Generalization of a field norm to the linear group

Question

I've been struggling with this question for some days now. Let $K$ be a field extension of $k$, and $x$ an invertible linear transformation of the $K$-vector space $V$. If we consider $V$ as a $k$-vector space and $x$ as $k$-linear then $$ det_kx = N(det_Kx) $$ where $det_Kx$ denotes the determinant of $K$ viewed as $K$-linear (same for $k$). This in fact generalizes the definition of field norm.

I have showed this is true for diagonal matrices, but not in general. In fact, I'm not sure this is the right equality, but I think so. I hope someone can give me a hint (or a reference :) I would be most thankful.

Greetings.

J.

P.S. In my case $K=\mathbb{F}_{q^2}$ and $k=\mathbb{F}_q$

Answer 1

This is very classical and standard exercise in linear algebra, and im sure there are many references. Still, there is something interesting to say about it.

The determinant of a matrix $A$ can be seen as the induced map $\wedge^n A : \wedge^n V \to \wedge^n V$. So, in some sense, a slight categorification of this formula is the natural isomorphism $\wedge^{[K:k]}((\wedge_K^n V)|_k) \cong \wedge^{[K:k] n}_k (V|_k)$, which once guessed is really easy to construct. The naturality of this isomorphism implies that applying the two functors written in both sides of this equation (with $V$ as an input) on an operator $A$ gives us the equality $det_k(det_K(A)|_k) = det_k(A|_k)$. So the problem is reduced to the statement on scalars, namely with $a = det_K(A)$ we have to show that $det_k(a|_k) = N_{K/k}(a)$ which is more or less the definition on the norm.