Left-multiplication of a finite field element - Matrix representation Can someone explain me why and how a left multiplication of an element of a finite field GF(2^k) can be seen as a linear transformation on GF(2^k) over GF(2)? 
I read this https://www.maa.org/sites/default/files/Wardlaw47052.pdf but it is not clear to me.
Thank you! 

The field $\operatorname{GF}(2^k)$ is a finite dimensional vector space over $\operatorname{GF}(2)$ for $k$ any integer greater zero, so we can talk about linear transformations on this vector space. Let $\alpha \in \operatorname{GF}(2^k)$ and define the following map
$$
\begin{align}
T_\alpha: \operatorname{GF}(2^k) &\to \operatorname{GF}(2^k) \\ x & \mapsto \alpha x
\end{align}
$$
That is, $T_\alpha$ is a map which takes an element of $x \in \operatorname{GF}(2^k)$ and maps it to $\alpha x \in \operatorname{GF}(2^k)$. This is the multiplication map you are talking about. This is indeed a linear transformation since for $x, y \in \operatorname{GF}(2^k)$ and $c \in \operatorname{GF}(2)$ we have that 
$$
T_\alpha(x + y) = \alpha(x + y) = \alpha x + \alpha y = T_\alpha(x) + T_\alpha(y) 
$$
and 
$$
T_\alpha(cx) = \alpha(cx) = c(\alpha x) = cT_\alpha(x)
$$

Edit: As mentioned in the comment by lhf, all of the above argument still holds if the field extension $\operatorname{GF}(2^k)/\operatorname{GF}(2)$ is replaced by an arbitrary field extension $L/K$. It is not necessary that the field $K$ be finite, nor is it necessary that the degree of the extension be finite. This is readily seen because the fact that $L$ is a vector space over $K$, as well as linearity of $T_\alpha$ follows simply from the field axioms and the fact that $K$ is a subfield of $L$.
