Let $\mathbb{F}$ be a finite field with $q$ elements and $\mathbb{E}$ an extension field of degree $n$ of $\mathbb{F}$. Let $S:\mathbb{F}^n\rightarrow \mathbb{F}^n$ be a affine transformation and define the following:
Definition: Let $0\leq i < n$ and $A,B_i \in \mathbb{E}$. Then we call the polynomial $S(X)=\sum_{i=0}^{n-1}B_i X^{{q}^i}+A$ the univariate representation of the affine transformation $S(X)$.
Definition: Let $M_S \in \mathbb{F}^{n\times n}$ be a matrix and $v_s \in \mathbb{F}^n$ a vector and let $S(x):=M_Sx+v_s$. We call this a matrix representation of the affine transformation $S$ .
Definition 3 Let $\mathbb{E}$ be an $n^{th}$ degree extension of the ground field $\mathbb{F}$ and $\mathbb{F}^n$ the corresponding vector space. Then we call $\phi:\mathbb{E}\rightarrow \mathbb{F}^n$ with $\phi(a):=b$ and $b_i=a_{i-1}$.
How I will able to demonstrate that an affine transfromation in univariate representation can be efficiently transfered into matrix representation?
In the book say that the rows of $M$ are $\phi(P(\phi^{−1}(e_k)))$ where $\phi$ is a canonical map between $\mathbb{E}$ and $\mathbb{F}^n$ and $e_k$ is a canonical basis of $\mathbb{F}^n$. Then, the expression is: $\phi(P(X))=M\phi(X)+v$.
I'm trying verify that expression for every row, then first I want calculate $P(\phi^{−1}(e_k))$, using the basis $V=\{t, t^q, t^{q^2},\ldots,t^{q^{n-1}}\}$ for $\mathbb{E}$ (Then any element of $\mathbb{E}$ I will be able to write than $a_0t^{q^0}+a_1t^{q^1}+\cdots+a_{n-1}t^{q^n-1}$) $$P(\phi^{−1}(e_k)) = \sum_{i=0}^{n-1} B_i(\phi^{−1}(e_k))^{q^i}$$ $$ = \sum_{i=0}^{n-1} B_i(t^{q^{k-1}})^{q^i}$$ $$\sum_{i=0}^{n-1} \left(\sum_{j=0}^{n-1} b_{i,j} \right) t^{q^{i}+q^{k-1+j}}$$
How I will be able to express that expression using the basis $V$ to secondly apply $\phi$ on the result of $P(\phi^{−1}(e_k))$.?
An affine transformation is a linear map plus a translation. Since $\mathbb{E}$ is an extension of degree n, we can write an $e \in \mathbb{E}$ as $e = e_0 + e_1 t + \ldots e_{n-1} t^{n-1}$, with $e_i \in \mathbb{F}$. So as a vector space, we also have that $\mathbb{E} \cong \mathbb{F}^n$. In particular, $A \in \mathbb{E}$ has a form $A = a_0 + a_1 t + \ldots + a_{n-1} t^{n-1}$. The set $\{1,t,\ldots,t^{n-1}\}$ forms a basis for $\mathbb{E} \cong \mathbb{F}^n$ and we can think of $A$ as just a point in the vector space. In this way, addition by $A$ and addition by $v_s\, (\, = A)$ is the translation.
The trickier bit is noticing that the map
$$X \mapsto \sum_{i=0}^{n-1} B_i X^{q^i}$$
is a linear transformation on $\mathbb{F}^n$, and that any linear transformation on that space can be expressed this way.
To see that the sum is linear, simply try it out. Let $f,g \in \mathbb{E}$ and $\lambda \in \mathbb{F}$. Then
$$ (f + \lambda g) \mapsto B_i (f+\lambda g)^{q^i} = B_i f^{q^i} + \lambda B_i g^{q^i}$$
This is because exponentiation by $q$ is a power of the Frobenius automorphism for a finite field, so $(a+b)^q = a^q + b^q$. Moreover, the base field is fixed by exponentiation by $q$. The sum of linear maps is still linear, so the whole thing is a linear map. Linear algebra says we can always represent linear maps by a matrix with respect to some basis.
Now, to get the other way.. Chose $t \in \mathbb{E}$ such that $\{t, t^q, t^{q^2},\ldots,t^{q^{n-1}}\}$ is a basis for $\mathbb{E} \cong \mathbb{F}^n$, and take the matrix $M = (m_{i,j})$ with respect to that basis. Then the matrix/vector product $Mt^{q^i}$ on each basis vector will be
$$\sum_{i=0}^{n-1} \left(\sum_{j=0}^{n-1} m_{i,j} \right) t^{q^i}$$
Take $B_i = \sum_{j=0}^{n-1} m_{i,j}$. Note that each $m_{i,j}\in \mathbb{E}$, so that $B_i$ is in $\mathbb{E}$ as well.