Position of complete flags $E$ is a vector space of dimension $n \geq 2$. $\mathbb{F}=(F_1,F_2,\dots,F_n)$ and $\mathbb{G}=(G_1,G_2,\dots,G_n)$ are two complete flags of $E$.
We say that $(\mathbb{F},\mathbb{G})$ is in position $\sigma \in \mathfrak{S}_n$ if and only if there exists a basis $(e_1,\dots,e_n)$ of  $E$ such that for each $(i,j) \in [1,n]^2$, the sub-space $F_i$ is generated by $\{e_1,\dots,e_i\}$ and the sub-space $G_j$ by $\{e_{\sigma(1)},\dots,e_{\sigma(j)}\}$.
Suppose that $(\mathbb{F},\mathbb{G})$ is an ordered pair of complete flags in position $\sigma \in \mathfrak{S}_n$, that $k$ is an integer $1 \leq k \leq n-1$ and that $\sigma^{-1}(k) < \sigma^{-1}(k+1)$. $\mathbb{F^\prime}=(F^\prime_1,F^\prime_2,\dots,F^\prime_n)$ is another complete flag differing from $\mathbb{F}$ only for the sub-space of index $k$, i.e. $F_i=F^\prime_i$ for $i \in [1,n] \setminus \{k\}$ and $F_k \neq F^\prime_k$.
How can we prove that $(\mathbb{F^\prime},\mathbb{G})$ is in position $\tau_k \circ \sigma$ where $\tau_k$ is the transposition which swaps $k$ and $k+1$?
I processed the case $n=2$, $k=1$... but I'm not able to extrapolate to the general case.
 A: Another way to characterize relative position is by looking at intersections: it is equivalent asking that $\dim(F_i\cap G_j)=\#\big([1,i]\cap\sigma([1,j])\big)$.  How does this number change when we replace $\sigma$ by $\tau_k\sigma$?  The RHS stays the same unless $i=k$, and in that case it drops if $k\in \sigma([1,j])$ and $k+1\notin \sigma([1,j])$. (It would increase if $k+1\in \sigma([1,j])$ and $k\notin \sigma([1,j])$ but this can't happen by assumption).  
So, we just have to see that the behavior of the LHS is the same.  If $i\neq k$, then $F_i=F_i'$, so that's good.  On the other hand, $F_k\neq F_k'$, and in fact $F_k+F_k'=F_{k+1}$ and $F_k\cap F_k'=F_{k-1}$.  Thus, 
$\dim(F_k\cap G_j)+\dim(F_k'\cap G_j)=\dim(F_{k-1}\cap G_j)+\dim(F_{k+1}\cap G_j),\qquad (*)$  
since $(F_k\cap G_j)\cap(F_k'\cap G_j)=F_{k-1}\cap G_j$ and $(F_k\cap G_j)+(F_k'\cap G_j)=F_{k+1}\cap G_j$.  Let $m=\dim(F_{k-1}\cap G_j)$; in this case $m\leq \dim(F_{k+1}\cap G_j)\leq m+2$.
This naturally divides us into 3 cases:


*

*$k,k+1\notin \sigma([1,j])$: in this case, $m=\dim(F_{k+1}\cap G_j)$.  Necessarily, $\dim(F_k\cap G_j)=\dim(F_k'\cap G_j)=m$.

*$k,k+1\in \sigma([1,j])$: in this case, $m+2=\dim(F_{k+1}\cap G_j)$.  This is only possible if $\dim(F_k\cap G_j)=\dim(F_k'\cap G_j)=m+1$.

*$k\in \sigma([1,j])$ and $k+1\notin \sigma([1,j])$: in this case  $\dim(F_{k+1}\cap G_j)=m+1$; by assumption, we must have $\dim(F_{k+1}\cap G_j)=\dim(F_{k}\cap G_j)$, so $\dim(F_{k}'\cap G_j)=\dim(F_{k-1}\cap G_j)$, and the dimension has dropped, as desired.


EDIT: I should mention that there's also an abstract way of thinking about this.  If we let $B$ be the group of matrices preserving $G_j$ for all $j$.  We may as well assume, $G_j$ is the span of the first $j$ coordinates, so these are upper triangular matrices.  In this case  $F_\bullet$ has relative position $\sigma$ if and only if $\sigma b(F_j)=G_j$, there $b\in B$ and $\sigma$ stands for the permutation matrix.  Now, consider $G'_\bullet =\sigma b(F_\bullet')$.  This is a flag that agrees with $G_\bullet$, except that $G_k\neq G_k'$.  You can easily check that $\tau_kb'(G_j')=G_j$ for some $b'\in B$.  Thus $\tau_kb'\sigma b(F_j')=G_j$.  It's a standard calculation that $B\tau_kB\sigma B=B\tau_k\sigma B$, so for some $b''\in B$, we have $\tau_k\sigma b''(F_j')=G_j$, and we're done.
