Code for locating pivots in LU decomposition

It is not possible to write a code to locate the pivot required for partial pivot in LU decomposition. Help me.

from for j=i:n-1 to if j~=k

function [L,U,P] = mylu(A)

[m,n] = size(A);

if m ~= n

fprintf('***입력된 행렬이 정사각행렬이 아님! ***\n');

L = []; U = []; P = []; return;

end

L=eye(n,n);

U=A;

p=1:n;

for j = i:n-1

if j~=k

U([j k],j:n)=U([k j],j:n);

L([j k],1:j-1)=L([k j],1:j-1);

p([j k]) = p([k j]);

end

for i=j+1:n

L(i,j) = U(i,j) / U(j,j);

U(i,j) = 0;

U(i,j+1:n) = U(i,j+1:n) - L(i,j)*U(j,j+1:n);

end

end

P=eye(n.n);

P = P(p,:);

Answer 1

[Shantanu Dixit]

Hi Ye,

The missing logic in your code is the part that identifies the pivot element needed for partial pivoting. Specifically, the steps to determine the pivot index 'k' are absent. Here's how you can implement it:

• Locate the Pivot: Before the if j~=k statement, you need to find the row index 'k' of the largest absolute value in the current column from row 'j' to 'n'.
• Implement Partial Pivoting: Use the pivot index k to swap rows in both 'U' and 'L' matrices if necessary.

Below is the complete implementation for LUP decomposition:

function [L, U, P] = mylu(A)
    [m, n] = size(A);
    if m ~= n
        fprintf('The input matrix is not square!');
        L = []; U = []; P = []; return;
    end
    L = eye(n, n);
    U = A;
    p = 1:n;

    for j = 1:n-1
        % Locate the pivot
        % find row index k with the largest absolute
        % value in the jth column
        [~, k] = max(abs(U(j:n, j)));
        k = k + j - 1; % Adjust index

        if j ~= k
            U([j k], j:n) = U([k j], j:n);
            L([j k], 1:j-1) = L([k j], 1:j-1);
            p([j k]) = p([k j]);
        end

        for i = j+1:n
            L(i, j) = U(i, j) / U(j, j);
            U(i, j) = 0;
            U(i, j+1:n) = U(i, j+1:n) - L(i, j) * U(j, j+1:n);
        end
    end

    P = eye(n);
    P = P(p, :);
end