[proofplan]
We first verify that an invertible [linear map](/page/Linear%20Map) sends an ordered basis to an ordered basis, so the displayed formula is a well-defined action. Given two ordered bases, we then construct the unique linear map sending each vector of the first basis to the corresponding vector of the second by using coordinate expansion in the first basis. A direct inverse is obtained by reversing the roles of the two bases, so the constructed map lies in $GL(V)$. Finally, uniqueness follows because a linear map is determined by its values on a basis, giving exactly one group element carrying any ordered basis frame to any other.
[/proofplan]
[step:Verify that invertible linear maps act on ordered bases]
Let $\mathcal{B}=(v_1,\dots,v_n)\in \operatorname{Fr}(V)$ and let $T\in GL(V)$. Define the ordered $n$-tuple $T\mathcal{B}:=(T(v_1),\dots,T(v_n))$.
We first show that $T\mathcal{B}$ is a basis of $V$. Suppose $a_1,\dots,a_n\in k$ satisfy
\begin{align*}
\sum_{i=1}^{n} a_i T(v_i)=0.
\end{align*}
By linearity of $T$,
\begin{align*}
T\left(\sum_{i=1}^{n} a_i v_i\right)=0.
\end{align*}
Since $T$ is injective, this gives
\begin{align*}
\sum_{i=1}^{n} a_i v_i=0.
\end{align*}
Because $(v_1,\dots,v_n)$ is linearly independent, $a_i=0$ for every $i\in\{1,\dots,n\}$.
Next let $y\in V$. Since $T$ is surjective, there exists $x\in V$ such that $T(x)=y$. Since $\mathcal{B}$ spans $V$, there are scalars $b_1,\dots,b_n\in k$ such that
\begin{align*}
x=\sum_{i=1}^{n} b_i v_i.
\end{align*}
Applying $T$ and using linearity gives
\begin{align*}
y=T(x)=\sum_{i=1}^{n} b_i T(v_i).
\end{align*}
Thus $(T(v_1),\dots,T(v_n))$ spans $V$. It is therefore an ordered basis of $V$.
The identity map $\operatorname{id}_V:V\to V$ satisfies $\operatorname{id}_V\mathcal{B}=\mathcal{B}$. If $S,T\in GL(V)$, then for every $i\in\{1,\dots,n\}$,
\begin{align*}
(ST)(v_i)=S(T(v_i)).
\end{align*}
Hence $(ST)\mathcal{B}=S(T\mathcal{B})$. Therefore the displayed formula defines a left [group action](/page/Group%20Action) of $GL(V)$ on $\operatorname{Fr}(V)$.
[/step]
[step:Construct the linear map carrying one ordered basis to another]
Let $\mathcal{B}=(v_1,\dots,v_n)$ and $\mathcal{C}=(w_1,\dots,w_n)$ be elements of $\operatorname{Fr}(V)$. Define a map $T_{\mathcal{C}\leftarrow \mathcal{B}}:V\to V$ as follows. For each $x\in V$, let $a_1,\dots,a_n\in k$ be the unique scalars such that
\begin{align*}
x=\sum_{i=1}^{n} a_i v_i.
\end{align*}
Set
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(x)=\sum_{i=1}^{n} a_i w_i.
\end{align*}
The uniqueness of the coordinates of $x$ in the basis $\mathcal{B}$ makes this definition well-defined.
We verify linearity. Let $x,y\in V$ and let $\lambda\in k$. Write
\begin{align*}
x=\sum_{i=1}^{n} a_i v_i
\end{align*}
and
\begin{align*}
y=\sum_{i=1}^{n} b_i v_i
\end{align*}
with $a_i,b_i\in k$. Then
\begin{align*}
x+\lambda y=\sum_{i=1}^{n} (a_i+\lambda b_i)v_i.
\end{align*}
Using the definition of $T_{\mathcal{C}\leftarrow \mathcal{B}}$,
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(x+\lambda y)=\sum_{i=1}^{n} (a_i+\lambda b_i)w_i.
