TITLE: Using permutation matrix to get LU-Factorization with $A=UL$
QUESTION [8 upvotes]: Let $Q$ be the $n$x$n$ permutation matrix
$$Q=
\begin{bmatrix}
0&0&...&0&1\\
0&0&...&1&0\\
.& \\
.&\\
.&\\
0&0&...&0&0\\
1&0&...&0&0\\
\end{bmatrix}$$
If $L \in \mathbb{R}$ is lower triangular, what is the structure of $QLQ$? Use this to show that one can factorize $A=UL$ where $U$ is unit upper triangular and $L$ is lower triangular.
I can see that $QLQ = L^T$
And $Q^2=I$
So here's what I am doing
$A=LU$
$QAQ=QLUQ = QLQ^2UQ =QLQQUQ = UL$
But now I am left with $QAQ = UL$ rather than $A=UL$
But does that matter?
It seems like it does as the factorization I get would be for solving $A^Tx=b$ rather than $Ax=b$
So have I missed something, is there a way to get the factorization $A=UL$ or have I actually got it but it's the case that I am misinterpreting my answer?

REPLY [8 votes]: You know how to get $A=LU,$ so what if you instead do LU decomposition of $QAQ$?
Then, $QAQ=LU$ for some $L,U$ and by multiplying both sides with $Q$ from left and right, and inserting $Q^2$ between $L$ and $U$, you arrive at $A=U'L'$ for some upper- and lower-triangular matrices $U',L'$