LU factorization with decreasing elements on the main diagonal of U

1 回表示 (過去 30 日間)
Sara
Sara 2023 年 9 月 25 日
コメント済み: Bruno Luong 2023 年 9 月 25 日
Does select P so that is decreasing?
If not, how can I request it?

回答 (1 件)

Bruno Luong
Bruno Luong 2023 年 9 月 25 日
編集済み: Bruno Luong 2023 年 9 月 25 日
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 件のコメント
Sara
Sara 2023 年 9 月 25 日
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?
Bruno Luong
Bruno Luong 2023 年 9 月 25 日
No

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeOperating on Diagonal Matrices についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by