Southwest rank matrices determine Bruhat cells

Left invariance: for $b\in B$ and any $x\in M_n({\mathbb{F}}_q)$ the submatrix of $bx$ on the rows $i,\dots ,n$ and columns $1,\dots ,j$ equals the product of the invertible upper triangular block $\beta :=b[i,\dots ,n;i,\dots ,n]$ with the submatrix $x[i,\dots ,n;1,\dots ,j]$. Indeed, for $k\ge i$ the entry $b_{km}$ vanishes whenever $m<i\le k$, by [L2], so only the columns $m\ge i$ of $b$ contribute, and the product formula of [L5] applies blockwise. The block $\beta$ is invertible: since $b$ is invertible and upper triangular, all its diagonal entries are nonzero, and the trailing block $\beta$ is upper triangular with those same nonzero diagonal entries, so triangular back substitution gives $N(\beta )=\{0\}$ and [L6] applies. Left multiplication by an invertible matrix does not change the row space, since the rows of $YX$ are linear combinations of the rows of $X$ while the rows of $X$ are those of $Y^{-1}(YX)$; hence $\mathrm{Row}(bx[i\mathrm{.}\mathrm{.}n,\mathrm{1..}j])=\mathrm{Row}(x[i\mathrm{.}\mathrm{.}n,\mathrm{1..}j])$ and $r_{i,j}(bx)=r_{i,j}(x)$.

Right invariance: for $b\in B$ and any $x$ the submatrix of $xb$ on the rows $i,\dots ,n$ and columns $1,\dots ,j$ equals the product of $x[i,\dots ,n;1,\dots ,j]$ with the invertible upper triangular block $\gamma :=b[1,\dots ,j;1,\dots ,j]$: since $b_{ml}=0$ whenever $m>l$, the sum ${\sum }_m{x}_{km}{b}_{ml}$ of [L5] runs over the indices $m\le l\le j$ only. The block $\gamma$ is invertible, because a nonzero kernel vector $v$ with $\gamma v=0$ gives $b\left(\genfrac{}{}{0px}{}{v}{0}\right)=0$ by [L2] and hence $v=0$ by [L6]. Right multiplication by an invertible matrix does not change the column space: $\mathrm{Col}(X\gamma )=\{X\gamma w:w\in F^j\}=X(F^j)=\mathrm{Col}(X)$ because $\gamma$ is bijective. By [L4] the rank equals the column rank, so $r_{i,j}(xb)=r_{i,j}(x)$.

Permutation matrices: the submatrix of $P_{\sigma }$ on the rows $i,\dots ,n$ and columns $1,\dots ,j$ has, in column $k\le j$, the single nonzero entry $1$ in row $\sigma (k)$ when $\sigma (k)\ge i$, and is the zero column otherwise, by [L7]. Its nonzero columns are the distinct standard basis vectors $e_{\sigma (k)}$ of the coordinate space on the rows $i,\dots ,n$, one for each $k\le j$ with $\sigma (k)\ge i$; they form a linearly independent spanning set of the column space, hence a basis, so by [L3] and [L8] the column rank, and therefore by [L4] the rank of this submatrix, equals the number of such columns, that is $r_{i,j}(P_{\sigma })=\mathrm{\#}\{k\le j:\sigma (k)\ge i\}$.

Let $g\in BwB$ with $w=P_{\sigma }$. Writing $g=b_{1}w{b}_2$ with $b_1,b_2\in B$ and applying step 1.1 to $b_1$ and step 1.2 to $b_2$, then step 1.3 to $w$, gives $r_{i,j}(g)=r_{i,j}(w)=\mathrm{\#}\{k\le j:\sigma (k)\ge i\}$ for all $i,j$. In particular, subtracting consecutive columns of the rank matrix, $r_{i,j}(g)-r_{i,j-1}(g)$ equals $1$ exactly when $\sigma (j)\ge i$, where $r_{i,0}(g):=0$; hence the set of $i$ with $r_{i,j}(g)>r_{i,j-1}(g)$ is $\{1,\dots ,\sigma (j)\}$ and $\sigma (j)=\max \{i\le n:r_{i,j}(g)>r_{i,j-1}(g)\}$ for every $j$. Thus the rank matrix of $g$ determines $\sigma$. By [L9] every element of $G$ admits such a factorisation, so two elements of $G$ lie in the same double coset $BwB$ exactly when their southwest rank matrices agree. ∎