How to show $SL_{n}(\mathbb{R})=\bigsqcup_{w\in W}LwU$ where L (or U) are lower(or upper) triangular matrix? I want to ask you the following problem.
Let W denote the subgroup of permutation matrices in $SL_{n}(\mathbb{R})$. Show the following decomposition.
$SL_{n}(\mathbb{R})=\bigsqcup_{w\in W}LwU$
where L denotes all lower triangular matrices, and U denotes all upper triangular matrices.
I've found some similar statements in Decompose invertible matrix $A$ as $A = LPU$. (Artin, Chapter 2, Exercise M.11) and the http://en.wikipedia.org/wiki/Bruhat_decomposition. Especially, the first one is exactly the same with the problem above except the group the problem deal with : in the link, they deal with $GL_{n}(\mathbb{R})$.
I thought the problem has something wrong by the argument below :
When considering n=2 case, there exists only one permutation matrix - the identity matrix $I$ (if we regard a "permutation matrix" as the one earned by changing the rows of the identity matrix). Then, the decomposition says all the matrices with determinant 1(pick one of those, say A) can be decomposed as
$A = LU$ where  $L = \begin{pmatrix} a_{1} & 0 \\ a_{2} & a_{3} \end{pmatrix} 
$ and $U = \begin{pmatrix} b_{1} & b_{2} \\ 0 & b_{3} \end{pmatrix}$ and $a_{1}, a_{3}, b_{1}, b_{3}$ are not zero.
Then,  $A = \begin{pmatrix} a_{1}b_{1} & a_{1}b_{2} \\ a_{2}b_{1} & a_{2}b_{2}+a_{3}b_{3} \end{pmatrix}$, where $a_{1}b_{1}$ should not be zero. But there exists elements in $SL_{n}(\mathbb{R})$ whose first row, frist column entry is zero. After that I leanred that $S_{n}$ can be embedded into $SL_{n}$ natrually. Using this, I examined the case n=2 again. By an elementary eigenvalue argument says that $S_{2}$ can be embedded into $SL_{2}(\mathbb{R})$ in the only one way i.e. to {I, -I}. The only different thing is -I, but by the same argument above, there are elements that cannot be represented by above decomposition even we consider -I.(just the problem of sign, huh?)
After that I rather considered $PSL_{2}(\mathbb{R})$, not $SL_{2}(\mathbb{R})$ when n=2, and when n is odd, but I still don't get it, and at this moment I gave up.
 A: The other $2 \times 2$ permutation matrix is $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.
Here's one way to come up with a proof.  Given any $n \times n$ matrix $A$, starting in the first column and moving rightward, find the highest non-zero entry that is in a row that does not contain a pivot in an earlier column, and this will be your pivot for this column.  Using the two elementary row operations that are not swapping two rows, you can make the pivot in that column equal to $1$ and then clear the entries below this pivot.  The product of these elementary matrices $L$ is lower triangular, and $LA$ has $1$'s in pivot columns, say the pivot in column $j$ occurs in row $w(j)$, and zeroes below all the pivots.  Let $P$ be the permutation matrix with its $1$'s in the entries $(w(j), j)$.  Then $LA = PU$ with $U$ upper triangular.  So $A = L^{-1} P U$.
You also need to check that the union is disjoint, that is if $A = L P U = L' P' U'$, then $P = P'$.  The way to check this is that if you consider the upper left $p \times q$ submatrix $A_{p,q}$ of $A$, then $\text{rank}(A_{p,q}) = \text{rank}(P_{p,q})$.  Moreover, a permutation matrix $P$ is uniquely determined by the values $P_{p,q}$ for all $1 \leq p, q \leq n$.
