Let $K$ be a field, $V$ a $n$-dimensional vector space over $K$. Let $\mathcal{B}$ be the set of all the bases of $V$, $G=GL(V)$ and $X:=\{\bar{x}=(x_1,\ldots,x_{n+1})\in V^{n+1} : (x_1,\ldots,x_{n})\in \mathcal{B}\}$.
I’d like to show that the group $G$ acts naturally over the set $X$, and that the $X\rightarrow K^n$ map that assigns to each $\bar{x}$ the coordinates of $x_{n+1}$ on the basis $(x_1,\ldots,x_{n})$ is constant over the orbits of $G$.
To be honest, I’m not totally sure I understand the statement of this problem. It’s part of some lecture notes I’m following (it’s not a graded assignment). I’ll try to describe what I understood so far.
For the first part, I’ve first observed that if $B\in \mathcal{B}$ is a basis of $V$ and $T\in GL(V)$, then $T.B=TB=B’$ will be another basis because $T$ is invertible. Similarly, the associative condition and existence of a neutral element conditions are satisfied, so I conclude $G$ acts naturally over $\mathcal{B}$. To see how it would act over $X$, I was guessing I could use the fact that $GL(n,K)$ is a subgroup of $GL(n+1,K)$, so $T\in GL(n,K)$. Then, I guess $T$ would act on $\bar{x}$ as $$T \bar{x}=((T\bar{x})_1,\ldots,(T\bar{x})_n, x_{n+1} )$$
and the vector formed by the first $n$ entries of this vector would be a basis of $V$, and hence $T\bar{x}\in X$. The rest of the properties of the group action would also hold.
But for the second part of the question, I have no idea how to make sense of what the problem is asking. What does “the coordinates of $x_{n+1}$” mean here? I thought $x_{n+1}$ was just an element of $K$, not a vector. There’s definitely something I’m missing here.
Note that the set $X$ consists of tuples of vectors, not tuples of elements of the field $K$. That is, $X\subset V^{n+1}$, where $V^{n+1}$ is a Cartesian product of $V$ with itself $n+1$ times, in the same way that the vector space $V$ can be associated with the set of $n$-tuples $K^n$ whose elements are from the field $K$.
Under this interpretation, the natural action of $G$ is as $$ T(\bar{x})=(Tx_1,Tx_2,\dots,Tx_n,Tx_{n+1}), $$ Since $T$ is invertible, and $\{x_1,\dots,x_n\}$ forms a basis for $V$, then so does $\{Tx_1,Tx_2,\dots,Tx_n\}$, which shows that $G$ at leasts maps $X$ to itself. Showing that this is a group action is straight-forward.
Finally, the map $f:X\to K^n$ assigns to $\bar{x}$ the coefficients of $x_{n+1}$ in the basis $\{x_1,x_2,\dots,x_n\}$, i.e., $$ f:\bar{x}\mapsto(\alpha_1,\dots\alpha_n), $$ where $$ x_{n+1}=\sum_{i=1}^n\alpha_ix_i. $$ To show that this function is constant over an orbit of $G$, it suffices to calculate the action of an element $T$ of $G$ on both sides of the equation above, i.e., $$ Tx_{n+1}=T\left(\sum_{i=1}^n\alpha_ix_i\right) =\sum_{i=1}^n\alpha_iTx_i. $$ One can fill in the details from there.