qr - QR decomposition
From Q*R = X , it follows that the kth column of the matrix X , is expressed as a linear combination of the k first columns of Q (with coefficients R(1,k), ..., R(k,k) ). The k first columns of Q make an orthogonal basis of the subspace spanned by the k first comumns of X . If column k of X (i.e. X(:,k) ) is a linear combination of the first p columns of X , then the entries R(p+1,k), ..., R(k,k) are zero. It this situation, R is upper trapezoidal. If X has rank rk , rows R(rk+1,:), R(rk+2,:), ... are zeros.
produces a (column) permutation matrix E , an upper triangular R with decreasing diagonal elements and an orthogonal (or unitary) Q so that X*E = Q*R . If rk is the rank of X , the rk first entries along the main diagonal of R , i.e. R(1,1), R(2,2), ..., R(rk,rk) are all different from zero. [Q,R,E] = qr(X,"e") produces an "economy size": If X is m-by-n with m > n, then only the first n columns of Q are computed as well as the first n rows of R .
// QR factorization, generic case // X is tall (full rank) X=rand(5,2);[Q,R]=qr(X); [Q'*X R] //X is fat (full rank) X=rand(2,3);[Q,R]=qr(X); [Q'*X R] //Column 4 of X is a linear combination of columns 1 and 2: X=rand(8,5);X(:,4)=X(:,1)+X(:,2); [Q,R]=qr(X); R, R(:,4) //X has rank 2, rows 3 to $ of R are zero: X=rand(8,2)*rand(2,5);[Q,R]=qr(X); R //Evaluating the rank rk: column pivoting ==> rk first //diagonal entries of R are non zero : A=rand(5,2)*rand(2,5); [Q,R,rk,E] = qr(A,1.d-10); norm(Q'*A-R) svd([A,Q(:,1:rk)]) //span(A) =span(Q(:,1:rk))
rankqr , rank , svd , rowcomp , colcomp ,
qr decomposition is based the Lapack routines DGEQRF, DGEQPF, DORGQR for the real matrices and ZGEQRF, ZGEQPF, ZORGQR for the complex case.