Let $L/K$ be a separable extension of fields. Let $x \in L$ and let $[L : K(x)] = d$ and $[K(x) : K] = m$ so that if $a_1, \dots, a_d$ is a basis for $L/K(x)$ then:
$$ a_1, a_1 x, \dots, a_1 x^{m-1}, \\ a_2, a_2 x, \dots, a_2 x^{m-1}, \\ \vdots \\ a_d, a_d x, \dots, a_d x^{m-1} $$
a list of $m d$ numbers is a basis for $L/K$. Now let $T_x : L \to L$ be multiplication by $x$: $T_x(y) = xy$. Then the matrix for the vector space endomorphism over $K$, $T_x$ is a block-diagonal matrix with diagonal entries:
$$ \begin{pmatrix} 0 & 1 & 0 & 0 &\dots & 0 \\ 0 & 0 & 1 & 0 &\dots & 0 \\ \vdots \\ 0 & 0 &0 & 0 & \dots & 1 \\ -c_m & -c_{m-1} & -c_{m-2} & -c_{m-3} & \dots & -c_1 \end{pmatrix} $$
where $p_x(t) = t^n + c_1 t^{n-1} + \dots + c_m$ is the minimal polynomial for $x$ over $K(x)/K$.
I'm having trouble seeing this as the case, but I know that $T_x(a_i x^k) = a_i x^{k+1}$.
This is out of Neukirch's book, page 25 on Algebraic Number Theory.
I did some of the work: Let $y = \sum\limits_{i=1 \\ j=1}^{d,m} y_{ij} \alpha_{i} x^{j-1} \in L/K$ be a general element over $K$ and let $\overline{y} = (y_{1,1}, \dots, y_{1,m-1}; y_{2, 1}, \dots, y_{2,m-1}; \dots;y_{d, 1}, \dots, y_{d, m-1})$ show the mapping of vector coordinates to columns. Let $T = $ the given block matrix:
$$ T = \begin{pmatrix} P & 0 & \dots & 0 \\ 0 & P &\dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & P \\ \end{pmatrix} $$
Write the vector $\overline{y} = (\overline{y_1}, \cdots, \overline{y_d})$ be a vector of vectors. Then $T\overline{y} = (P\overline{y_1}, \dots, P\overline{y_d})$. And
$$ P \overline{y_i} = \sum_{j=1}^{m} y_{i,j} P(\alpha_i x^{j-1}) $$
where, $P(\alpha_i x^{j-1}) =$ ?
Not sure if you're still looking for an answer but a slightly easier approach is as follows. Using the basis you gave as $$ a_1,a_1x,...,a_1x^{m-1}$$ $$.$$ $$.$$ $$a_d,a_dx,...a_dx^{m-1}$$ we want a matrix to represent the action of $T_x$ on this basis. As an intuitive explanation, I will leave the precise details to you notice that, since $$xa_jx^{l} = a_jx^{l+1}$$ so for $l$ not equal to $m-1$ we just get a shift up by 1 on these bases hence the line of ones above the diagonal. If $l$ $=$ $m-1$ then we want to represent $a_jx^{m}$ in terms of the bases given before. But if $p_x(t) = t^m + c_1t^{n-1} + ... + c_m$ is the minimal polynomial of x then finally $$a_jx^{m} = a_j(-c_1x^{m-1} -c_2x^{n-2} -... -c_m).$$ This is the intuition of the answer. The details are simply converting this logic into a statement of matrices.