QR Factorization Using Householder Transformations
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function [Q,R]=QRfactor(A)
[m,n]=size(A);
R=A; %Start with R=A
Q=eye(m); %Set Q as the identity matrix
for k=1:m-1
x=zeros(m,1);
x(k:m,1)=R(k:m,k);
g=norm(x);
v=x; v(k)=x(k)+g;
%Orthogonal transformation matrix that eliminates one element
%below the diagonal of the matrix it is post-multiplying:
s=norm(v);
if s~=0, w=v/s; u=2*R'*w;
R=R-w*u'; %Product HR
Q=Q-2*Q*w*w'; %Product QR
end
end
for A=[-2 2 3; 1 3 5; -3 -1 2]
I got the answers Q and R different from when I use [Q,R]=qr(A). Where am I wrong with code.
1 个评论
John D'Errico
2015-1-12
Please learn to use the code button when you post code. It takes only one click of the mouse to do so.
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Francesco Onorati
2016-6-9
2 个投票
The reason why there are differencies in the sign is that for numerical stability the 2-norm of each vector is taken with the opposite sign of the pivotal element of the vector itself. As you take always the norm as positive, sometimes it is in agreement with MATLAB code, sometimes it is not (here I'm supposing MATLAB uses Housolder transformation to do QR decomposition).
Davide Poggiali
2020-4-20
You just have to change two lines
g=-sign(x(k))*norm(x);
v=x; v(k)=x(k)-g;
to get what you're looking for. source: wiki
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