QR Factorization Using Householder Transformations
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Hüseyin
am 12 Jan. 2015
Beantwortet: Davide Poggiali
am 20 Apr. 2020
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 Kommentar
John D'Errico
am 12 Jan. 2015
Please learn to use the code button when you post code. It takes only one click of the mouse to do so.
Akzeptierte Antwort
Titus Edelhofer
am 12 Jan. 2015
Hi Hüseyin,
I don't think something is wrong. Q*R gives A (at least for your matrix A). Having different Q and R from MATLAB's implementation does not necessarily mean something is wrong (as long as Q*R=A and Q is orthogonal, i.e. Q'*Q = identity).
Titus
3 Kommentare
John D'Errico
am 12 Jan. 2015
Bearbeitet: John D'Errico
am 12 Jan. 2015
You are not listening. Q and R are not unique. Your code is fine. That it produces elements with sign differences in some cases merely means that an arbitrary choice was made about sign in the MATLAB code that differs from your choice. And since the MATLAB code for QR is proprietary, you can NEVER know exactly what they did.
Weitere Antworten (2)
Francesco Onorati
am 9 Jun. 2016
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).
0 Kommentare
Davide Poggiali
am 20 Apr. 2020
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
0 Kommentare
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