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Consider the permutation matrix

$P= \begin{pmatrix} 0 & \cdots & 0 & 1 \\ 0 & \cdots & 1 &0 \\ \vdots & \ddots & \vdots & \vdots \\ 1 & \cdots & 0 & 0 {} \end{pmatrix}$

SoI showed that if $L$ is lower triangular then $PLP$ is upper triangular.

Now the problem asks me to descrive how to calculate a factorisation $A=\overline U \ \overline L $

where $\overline U$ is unit upper triangular and $\overline L$ is lower triangular.

I am finding it difficult because I think the previous part was an hint to do the following:

$PAP=PLUP=PLP^2UP$ as $P^2=I$

And then simply define $\overline U = PLP$ and $\overline P = PUP$

How can I get rid of the $P$ s on the rhs?

Any hint would be appreciated, I think I am missing something!

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  • $\begingroup$ actually, writing it down made it clearer to me. is the way just considering the LU factorisation of the transpose of A? $\endgroup$
    – Moritzplatz
    Dec 12, 2012 at 20:09
  • $\begingroup$ If $A^T=LU$, then $A=U^TL^T$. $U^T$ is lower triangular and $L^T$ is upper triangular. It is a method to 'convert' between the Doolittle and Crout LU factorisations, if you want too. $\endgroup$
    – Daryl
    Dec 12, 2012 at 20:52
  • $\begingroup$ ah yes, even easier, I think the question wasn't really clear! just to use P I'll do the same as what I wrote in the question substituting A by $A^{T}$ thank you! $\endgroup$
    – Moritzplatz
    Dec 12, 2012 at 20:56
  • $\begingroup$ Also, $PAP=A^T$. You may have already figured that out, though. :-) $\endgroup$
    – Daryl
    Dec 12, 2012 at 21:53

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