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! 

Let's look at something you are hopefully more familiar with and that is the complex and real numbers. Every complex number can be written uniquely as $a + bi$ with $a, b \in \mathbf R$. This means that $\{1, i\}$ is linearly independent over $\mathbf R$ (it is a basis of $\mathbf C$ over $\mathbf R$). A linear transformation of $\mathbf C$ over $\mathbf R$ is a function $T : \mathbf C \to \mathbf C$ such that


*

*$T(w + z) = T(w) + T(z)$ for every $w, z \in \mathbf C$

*$T(az) = aT(z)$ for every $z \in \mathbf C$ and $a \in \mathbf R$


Every such transformation takes the form
$$ T(a + bi) = (ar + bs) + (at + bu)i, $$
for some real numbers $r,s,t,u$. We can write this equation in matrix form as
$$ \begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} =  \begin{pmatrix} ar + bs \\ at + bu \end{pmatrix}. $$
For instance, the complex conjugation map $T(a + bi) = a - bi$ corresponds to the matrix equation
$$ \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} =  \begin{pmatrix} a \\ -b \end{pmatrix}. $$
We can also consider the map $T(a + bi) = (c + di)(a + bi)$ (that is, $T$ multiplies on the left by $c + di$). Expanding this out we have
$$T(a + bi) = (c + di)(a + bi) = (ac - bd) + (ad + bc)i, $$
which corresponds to the matrix equation
$$ \begin{pmatrix} c & -d \\ d & c \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} =  \begin{pmatrix} ac - bd \\ ad + bc \end{pmatrix}. $$
For ${\rm GF}(2^k)$ over ${\rm GF}(2)$ it is the same idea but instead of $\{1, i\}$ as a basis you have some other basis.
