Let $L$ be a field extension of $K$. Consider the set $\operatorname{End}_KL$ of all functions from $L$ to $L$ which are linear over $K$. A subset of $\operatorname{End}_KL$ is $\Gamma(L:K)$, the group (under composition) of all automorphisms on $L$ which fix $K$.
So, we can consider $\operatorname{End}_KL$ to be a vector space over $K$. We have addition of functions in the natural sense, and multiplication of functions by scalars, and all the other properties of a vector space.
I have been working for over a couple days on trying to show that $\Gamma(L:K)$ is linearly independent over $\operatorname{End}_KL$.
In the usual way, we let $\{\sigma_i\} \subset \Gamma(L:K)$ and $\{k_i\} \subset K$, both for $i = 1 \ldots n$ and suppose that $\sum_1^n k_i\sigma_i = 0$. This is an equation in $\operatorname{End}_KL$. Without loss of generality, suppose that $k_i \neq 0$ for each $i = 1 \ldots n$.
I have been wanting to find some contradiction, though unsuccessfully in the general case.
If $n=1$, $k_1 \sigma_1=0 \Rightarrow \sigma_1 = 0$ and this is an immediate contradiction as $\sigma_1$ is not surjective, and therefore not an automorphism.
If $n=2$, $k_1\sigma_1 + k_2\sigma_2=0$. Applying $\sigma_1^{-1}$ to both sides yields $k_1\tau + k_2 \sigma_1^{-1}\sigma_2 = 0$, where $\tau$ is the trivial automorphism. Solving for $\tau$, we have $\tau = \frac{-k_2}{k_1}\sigma_1^{-1}\sigma_2$. Now, evaluating both sides at $-\dfrac{k_2}{k_1}$ gives the equation $-\frac{k_2}{k_1}= (-\frac{k_2}{k_1})^2$. This must imply that $-\frac{k_2}{k_1} = 1$, since $k_2 \neq 0 $. Thus we have that $\sigma_1=\sigma_2$, a contradiction.
Again, any help with solving this in general would be much appreciated.
Thanks.