[CSC 335] PLU

[Andrew J. Pounds]

Wow -- the last 36 hours has been BUSY... Anyway, there were a few 
questions about the PLU decomposition at the end of class yesterday and 
I wanted to make sure we were all on the same page.

If $\hat{A}\vec{x}= \vec{b}$ then I can get the same result by 
premultiplying both sides with the permutation matrix 
$\hat{P}\hat{A}\vec{x}=\hat{P}\vec{b}$ and if $\hat{A}$ has been broken 
down into its LU decomposition then
$\hat{P}\hat{L}\hat{U}\vec{x}=\hat{P}\vec{b}$.

So, if I factor matrix $\hat{A}$ into an PLU decomposition and then give 
the algorithm the vector $\vec{b}$, I recover the solution vector 
$\vec{x}$ by first permuting the vector $\vec{b}$ and solving the system

$\hat{L}\vec{y}=\hat{P}\vec{b}$

and then follow this with

$\hat{U}\vec{x} = \vec{y}$.

Before you ask "hey -- where is the permutation matrix on the left side 
of those last two equations" remember that it was already applied to get 
the LU factorization.  We simply use the $\hat{P}$ on the right above to 
bring the $\vec{b}$ into the same permuted order.

Does that help clarify things?

-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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