I am trying to prove the existence of a bijection between the set of ordered bases for $F^n$, where $F$ is an arbitrary field, and $\text{GL}_n (F)$, the group of invertible $n \times n$ matrices over $F$. My attempted proof is below :
Denote by $\mathcal{B}$ the set of bases of $\mathbb{F}^n$. Define $\varphi: \mathcal{B} \to \text{GL}_n (\mathbb{F})$ sending an ordered basis $(v_1, \ldots, v_n)$ to the matrix $A = \begin{pmatrix} v_1 & v_2 & \ldots & v_n \end{pmatrix}$ whose $i$th column is $v_i$ for each $i$. As $(v_1, \ldots, v_n)$ form a basis for $\mathbb{F}^n$, they are linearly independent, so the matrix $A$ is invertible and hence an element of $\text{GL}_n (\mathbb{F})$. If $\varphi(B_1) = \varphi(B_2)$ for $B_1, B_2 \in \mathcal{B}$, then each column of the resulting matrix is equal, hence the ordered basis $B_1$ and $B_2$ are equal, so $\varphi$ is injective. Furthermore, let $A \in \text{GL}_n (\mathbb{F})$. As $A$ is invertible, its columns are linearly independent and hence forms a basis for $\mathbb{R}^n$. Forming these columns into an ordered basis then gives a preimage for $A$, so $\varphi$ is surjective and hence bijective.
The biggest problem I have with this proof is that I don't know for sure what bases of $F^n$ I'm using for the domain and codomain of the transformation. I want to view a matrix as an encoding of a linear transformation, rather than a rectangular array of numbers, but the exact bases aren't clear. I can't say that the basis is $(v_1, \ldots, v_n)$, because in that case, I'd have $T(v_i) = v_i$, which would give me the identity matrix. Since I'm giving a recipe for the $i$th column, it seems I'm letting $T(e_i) = v_i$ (where $e _i$ is the $i$th standard basis vector), and I'm writing both the domain and codomain relative to the standard basis. It seems that the definition of $\text{GL}_n (F)$ does not take the choice of bases into account, and views any "rectangular array" as the same matrix.
Is this a correct understanding of what's going on?