Dimension of fixed vectors of a semi-linear operator

Question

Let $L$ be a field with a field embedding $\sigma:L \rightarrow L$, and $K=L^{\sigma}$ be the fixed field of $\sigma$. For $A \in M_n(L)$ a matrix, consider the set $X=\{x \in L^n|Ax=\sigma(x) \}$ which is a $K$-vector space where the action of $\sigma$ on $L^n$ is just applying $\sigma$ on each coordinate.

Do we always have $\dim_{K}X \leq n$ (at least in the case $L/K$ is algebraic)? When is $\dim_{K}X = n$?

The case $n=1$ is clear. If $a \not =0 $, then if $ax_1=\sigma(x_1)$ and $ax_2=\sigma(x_2)$ for non-zero $x_1, x_2$, we get $\sigma(x_1/x_2)=x_1/x_2$ hence $x_1 \in Kx_2$ so the dimension is no bigger than $1$. And the dimension is $1$ iff $a=\sigma(x)/x$ for some no-zero $x \in L$. If $L/K$ is finite cyclic, then this is equivalent to the norm $Na=1$.

The case $char L=p>0$ and $\sigma=(x \rightarrow x^p)$, the dimension is $n$ when $L$ is large enough (separably closed) and $A \in GL_n(L)$, which can be proved using Jacobian criterion for smooth morphisms. This is a classical lemma used in the theory of $p$-adic Galois representations.

How about the case $L=\mathbb C$ and $\sigma$ is the complex conjugation?

Answer 1

This is only a partial answer that shows $\dim_K X \leq n$ for $A \in GL_n(L)$.

It suffices to show that if $x_1, \dots, x_m \in X$ are $K$-linearly independent, then they are $L$-linearly independent. If they weren't, we could assume (up to reordering the $x_i$) that there exists a relation $\lambda_1 x_1+\dots+\lambda_l x_l = 0$ such that all the $\lambda_i$ are non-zero, $\lambda_1 = 1$ and that there exists no such relation involving fewer than $l$ of the $x_i$.

Applying $\sigma$ yields $\sigma(\lambda_1)\sigma(x_1)+\dots+\sigma(\lambda_l)\sigma(x_l) = 0$ and hence $A(\sigma(\lambda_1)x_1+\dots+\sigma(\lambda_l)x_l) = 0$. As $A$ is invertible, it follows that $\sigma(\lambda_1)x_1+\dots+\sigma(\lambda_l)x_l = 0$. Subtracting this relation from the first one contradicts the minimality of $l$ (since $1 = \lambda_1 = \sigma(\lambda_1)$) unless $\lambda_i = \sigma(\lambda_i)$ for all $i$. But then all $\lambda_i$ are in $K$, contradicting the $K$-linear independence of $x_1, \dots, x_m$.