Show that as groups, $\mathrm{GL}(V)\cong\mathrm{GL}_n(\mathbb{F})$, with the isomorphism given by $\theta \mapsto A_\theta$. The notation can be found in my attempt.
1st Attempt: If $\dim_\mathbb{F} V = n < \infty$, we can choose a basis $\mathbf{e}_1, \cdots, \mathbf{e}_n$ of $V$ over $\mathbb{F}$. So we can identify $V$ with $\mathbb{F}^n$. Then every endomorphism $\theta \in \mathrm{GL}(V)$ corresponds to a matrix $A_\theta = (a_{ij}) \in M_n(F)$ given by: $$\theta(\mathbf{e}_j) = \sum_i a_{ij} \mathbf{e}_i.$$ Based on the hints provided, I still can't figure out how do I rigorously show $\mathrm{GL}(V)\cong\mathrm{GL}_n(\mathbb{F})$?
2nd attempt: Consider the linear transformation $\theta$ in $\mathrm{GL}(V)$ such that $A_\theta$ is its matrix representation with respect to the chosen basis ${\mathbf{e}_1, \ldots, \mathbf{e}n}$. Assume that $\theta$ is in the kernel of the map, i.e., $A\theta$ is the identity matrix. Now, for any vector $\mathbf{v} \in V$, the action of $\theta$ on $\mathbf{v}$ corresponds to the multiplication by $A_\theta$. Since $A_\theta$ is the identity matrix, $\theta(\mathbf{v}) = \mathbf{v}$ for all $\mathbf{v} \in V$. This implies that $\theta$ is the identity transformation, and thus, the kernel of the map is trivial.
Given any invertible matrix $A \in \mathrm{GL}n(\mathbb{F})$, we want to find an invertible linear transformation $\theta$ such that $A\theta = A$. Consider the standard basis ${\mathbf{e}_1, \ldots, \mathbf{e}_n}$ in $\mathbb{F}^n$. Define the linear transformation $\theta$ by specifying its action on the basis vectors: $$\theta(\mathbf{e}_j)=\sum_{i=1}^na_{ij}\mathbf{e}_i$$ This defines a linear transformation $\theta: V \rightarrow V$ such that the matrix representation of $\theta$ with respect to the chosen basis is precisely $A$: $$A_\theta=\begin{bmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\a_{21}&a_{22}&\cdots&a_{2n}\\\vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\cdots&a_{nn}\end{bmatrix}=A$$
This shows that for any invertible matrix $A$, there exists an invertible linear transformation $\theta$ such that $A_\theta = A$. Hence, the map is surjective.
Since the map is injective and surjective, it is a bijective homomorphism, establishing the isomorphism between $\mathrm{GL}(V)$ and $\mathrm{GL}_n(\mathbb{F})$.
Is it correct and rigorous in any way? Have I made any mistakes or notation errors here?