^k$.
Hope this helps. I'll add in this next part if I get around to thinking about it.
