LU factorization with decreasing elements on the main diagonal of U

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Does select P so that is decreasing?
If not, how can I request it?

Answers (1)

Bruno Luong
Bruno Luong on 25 Sep 2023
Edited: Bruno Luong on 25 Sep 2023
Obviously not
A=[1 10 9;
5 1 9;
2 8 1]
A = 3×3
1 10 9 5 1 9 2 8 1
[L,U,P]=lu(A)
L = 3×3
1.0000 0 0 0.2000 1.0000 0 0.4000 0.7755 1.0000
U = 3×3
5.0000 1.0000 9.0000 0 9.8000 7.2000 0 0 -8.1837
P = 3×3
0 1 0 1 0 0 0 0 1
But you can fix the progression of abs(diag(U)) in any arbitray decrasing sequance you want, just scale appropiately L.
Here I select the sequene of U(1,1).*2.^(-(1:n-1))
% Fix it, assuming A is not singular
Ukk = abs(U(1,1));
for k=2:size(U,1)
s = Ukk/(2*abs(U(k,k)));
U(k,:) = s*U(k,:);
L(:,k) = L(:,k)/s;
Ukk = abs(U(k,k));
end
L
L = 3×3
1.0000 0 0 0.2000 3.9200 0 0.4000 3.0400 6.5469
U
U = 3×3
5.0000 1.0000 9.0000 0 2.5000 1.8367 0 0 -1.2500
P'*L*U % close to A
ans = 3×3
1 10 9 5 1 9 2 8 1
In some sense the question of scaling U alone is trivial and sort of useless if you don't specify what L should be.
Note that MATLAB returns L such that diag(L) are 1.
  5 Comments
Sara
Sara on 25 Sep 2023
Thank you, you are right.
Last question, is there any option in matlab to generate a PLU factorization such that is decreasing without having to modify a lu factorization previously calculated by matlab?

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