I've been learning about Galois theory and in a lot of proofs, an automorphism $\sigma\in Gal(L/K)$ is extended to $\hat{\sigma}\in Gal(M/K)$, when $K\subseteq L\subseteq M$ and $M/K$ and $L/K$ are both Galois extensions.
I don't have a great intuition for this. On the one hand, since $L$ is normal, it seems like $\sigma\in Gal(L/K)$ will permute roots of polynomials that split in $L$, and so we can extend $\sigma$ to do anything we want to elements in $M$ outside of $L$ and it won't cause any inconsistenciees with $\sigma$. On the other hand, if $\alpha\in M$ is a root of some polynomial $p(x)\in L[x]$, then $\sigma(p(x))\neq p(x)$, so $\sigma(\alpha)$ must change in some non-trivial way.
The only proof I could come up with is to argue that we can embed $M/L$ into $M_n(L)$ for some $n$ by expressing $M$ as an $n$-dimensional $L$-vector space and mapping each $\alpha$ to a linear operator $L_\alpha$ as a matrix. Then any $\sigma\in Gal(L/K)$ can be applied element-wise to these matrices, and this will by a ring homomorphism. Since the image of $M$ under this map will (a) contain a subfield isomorphic to $L$ (the diagonal matrices) and (b) be isomorphic to $M$, we can pick any isomorphism that fixes $L$ and map back to $M$. This forms an automorphism of $M$, and the action on $L$ will be the same as $\sigma$.
This seems like a very roundabout way to prove the result. Is there a simpler proof?