Talk:LU decomposition

Python example contains error

The Python example has a divide by zero error for the matrix

b = [[1, 1, 1, 1], [1, 1, -1, -1], [1, -1, 0, 0], [0, 0, 1, -1]]

although a LUP decomposition exists: [1]

The permutation matrix has to be updated at each step, but that will make the code a lot more complicated.

Example 2 pivot matrix seems to be wrong

The pivot matrix in example 2 should also swap the last two rows of the current resulting pivoted matrix: A'(3, 3) = 2 while there is a 7 right beneath it. Here is the result of the multiplication of the two matrices on wolfram alpha

The pivot matrix I propose is {{1,0,0,0},{0,0,1,0},{0,0,0,1},{0,1,0,0}}.

I did not change the article because it seemed very strange that I would be the first to see this and I therefore wonder if I'm not wrong.

Re: Example 2 pivot matrix seems to be wrong

I noticed this too. In the implementations it appears that the max pivot values are taken from the original A matrix without swapping. When the max pivot is being found for row 3 of A the choices presented are row 3 [3, 7, 18, 1] and row 4 [2, 5, 7, 1], since row 3 contains the largest pivot value the third row of the current permutation matrix is swapped with itself, which contains [0, 1, 0, 0] from the initial swap with row 2. It's strange, but a lot of the algorithm implementations seem to follow that trend and multiplying by the permutation matrix instead of swapping inline with the permutation matrix.

I think the permutation matrix you proposed makes more sense as well.

Matlab code is wrong

Please lock at the following example A=[1 2 -1 0;2 4 -2 -1; -3 -5 6 1; -1 2 8 -2] the code returns nonsense for U, namely

U =

        0         0         0   -1.0000
   0.5000    1.0000   -0.5000         0
  -0.1429         0    1.0000    0.2857
   0.1563         0         0    1.0000

I presume the reason is the permutation matrix. I will try to correct it.

Different output as per common literature

If one sees the document at: file:///C:/Users/Gagandeep%20S.%20Datta/Downloads/New%20folder%20(2)/30_3_lu_decmp.pdf

For A= [{1,2,4;3,8,14;2,6,13}] L= [{1,0,0;3,1,0;2,1,1}] and U= [{1,2,4;0,2,2;0,0,3}].

But on applying the LU VBA code the output is different. We get, L=[{1,0,0;0.333333333333333,1,0;0.666666666666667,-1,1}] and H=[{3,8,14;0,-0.666666666666667,-0.666666666666666;0,0,3}].

Why is this so?