MATLAB LU分解部分旋转 [英] MATLAB LU Decomposition Partial pivoting

问题描述

我正在尝试主要基于部分分解的LU分解来处理lu分解旋转Matlab

function [L,U,P] = lup(A)
n = length(A);
L = eye(n);
U = zeros(n);
P = eye(n);

for k=1:n-1
% find the entry in the left column with the largest abs value (pivot)
[~,r] = max(abs(A(k:end,k)));
r = n-(n-k+1)+r;

A([k r],:) = A([r k],:);
P([k r],:) = P([r k],:);
L([k r],:) = L([r k],:);

% from the pivot down divide by the pivot
L(k+1:n,k) = A(k+1:n,k) / A(k,k);

U(k,1:n) = A(k,1:n);
A(k+1:n,1:n) = A(k+1:n,1:n) - L(k+1:n,k)*A(k,1:n);

end
U(:,end) = A(:,end);

end

它似乎适用于大多数矩阵(等于matlab lu函数)，但是以下矩阵似乎会产生不同的结果:

It seems to work for most matrices (equal to the matlab lu function), however the following matrix seems to produce different results:

A = [
 3    -7    -2     2
-3     5     1     0
 6    -4     0    -5
-9     5    -5    12
];

我只是不知道出了什么问题.在链接的文章中提到的矩阵上似乎可以正常工作

I just can't figure out what is going wrong. It seems to work fine on the matrices mentioned in the linked post

推荐答案

您非常接近.我一共换了三行

you were pretty close. I changed three lines total

for k=1:n-1变为for k=1:n，我们不执行-1，因为我们也希望使用您的方法来获取L(n,n)=u(n,n)/u(n,n)=1，因此我们将其排除在外

for k=1:n-1 became for k=1:n we don't do the -1 because we also want to get L(n,n)=u(n,n)/u(n,n)=1 with your method we were leaving this out

L(k+1:n,k) = A(k+1:n,k) / A(k,k);成为L(k:n,k) = A(k:n,k) / A(k,k);，因为您遗漏了L(k,k)=A(k,k)/A(k,k)=1

L(k+1:n,k) = A(k+1:n,k) / A(k,k); became L(k:n,k) = A(k:n,k) / A(k,k); because you were leaving out L(k,k)=A(k,k)/A(k,k)=1

由于k+1的变化，我们不需要从L的单位矩阵开始，因为我们现在在对角线上重现1，因此L=eyes(n);变为L=zeros(n);

because the k+1 change we dont need to start with an identity matrix for L since we are now reproducing the 1's on the diagonals so L=eyes(n); became L=zeros(n);

和完整的代码

function [L,U,P] = lup(A)
% lup factorization with partial pivoting
% [L,U,P] = lup(A) returns unit lower triangular matrix L, upper
% triangular matrix U, and permutation matrix P so that P*A = L*U.
    n = length(A);
    L = zeros(n);
    U = zeros(n);
    P = eye(n);


    for k=1:n
        % find the entry in the left column with the largest abs value (pivot)
        [~,r] = max(abs(A(k:end,k)));
        r = n-(n-k+1)+r;    

        A([k r],:) = A([r k],:);
        P([k r],:) = P([r k],:);
        L([k r],:) = L([r k],:);

        % from the pivot down divide by the pivot
        L(k:n,k) = A(k:n,k) / A(k,k);

        U(k,1:n) = A(k,1:n);
        A(k+1:n,1:n) = A(k+1:n,1:n) - L(k+1:n,k)*A(k,1:n);

    end
    U(:,end) = A(:,end);

end

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