TITLE: A slick proof of the Bruhat Decomposition for GL_n(k)?
QUESTION [39 upvotes]: On one of my exams last year, we were given a problem (we chose five or six out of eight problems) on an exam, the goal of which was to prove the Bruhat decomposition for $GL_n(k)$.  I was one of the two people to choose said problem.  I gave a very long convoluted argument which although correct was really inelegant.  I proved it more than once because I wasn't satisfied with my proof, and I figured out a somewhat slick contradiction argument based on maximizing leading zeroes of rows (number of zeroes before the pivot), but the proof was still a real mess.
Statement of the problem:
Let $G:=GL(V)$ for $V$ a finite dimensional $k$ vector space.  Let $B$ be the stabilizer of the standard flag (these will be invertible upper triangular matrices), and let $W$ be the subgroup of permutation matrices.
Show that $G=\coprod_{w\in W} BwB$, where the $BwB$ are double cosets.  That is, show that $G=BWB$.  
Question:
Is there a slick proof of this fact maybe using "more machinery"? In particular, is there any sort of "coordinate-free" proof (Can we even define the Borel and Weyl subgroups without coordinates?)?

REPLY [17 votes]: One can prove the Bruhat decomposition by applying the theorem of Jordan-Hölder.
This theorem shows that two chains of submodules of a module of finite length
 whose successive quotients
are simple have the same lengths, and that the same  quotients appear.
But it is slightly more precise, because it gives a precise recipe
for a bijection between the two lists. Here we apply it for modules of finite length over a field,
aka finitely dimensional vector spaces.
Let $E=(e_1,\dots,e_n)$ and $F=(f_1,\dots,f_n)$ be two bases of  a vector space $M$
over a field $K$.
For $0\leq i\leq n$, define
 $M_i=\langle e_1,\dots,e_i\rangle $ and $N_i=\langle f_1,\dots,f_i\rangle$.
The proof of the theorem of Jordan-Hölder furnishes
a (unique) permutation $\sigma$ of $\{1,\dots,n\}$
such that $M_{i-1}+M_i\cap N_{\sigma(i)-1}=M_{i-1}$
and  $M_{i-1}+M_i\cap N_{\sigma(i)}=M_i$,
for every $i\in\{1,\dots,n\}$. 
For any $i$, let $x_i$
be a vector belonging to $M_{i}\cap N_{\sigma(i)}$
but not to $M_{i-1}$.
For every $i$, one has $\langle x_1,\dots,x_i\rangle =M_i$;
it follows that $X=(x_1,\dots,x_n)$ is a basis of $M$;
moreover, there exists a matrix $B_1$, in upper triangular form,
such that $X=E B_1$.
Set $\tau=\sigma^{-1}$.
Similarly, one has $\langle x_{\tau(1)},\dots,x_{\tau(i)}\rangle=N_i$
for every $i$. Consequently, there exists a matrix $B_2$,
still in upper-triangular form, such that
$(x_{\tau(1)},\dots,x_{\tau(n)})=F B_2$.
Let $P_\tau$ be the permutation matrix associated to $\tau$,
we have $(x_{\tau(1)},\dots,x_{\tau(n)})=(x_1,\dots,x_n)P_\tau$.
This implies that $FB_2=EB_1P_\tau$, hence
$F=E B_1 P_{\tau} B_2^{-1} $.
Therefore, the matrix $A=B_1P_{\tau}P_2^{-1}$
that expresses the coordinates of the vectors of $F$
in the basis $E$
is the product of an upper-triangular matrix, a permutation matrix
and another upper-triangular matrix.
In the group $\mathop{\rm GL}(n,K)$, let  $B$ be the subgroup consisting
of upper-triangular matrices, and let $W$ be the subgroup
consisting of permutation matrices.
We have proved that $\mathop{\rm GL}(n,K)=BWB$: this is precisely the Bruhat decomposition.