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?