\end{align*}
Separating the sum gives
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(x+\lambda y)=\sum_{i=1}^{n} a_iw_i+\lambda\sum_{i=1}^{n} b_iw_i.
\end{align*}
Therefore
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(x+\lambda y)=T_{\mathcal{C}\leftarrow \mathcal{B}}(x)+\lambda T_{\mathcal{C}\leftarrow \mathcal{B}}(y).
\end{align*}
Thus $T_{\mathcal{C}\leftarrow \mathcal{B}}$ is $k$-linear. For each $j\in\{1,\dots,n\}$, the coordinate expansion of $v_j$ in $\mathcal{B}$ has coefficient $1$ at $j$ and coefficient $0$ elsewhere, so
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(v_j)=w_j.
\end{align*}
If $n=0$, then $V=\{0\}$ and both $\mathcal{B}$ and $\mathcal{C}$ are the empty ordered basis. The same definition gives the unique map $V\to V$, namely $\operatorname{id}_V$.
[guided]
The construction uses the defining feature of a basis: every vector has unique coordinates. Since $\mathcal{B}=(v_1,\dots,v_n)$ is a basis, each $x\in V$ has a unique coordinate expansion
\begin{align*}
x=\sum_{i=1}^{n} a_i v_i
\end{align*}
with scalars $a_1,\dots,a_n\in k$. We use those same coordinates, but reinterpret them in the target basis $\mathcal{C}=(w_1,\dots,w_n)$. Thus we define the map $T_{\mathcal{C}\leftarrow \mathcal{B}}:V\to V$ by
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(x)=\sum_{i=1}^{n} a_i w_i,
\end{align*}
where the scalars $a_i$ are determined by the preceding expansion of $x$ in the basis $\mathcal{B}$. This is well-defined because those scalars are unique.
Now we verify that this coordinate rule is linear. Let $x,y\in V$ and $\lambda\in k$. Write
\begin{align*}
x=\sum_{i=1}^{n} a_i v_i
\end{align*}
and
\begin{align*}
y=\sum_{i=1}^{n} b_i v_i.
\end{align*}
Then, by vector-space addition and scalar multiplication,
\begin{align*}
x+\lambda y=\sum_{i=1}^{n} (a_i+\lambda b_i)v_i.
\end{align*}
Applying the definition of $T_{\mathcal{C}\leftarrow \mathcal{B}}$ to this coordinate expansion gives
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(x+\lambda y)=\sum_{i=1}^{n} (a_i+\lambda b_i)w_i.
\end{align*}
Distributing in the [vector space](/page/Vector%20Space) $V$,
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(x+\lambda y)=\sum_{i=1}^{n} a_iw_i+\lambda\sum_{i=1}^{n} b_iw_i.
\end{align*}
The two sums on the right are precisely $T_{\mathcal{C}\leftarrow \mathcal{B}}(x)$ and $T_{\mathcal{C}\leftarrow \mathcal{B}}(y)$, so
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(x+\lambda y)=T_{\mathcal{C}\leftarrow \mathcal{B}}(x)+\lambda T_{\mathcal{C}\leftarrow \mathcal{B}}(y).
\end{align*}
Hence $T_{\mathcal{C}\leftarrow \mathcal{B}}$ is $k$-linear.
Finally, the construction sends each vector of $\mathcal{B}$ to the corresponding vector of $\mathcal{C}$. For $j\in\{1,\dots,n\}$, the vector $v_j$ has coordinate expansion
\begin{align*}
v_j=\sum_{i=1}^{n} \delta_{ij}v_i,
\end{align*}
where $\delta_{ij}\in k$ denotes the Kronecker delta, equal to $1$ when $i=j$ and equal to $0$ otherwise. Therefore
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}(v_j)=\sum_{i=1}^{n} \delta_{ij}w_i=w_j.
\end{align*}
If $n=0$, then $V=\{0\}$, the ordered basis is empty, and the empty coordinate sum defines the unique linear map $V\to V$, which is $\operatorname{id}_V$.
