class QRSolver

(Return to index)

Least-square solution of systems of linear equations.

Given an m by n matrix A, an n by n diagonal matrix D, and an m-vector B, two problem can be solved:

1. Solve the the system A*X = B in the least squares sense. The first step to solve this problem is:

   QRSolver solver(A, pivot);
   

   The second step is then

   solver.solveR(X);
   

2. Solve the the two systems A*X = B, D*X = 0 in the least squares sense. The first step to solve this problem is

   QRSolver solver(A, pivot);
   

   The second step is

   solver.solveS(D, X);
   

   The second step can be repeated as many times as required for different diagonal matrices D .

solver.getColNorm(C);

Type:	Instantiable
Include file:	./Algebra/QRSolver.hh

Synopsis (including inherited members):

• Public members:
  • QRSolver (const Matrix<double>&,const Vector<double>&,bool) ;
  • ~QRSolver () ;
  • void solveR (Vector<double>&)const ;
  • void solveS (const Array1D<double>&,double,Vector<double>&) ;
  • void solveRT (Vector<double>&)const ;
  • void solveST (Vector<double>&)const ;
  • void getColNorm (Array1D<double>&)const ;

Documentation:

• Public members:

  QRSolver (const Matrix<double>&,const Vector<double>&,bool) ;

  Constructor.
  Determine the QR-factorisation A = Q*R of the matrix A and transform the right-hand side B . If pivot is true, then pivot search is done.

  This method uses Householder transformations with optional column pivoting to compute a QR-factorization of the m-by-n matrix A . It determines an orthogonal matrix Q , a permutation matrix P , and an upper trapezoidal matrix R with diagonal elements of nonincreasing magnitude, such that A*P = Q*R . The Householder transformation for column k is of the form I - (1/U(k))*U*U^T , where U has zeros in the first k-1 positions. The form of this transformation and the method of pivoting first appeared in the corresponding LINPACK subroutine.

  void getColNorm (Array1D<double>&)const ;

  Return the original column norms of the matrix A .

  void solveR (Vector<double>&)const ;

  Solution of A*X = B in the least-squares sense.

  void solveRT (Vector<double>&)const ;

  Pre-multiply the vector V by R.transpose()^{-1 .}

  void solveS (const Array1D<double>&,double,Vector<double>&) ;

  Solution of A*X = B, D*X = 0 in the least-squares sense.

  void solveST (Vector<double>&)const ;

  Pre-multiply the vector V by S.transpose()^{-1 .}
  Requires prior execution of solveS.