How to understand elements of the scheme $GL_n$ as linear isomorphisms? Let as usual $GL_n$ be the scheme given by the equation $$det(\{x_{kl}\}_{1\leq k,l\leq n})y=1$$ in $\mathbb{A}^{{n^2}+1}$. I have seen that one considers elements of $GL_n$ likewise as linear isomorphisms $\mathbb{A}^n\to\mathbb{A}^n$ of schemes. I want to understand what this means formally and it would be nice if something like the following holds:
Perhaps the scheme $GL_n$ represents some functor of points $L:Sch^{op}\to Sets$ which in turn can easily be identified with linear isomorphisms $\mathbb{A}^n\to\mathbb{A}^n$. In other words, there is perhaps an isomorphism
$$
Hom_{Sch}(-,GL_n)\cong L(-)
$$
of functors from $Sch^{op}$ to $Sets$.
What could this functor $L$ be? Perhaps $L(V)$ is the set of linear isomorphisms $\mathbb{A}^n\times V\to\mathbb{A}^n\times V$ of schemes? This is a functor from $Sch^{op}$ to $Sets$ by pulling back along $\mathbb{A}^n\times f$ for a morphism $f:W\to V$ in $Sch$ but this is just a guess for $L$.
My question is:

Is there a way to understand $GL_n$ in these terms as the representing
  object of a certain functor $L(-)$ which can easily be
  identified with the linear isomorphisms $\mathbb{A}^n\to\mathbb{A}^n$?

I know that an $R$-point of $GL_n$ can be viewed as an invertible $(n\times n)$-matrix $\{a_{kl}\}_{1\leq k,l\leq n}$ with entries in $R$. This defines a linear morphism $\mathbb{A}^n\times V\to\mathbb{A}^n\times V$ where $V=\operatorname{Spec}(R)$ through an automomorphism on $R[x_1,\ldots,x_n]$ sending $x_k$ to $\sum_{l=0}^n a_{kl} x_l$ (I hope this is correct). The question is, how to express this fact precisely in the language above.
 A: This is standard and can be found in any book on algebraic groups. The scheme $\mathrm{GL}_n$ represents the contravariant functor $X \mapsto \mathrm{GL}_n(\Gamma(X,\mathcal{O}_X))$ (actually this is the functorial definition of $\mathrm{GL}_n$). For a commutative ring $R$, the group $\mathrm{GL}_n(R)$ is isomorphic to the group of linear automorphisms of $\mathbb{A}^n_R$.
If $X$ is an arbitrary scheme, then one has to be careful as for the definition of linear. In topology one usually demands this fiberwise, but this does not suffice here when $X$ is not reduced. A morphism $\mathbb{A}^n_X \to \mathbb{A}^m_X$ is called linear if for every $X$-scheme $T$ the induced map on $T$-valued points $\Gamma(T,\mathcal{O}_T)^n \to \Gamma(T,\mathcal{O}_T)^m$ is $\Gamma(T,\mathcal{O}_T)$-linear. Then it is induced by a matrix, and all these matrices have to be compatible. We see that the linear morphisms correspond to $M_{m \times n}(\Gamma(X,\mathcal{O}_X))$. In particular, the linear automorphisms of $\mathbb{A}^n_X$ correspond to $\mathrm{GL}_n(X)$.
