[proofplan]
We use the ordered basis $\mathcal{B}$ to convert vectors in $V$ into coordinate columns in $k^n$. Under this coordinate map, a linear automorphism $T:V\to V$ becomes an invertible [linear map](/page/Linear%20Map) of $k^n$, hence an invertible matrix. Composition of linear maps becomes multiplication of their coordinate matrices, and the inverse construction sends an invertible matrix $A$ back to the automorphism $c_{\mathcal{B}}^{-1}\circ L_A\circ c_{\mathcal{B}}$ of $V$.
[/proofplan]
[step:Define the coordinate map determined by the ordered basis]
Define the coordinate map $c_{\mathcal{B}}:V\to k^n$ by declaring that, for the unique scalars $a_1,\dots,a_n\in k$ satisfying
\begin{align*}
x=\sum_{i=1}^{n}a_i v_i,
\end{align*}
one has
\begin{align*}
c_{\mathcal{B}}(x)=(a_1,\dots,a_n)^\top.
\end{align*}
Because $\mathcal{B}$ is a basis, every $x\in V$ has a unique such expression. Therefore $c_{\mathcal{B}}$ is a well-defined $k$-linear bijection. Let $c_{\mathcal{B}}^{-1}:k^n\to V$ denote its inverse map.
For each matrix $A\in M_n(k)$, let $L_A:k^n\to k^n$ denote left multiplication by $A$, so $L_A(u)=Au$ for every column vector $u\in k^n$. By definition of the matrix of a linear map with respect to $\mathcal{B}$, $[T]_{\mathcal{B}}$ is the unique matrix satisfying
\begin{align*}
L_{[T]_{\mathcal{B}}}=c_{\mathcal{B}}\circ T\circ c_{\mathcal{B}}^{-1}.
\end{align*}
[/step]
[step:Show that $\Phi_{\mathcal{B}}$ is well-defined]
Let $T\in GL(V)$. Then $T:V\to V$ is a $k$-linear bijection, and its inverse $T^{-1}:V\to V$ is also $k$-linear. The composition
\begin{align*}
c_{\mathcal{B}}\circ T\circ c_{\mathcal{B}}^{-1}:k^n\to k^n
\end{align*}
is therefore a $k$-linear bijection. Hence its representing matrix $[T]_{\mathcal{B}}$ is invertible, with inverse matrix $[T^{-1}]_{\mathcal{B}}$. Thus $[T]_{\mathcal{B}}\in GL_n(k)$, so $\Phi_{\mathcal{B}}$ is well-defined as a map from $GL(V)$ to $GL_n(k)$.
[/step]
[step:Verify that composition becomes matrix multiplication]
Let $S,T\in GL(V)$. By the definition of coordinate matrices,
\begin{align*}
L_{[S\circ T]_{\mathcal{B}}}=c_{\mathcal{B}}\circ S\circ T\circ c_{\mathcal{B}}^{-1}.
\end{align*}
Insert the identity map $\operatorname{id}_V=c_{\mathcal{B}}^{-1}\circ c_{\mathcal{B}}$ between $S$ and $T$ to obtain
\begin{align*}
c_{\mathcal{B}}\circ S\circ T\circ c_{\mathcal{B}}^{-1}=(c_{\mathcal{B}}\circ S\circ c_{\mathcal{B}}^{-1})\circ(c_{\mathcal{B}}\circ T\circ c_{\mathcal{B}}^{-1}).
\end{align*}
Using the defining identities for $[S]_{\mathcal{B}}$ and $[T]_{\mathcal{B}}$, this becomes
\begin{align*}
L_{[S\circ T]_{\mathcal{B}}}=L_{[S]_{\mathcal{B}}}\circ L_{[T]_{\mathcal{B}}}=L_{[S]_{\mathcal{B}}[T]_{\mathcal{B}}}.
\end{align*}
The assignment $A\mapsto L_A$ from $M_n(k)$ to the space of $k$-linear maps $k^n\to k^n$ is injective, since a matrix is determined by its values on the standard basis vectors of $k^n$. Therefore
\begin{align*}
[S\circ T]_{\mathcal{B}}=[S]_{\mathcal{B}}[T]_{\mathcal{B}}.
\end{align*}
Thus $\Phi_{\mathcal{B}}(S\circ T)=\Phi_{\mathcal{B}}(S)\Phi_{\mathcal{B}}(T)$, so $\Phi_{\mathcal{B}}$ is a [group homomorphism](/page/Group%20Homomorphism).
[/step]
[step:Prove injectivity from equality of coordinate matrices]
Let $S,T\in GL(V)$ and suppose $\Phi_{\mathcal{B}}(S)=\Phi_{\mathcal{B}}(T)$. Then
\begin{align*}
L_{[S]_{\mathcal{B}}}=L_{[T]_{\mathcal{B}}}.
