Finding the norm map on number fields Let $K$ be a number field and $L$ be a finite extension of $K$ of degree $n$. I want to define a norm map on $L$. So that, for some $\alpha \in L$, I want to find the matrix $\mu_{\alpha}$, the map of the transformation $x \mapsto \alpha x$  for  for any $x\in K$.
For example let $L = \mathbb{Q}(\sqrt{2})$ and $K=\mathbb{Q}$. $L$ has basis $(1,\sqrt{2})$. Let $\alpha = a +b\sqrt{2}$.  If we check for the bais elements,we see that $1 \mapsto a + b\sqrt{2}$ and $\sqrt{2} \mapsto 2b+a\sqrt{2}$. Thus, I assume that the transformation matrix for $\mu_{\alpha}$ is given by:
$
   M=
  \left[ {\begin{array}{cc}
   a & 2b \\
   b & a\\
  \end{array} } \right]
$
However, $1 \mapsto a+b\sqrt{2}$ but we have $[1,0]  \left[ {\begin{array}{cc}
   a & 2b \\
   b & a\\
  \end{array} } \right]  = [a , 2b] = a + 2b\sqrt{2} \neq a+b\sqrt{2} .$
Where is the problem and how can we find the desired transformation matrix?
 A: You should multiply the matrix with the coordinate vector from the right side. Indeed you would have:
$$\left[ {\begin{array}{cc}
   a & 2b \\
   b & a\\
  \end{array} } \right]\left[ {\begin{array}{c}
   1  \\
   0 \\
  \end{array} } \right]=[a \enspace b]$$
On the other side if you want to keep left multiplication the transformation matrix should be:
$$M = \left[ {\begin{array}{cc}
   a & b \\
   2b & a\\
  \end{array} } \right]$$
In other words, instead of writing the values for the basis elements in the columns you are doing it in a row.

Let's determine the transformation matrix under left multiplication. We know that it's of the form: $M= \left[ {\begin{array}{cc}
   x & y \\
   z & t\\
  \end{array} } \right]$, where $x,y,z,t$ are yet to be determined.
Now we know that $[1,0]M = [a,b]$ by the definition of the function. Now just perform the multiplication on the left-hand side and comapre coefficients to get: $x=a$ and $y=b$.
Similarly from $[0,1]M = [2b,a]$ we get that $z=2b$ and $t=a$. Therefore the transformation matrix becomes $M = \left[ {\begin{array}{cc}
   a & b \\
   2b & a\\
  \end{array} } \right]$
