Documentation |
Orthogonal-triangular decomposition
[Q,R] = qr(A)
[Q,R] = qr(A,0)
[Q,R,E] = qr(A)
[Q,R,E] =
qr(A,'matrix')
[Q,R,e] = qr(A,'vector')
[Q,R,e] = qr(A,0)
X = qr(A)
X = qr(A,0)
R = qr(A)
R = qr(A,0)
[C,R] = qr(A,B)
[C,R,E] = qr(A,B)
[C,R,E]
= qr(A,B,'matrix')
[C,R,e] = qr(A,B,'vector')
[C,R] = qr(A,B,0)
[C,R,e] = qr(A,B,0)
[Q,R] = qr(A), where A is m-by-n, produces an m-by-n upper triangular matrix R and an m-by-m unitary matrix Q so that A = Q*R.
[Q,R] = qr(A,0) produces the economy-size decomposition. If m > n, only the first n columns of Q and the first n rows of R are computed. If m<=n, this is the same as [Q,R] = qr(A).
If A is full:
[Q,R,E] = qr(A) or [Q,R,E] = qr(A,'matrix') produces unitary Q, upper triangular R and a permutation matrix E so that A*E = Q*R. The column permutation E is chosen so that abs(diag(R)) is decreasing.
[Q,R,e] = qr(A,'vector') returns the permutation information as a vector instead of a matrix. That is, e is a row vector such that A(:,e) = Q*R.
[Q,R,e] = qr(A,0) produces an economy-size decomposition in which e is a permutation vector, so that A(:,e) = Q*R.
X = qr(A) and X = qr(A,0) return a matrix X such that triu(X) is the upper triangular factor R.
If A is sparse:
R = qr(A) computes a Q-less QR decomposition and returns the upper triangular factor R. Note that R = chol(A'*A). Since Q is often nearly full, this is preferred to [Q,R] = QR(A).
R = qr(A,0) produces economy-size R. If m>n, R has only n rows. If m<=n, this is the same as R = qr(A).
[Q,R,E] = qr(A) or [Q,R,E] = qr(A,'matrix') produces unitary Q, upper triangular R and a permutation matrix E so that A*E = Q*R. The column permutation E is chosen to reduce fill-in in R.
[Q,R,e] = qr(A,'vector') returns the permutation information as a vector instead of a matrix. That is, e is a row vector such that A(:,e) = Q*R.
[Q,R,e] = qr(A,0) produces an economy-size decomposition in which e is a permutation vector, so that A(:,e) = Q*R.
[C,R] = qr(A,B), where B has as many rows as A, returns C = Q'*B. The least-squares solution to A*X = B is X = R\C.
[C,R,E] = qr(A,B) or [C,R,E] = qr(A,B,'matrix'), also returns a fill-reducing ordering. The least-squares solution to A*X = B is X = E*(R\C).
[C,R,e] = qr(A,B,'vector') returns the permutation information as a vector instead of a matrix. That is, the least-squares solution to A*X = B is X(e,:) = R\C.
[C,R] = qr(A,B,0) produces economy-size results. If m>n, C and R have only n rows. If m<=n, this is the same as [C,R] = qr(A,B).
[C,R,e] = qr(A,B,0) additionally produces a fill-reducing permutation vector e. In this case, the least-squares solution to A*X = B is X(e,:) = R\C.