Re: lu decomposition without any permutations

[David Bateman]

Przemek Klosowski wrote:
> I think it is permuting the rows because you are asking it to permute the 
> rows by 
> providing the parameter p. If you don't it doesn't permute:
>
>         a=rand(5);  [l, u] = lu(a);  l*u-a
>   

No, it'll still pivot in this case. The only difference is that the
permutation matrix is incorporated in the L matrix.

a=rand(3);  [l, u] = lu(a)

l =

   0.87085   1.00000   0.00000
   0.72097  -0.43703   1.00000
   1.00000   0.00000   0.00000

u =

   0.97382   0.39181   0.47502
   0.00000   0.41088   0.07799
   0.00000   0.00000   0.28099

norm (l *u-a)

ans =  1.5732e-16


D.


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