Describing the double cosets $LgU$ for $g\in\operatorname{GL}_n$

Question

This is with reference to Artin's Algebra's Exercise M.11 of Chapter 2. I have managed to show that any invertible matrix can be decomposed as $lpu$ where $p$ is a permutation matrix, and $l$ and $u$ are lower and upper triangular matrices with $U$'s diagonal entries being $1$. What Artin asks next is to describe the double cosets $$LgU := \{lgu : g\in\operatorname{GL}_n\text{ and }l, u \text{ meaning the obvious matrices}\}.$$

In a previous exercise, I have proven that if $H$ and $K$ are subgroups, then their double cosets partition the group. So it is easily seen that each $LgU = LpU$ for some permutation matrix $p$.

Question: Is there anything more to this description? For instance, can two distinct permutation matrices be in a same double coset?

Answer 1

This is not the Bruhat decomposition (which is $LPL$) but close.

Since $g=lpu$ where $l\in L, u\in U, p\in P$, you have $LgU=LpU$. Suppose that $p,q$ are permutational matrices and $l_1pu_1=l_2qu_2$, so $LpU=LqU$. Then $lp=qu$ for some $l\in L, u\in U$. Then $lpe_i=que_i$ for every standard basic vector $e_i$: $le_j=q(u_i)$, where we denote by $x_i$ the $i$-th column of matrix $x$, $j=p(i)$.

Let $i$ be such that $p(i)=n$, so $le_n=q(u_i)$. So $l_{n,n}e_n=q(u_i)$. Since $q(u_i)$ has coordinate $1$, we see that this coordinate of $q(u_i)$ must be number $n$, other coordinates are $0$, all coordinates of $u(i)$ except coordinate number $i$ are $0$ and so $q(i)=n$.

Now let $p(i_1)=n-1$. Then $le_{n-1}=l_{n-1}=q(u_{i_1})$. Then $q(i_1)=$ either $n$ or $n-1$. The first option is impossible, so $p$ and $q$ coincide on $i,i_1$ (equivalently, $p^{-1}, q^{-1}$ coincide on $n,n-1$).

Continue this way, consider $n-2,...,1$ and prove that $p^{-1}=q^{-1}$. So $p=q$. Thus $LpU=LqU$ iff $p=q$.

A proof can also be found on this page of the book "Combinatorial matrix theory" (by Richard A. Brualdi, Ángeles Carmona, P. van den Driessche, Stephen Kirkland, Dragan Stevanović), theorem 1.1.10.

Answer 2

Hint.

It seems to me that the following things are better for solving this problem. Let $$ {t_{ij}(\alpha)={\begin{bmatrix}1&&&&&&\\&\ddots &&&&&\\&&1&&\alpha&&\\&&&\ddots &&&\\&&&&1&&\\&&&&&\ddots &\\&&&&&&1\end{bmatrix}}} $$

So $t_{ij}(\alpha)a$ is the matrix produced from $a$ by adding $\alpha$ times row $j$ to row $i$; and $at_{ij}(\alpha)$ is the matrix produced from $a$ by adding $\alpha$ times column $i$ to column $j$.

Multiplying on the left by a matrix of this form with $i>j$, and multiplying on the right by a matrix with $i<j$.