Let $G = \mathrm{GL}_{n}(k)$ for some field $k$. We have $G = \coprod_{w\in W}BwB$, where $B \leq G$ is a Borel subgroup (group of upper triangular matrices) and $W$ is a set of permutation matrices (or more generally, Weyl group of $G$).
I'm trying to prove this myself, without using any hard theory of algebraic groups. I proved that the double cosets are disjoint in a purely combinatorial way. I proved that $Bw_{1} \cap w_{2}B = \emptyset$ for any $w_1\neq w_2$, by showing that for any $b_1, b_2\in B$, there exists $i, j$ where $(i, j)$-th entry of $b_1 w_1$ is nonzero but the same entry $w_{2}b_{2}$ is zero. However, I can't show that their union is the whole $G$. I saw that for general reductive groups, one can show that the union forms a subgroup of $G$ by showing that $bw_{1}b'w_{2}b'' \in Bw_{3}B$ for some $w_{3}$, and consider their dimensions. However, I want to find some elementary proof without using any algebraic geometry, but only uses linear algebra or some combinatorial way. Is there any such proof? I know how to prove it for $n =2$, but this method (just case by case) is hard to be applied for $n \geq 3$.