How to prove that $GL(V)$ is a Lie-group without a choice of a basis. If we have a general linear group $GL(V)$ of a vector space $V$ (group of automorphisms), we may choose a basis in $V$, then there is an isomorphism (non-canonical) $GL(V) \cong GL(n, \mathbb{F})$ and latter is a manifold of dimension $n \times n$.
Question: Is there a way to prove that $GL(V)$ is a Lie-group (or just manifold) without a choice of a basis on $V$ (and thus without refering to any matrices whatsoever)?
 A: No, this is not possible, at least not in any remotely natural way.  Giving a manifold structure to an object is very much like choosing a basis: you are defining charts to identify your object (locally) with a subset of $\mathbb{R}^n$.  These charts are rarely unique or canonical, just like a basis for a vector space.  So, no matter what, if you want to find a manifold structure, you should expect to have to make some arbitrary choices which are analogous to choosing a basis.  You could probably find some convoluted way to do this by making different sorts of arbitrary choices that are not exactly equivalent to a basis on $V$, but why would you want to do that?
If you wanted to be able to prove something like this without using anything like a basis, you would need to change your definition of a manifold to not specifically reference $\mathbb{R}^n$.  For instance, you could define a manifold as a space that is locally homeomorphic to an open subset of a finite-dimensional topological vector space.  Then you can make $\operatorname{End}(V)$ into a manifold by just enhancing its vector space structure to the unique topological vector space structure (which can be defined without a basis; for instance, you can define it as the finest topology which makes all linear maps $\mathbb{R}^n\to\operatorname{End}(V)$ continuous for all $n$).
