any two sets of $n+2$ points are projectively equivalent in $\mathbb{P}^n$ Problem: Any two sets of $n+2$ points in general position in $\mathbb{P}^n$ are projectively equivalent.
In thinking about this problem, it is natural for me to reduce it to linear algebra considerations. So i have $n+2$ lines in $k^{n+1}$ spanned by the vectors $b_1,\dots,b_{n+2}$ and i want to map them with a linear transformation to another set of $n+2$ lines spanned by the vectors $c_1,\dots,c_{n+2}$. It is enough to map each vector $b_i$ to $c_i$. If we had $n+1$ lines (instead of $n+2$), then this is possible by linear algebra arguments. But i am failing to see how this would work for $n+2$ lines.
I would appreciate an answer which highlights why thinking in linear algebra terms does not work here.
 A: Let $(b_{i})_{i=1}^{n+2}$ be an ordered set of vectors in $k^{n+1}$ representing a set of points in general position in $\mathbf{P}^{n}$. Particularly, $(b_{i})_{i=1}^{n+1}$ is linearly independent, and $b_{n+2}$ cannot be expressed as a linear combination of any proper subset of $(b_{i})_{i=1}^{n+1}$. (Geometrically, $b_{n+2}$ does not lie in a linear space spanned by $n$ or fewer of the $(b_{i})_{i=1}^{n+1}$.)
Let $(e_{i})_{i=1}^{n+1}$, be the "standard basis vectors" in $k^{n+1}$, and $e_{n+2} = e_{1} + \dots + e_{n+1}$. To show two sets of $(n + 2)$ points in general position in $\mathbf{P}^{n}$ are projectively equivalent, it suffices to show $\bigl([b_{i}]\bigr)_{i=1}^{n+2}$ is projectively equivalent to $\bigl([e_{i}]\bigr)_{i=1}^{n+2}$.
The (unique, invertible) linear transformation $T:k^{n+1} \to k^{n+1}$ carrying $b_{i}$ to $e_{i}$ for $1 \leq i \leq n+1$ carries $b_{n+2}$ to some vector $b_{n+2}'$ with all coordinates non-zero. (Otherwise, $b_{n+2}$ would be a linear combination of some proper subset of $(b_{i})_{i=1}^{n+1}$. In other words, the $j$th coordinate hyperplane $H_{j} = \{x_{j} = 0\} \subset \mathbf{P}^{n}$ already contains $n$ points, namely the $[e_{i}]$ with $i \neq j$; since no projective hyperplane contains more than $n$ points $[T(b_{i})]$, we have $[b_{n+2}'] \not\in H_{j}$ for all $j$.)
Follow $T$ with the diagonal transformation $D$ whose $i$th diagonal entry is the reciprocal of the $i$th coordinate of $b_{n+2}'$, noting that $D$ fixes each coordinate axis (i.e., maps $[b_{i}]$ to $[e_{i}]$ for $1 \leq i \leq n + 1$)and sends $b_{n+2}'$ to $e_{n+2}$.
The fancy way to say this is, the diagonal (multiplicative) action of the "torus" $(k^{\times})^{n+1}$ is transitive on the complement of the coordinate hyperplanes in $k^{n+1}$, and fixes the coordinate axes (i.e., the points $[e_{i}]$ for $1 \leq i \leq n+1$).
A: Here is a similar argument (Lemma 19.3.), where points are in linear general position.