\end{align*}
By the defining identity for coordinate matrices,
\begin{align*}
c_{\mathcal{B}}\circ S\circ c_{\mathcal{B}}^{-1}=c_{\mathcal{B}}\circ T\circ c_{\mathcal{B}}^{-1}.
\end{align*}
Composing on the left with $c_{\mathcal{B}}^{-1}$ and on the right with $c_{\mathcal{B}}$ gives $S=T$. Hence $\Phi_{\mathcal{B}}$ is injective.
[/step]
[step:Construct a preimage for every invertible matrix]
Let $A\in GL_n(k)$. Define $T_A:V\to V$ by
\begin{align*}
T_A=c_{\mathcal{B}}^{-1}\circ L_A\circ c_{\mathcal{B}}.
\end{align*}
Since $c_{\mathcal{B}}$, $c_{\mathcal{B}}^{-1}$, and $L_A$ are $k$-linear, the map $T_A$ is $k$-linear. Since $A\in GL_n(k)$, the matrix $A^{-1}$ exists, and $L_{A^{-1}}$ is the inverse of $L_A$. Therefore
\begin{align*}
T_A^{-1}=c_{\mathcal{B}}^{-1}\circ L_{A^{-1}}\circ c_{\mathcal{B}}.
\end{align*}
Thus $T_A\in GL(V)$. Finally,
\begin{align*}
c_{\mathcal{B}}\circ T_A\circ c_{\mathcal{B}}^{-1}=L_A,
\end{align*}
so $[T_A]_{\mathcal{B}}=A$. Hence $\Phi_{\mathcal{B}}(T_A)=A$, proving that $\Phi_{\mathcal{B}}$ is surjective.
[guided]
We now prove surjectivity by reversing the construction of a coordinate matrix. Start with an arbitrary matrix $A\in GL_n(k)$. The symbol $L_A:k^n\to k^n$ denotes the $k$-linear map $u\mapsto Au$ on coordinate columns. To turn this coordinate-space map into a map on $V$, we first send a vector of $V$ to its coordinate column, apply $A$, and then convert the resulting coordinate column back into a vector of $V$. This defines
\begin{align*}
T_A=c_{\mathcal{B}}^{-1}\circ L_A\circ c_{\mathcal{B}}.
\end{align*}
Each map in this composition is $k$-linear: $c_{\mathcal{B}}$ and $c_{\mathcal{B}}^{-1}$ are linear because they are the coordinate isomorphism and its inverse, and $L_A$ is linear because matrix multiplication distributes over vector addition and scalar multiplication. Hence $T_A:V\to V$ is $k$-linear.
We must also check that $T_A$ is invertible, because the domain of $\Phi_{\mathcal{B}}$ is $GL(V)$, not the set of all linear endomorphisms of $V$. Since $A\in GL_n(k)$, the inverse matrix $A^{-1}\in GL_n(k)$ exists. The corresponding coordinate map $L_{A^{-1}}:k^n\to k^n$ is the inverse of $L_A$, because for every $u\in k^n$,
\begin{align*}
L_{A^{-1}}(L_A(u))=A^{-1}Au=u
\end{align*}
and
\begin{align*}
L_A(L_{A^{-1}}(u))=AA^{-1}u=u.
\end{align*}
Therefore the inverse of $T_A$ is obtained by replacing $L_A$ with $L_{A^{-1}}$:
\begin{align*}
T_A^{-1}=c_{\mathcal{B}}^{-1}\circ L_{A^{-1}}\circ c_{\mathcal{B}}.
\end{align*}
This proves $T_A\in GL(V)$.
It remains to verify that $T_A$ maps to the original matrix $A$. By the definition of coordinate matrix, $[T_A]_{\mathcal{B}}$ is the unique matrix whose left-multiplication map equals $c_{\mathcal{B}}\circ T_A\circ c_{\mathcal{B}}^{-1}$. Substituting the definition of $T_A$ gives
\begin{align*}
c_{\mathcal{B}}\circ T_A\circ c_{\mathcal{B}}^{-1}=c_{\mathcal{B}}\circ c_{\mathcal{B}}^{-1}\circ L_A\circ c_{\mathcal{B}}\circ c_{\mathcal{B}}^{-1}=L_A.
\end{align*}
Thus the coordinate matrix of $T_A$ is exactly $A$, so $\Phi_{\mathcal{B}}(T_A)=A$. Since $A\in GL_n(k)$ was arbitrary, $\Phi_{\mathcal{B}}$ is surjective.
[/guided]
[/step]
[step:Conclude that the coordinate map is a group isomorphism]
The map $\Phi_{\mathcal{B}}:GL(V)\to GL_n(k)$ is a group homomorphism, injective, and surjective. Therefore $\Phi_{\mathcal{B}}$ is an isomorphism of groups. This proves the theorem.
[/step]