[/guided]
[/step]
[step:Show that the constructed map is invertible]
Define $S_{\mathcal{B}\leftarrow \mathcal{C}}:V\to V$ by the same construction with the roles of $\mathcal{B}$ and $\mathcal{C}$ reversed. Thus $S_{\mathcal{B}\leftarrow \mathcal{C}}$ is the linear map satisfying
\begin{align*}
S_{\mathcal{B}\leftarrow \mathcal{C}}(w_i)=v_i
\end{align*}
for every $i\in\{1,\dots,n\}$.
For each $i\in\{1,\dots,n\}$,
\begin{align*}
(S_{\mathcal{B}\leftarrow \mathcal{C}}\circ T_{\mathcal{C}\leftarrow \mathcal{B}})(v_i)=S_{\mathcal{B}\leftarrow \mathcal{C}}(w_i)=v_i.
\end{align*}
Let $x\in V$ and write $x=\sum_{i=1}^{n} a_i v_i$ with $a_i\in k$. By linearity,
\begin{align*}
(S_{\mathcal{B}\leftarrow \mathcal{C}}\circ T_{\mathcal{C}\leftarrow \mathcal{B}})(x)=\sum_{i=1}^{n} a_i(S_{\mathcal{B}\leftarrow \mathcal{C}}\circ T_{\mathcal{C}\leftarrow \mathcal{B}})(v_i).
\end{align*}
Hence
\begin{align*}
(S_{\mathcal{B}\leftarrow \mathcal{C}}\circ T_{\mathcal{C}\leftarrow \mathcal{B}})(x)=\sum_{i=1}^{n} a_i v_i=x.
\end{align*}
Therefore $S_{\mathcal{B}\leftarrow \mathcal{C}}\circ T_{\mathcal{C}\leftarrow \mathcal{B}}=\operatorname{id}_V$. The same argument with $\mathcal{B}$ and $\mathcal{C}$ interchanged gives
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}\circ S_{\mathcal{B}\leftarrow \mathcal{C}}=\operatorname{id}_V.
\end{align*}
Thus $T_{\mathcal{C}\leftarrow \mathcal{B}}$ is invertible, with inverse $S_{\mathcal{B}\leftarrow \mathcal{C}}$. Hence $T_{\mathcal{C}\leftarrow \mathcal{B}}\in GL(V)$, and
\begin{align*}
T_{\mathcal{C}\leftarrow \mathcal{B}}\mathcal{B}=\mathcal{C}.
\end{align*}
This proves transitivity of the action.
[/step]
[step:Prove uniqueness of the element sending one ordered basis to another]
Let $R\in GL(V)$ satisfy $R\mathcal{B}=\mathcal{C}$. Then
\begin{align*}
R(v_i)=w_i
\end{align*}
for every $i\in\{1,\dots,n\}$.
Let $x\in V$ and write $x=\sum_{i=1}^{n} a_i v_i$ with $a_i\in k$. By linearity of $R$,
\begin{align*}
R(x)=\sum_{i=1}^{n} a_i R(v_i).
\end{align*}
Using $R(v_i)=w_i$ for every $i$ gives
\begin{align*}
R(x)=\sum_{i=1}^{n} a_i w_i.
\end{align*}
By the definition of $T_{\mathcal{C}\leftarrow \mathcal{B}}$,
\begin{align*}
R(x)=T_{\mathcal{C}\leftarrow \mathcal{B}}(x).
\end{align*}
Since this holds for every $x\in V$, we have
\begin{align*}
R=T_{\mathcal{C}\leftarrow \mathcal{B}}.
\end{align*}
Thus there is at most one element of $GL(V)$ sending $\mathcal{B}$ to $\mathcal{C}$.
[/step]
[step:Conclude simple transitivity]
For every pair $\mathcal{B},\mathcal{C}\in \operatorname{Fr}(V)$, the map $T_{\mathcal{C}\leftarrow \mathcal{B}}\in GL(V)$ constructed above satisfies $T_{\mathcal{C}\leftarrow \mathcal{B}}\mathcal{B}=\mathcal{C}$, and the uniqueness argument shows that no other element of $GL(V)$ has this property. Therefore the action is transitive and free, equivalently simply transitive.
[/step]
