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The preconditioned matrix or is rarely explicitly formed. Only the action of applying the preconditioner solve operation to a given vector may need to be computed. Typically there is a trade-off in the choice of .
In mathematics, more specifically in numerical linear algebra, the biconjugate gradient method is an algorithm to solve systems of linear equations A x = b . {\displaystyle Ax=b.\,} Unlike the conjugate gradient method , this algorithm does not require the matrix A {\displaystyle A} to be self-adjoint , but instead one needs to perform ...
To solve a linear system Ax = b with a preconditioner K = K 1 K 2 ≈ A, preconditioned BiCGSTAB starts with an initial guess x 0 and proceeds as follows: r 0 = b − Ax 0 Choose an arbitrary vector r̂ 0 such that ( r̂ 0 , r 0 ) ≠ 0 , e.g., r̂ 0 = r 0
The standard convergence condition (for any iterative method) is when the spectral radius of the iteration matrix is less than 1: ((+)) < A sufficient (but not necessary) condition for the method to converge is that the matrix A is strictly or irreducibly diagonally dominant. Strict row diagonal dominance means that for each row, the absolute ...
We seek the solution to a set of linear equations, expressed in matrix terms as =. The Richardson iteration is (+) = + (()), where is a scalar ...
The conjugate gradient method can be applied to an arbitrary n-by-m matrix by applying it to normal equations A T A and right-hand side vector A T b, since A T A is a symmetric positive-semidefinite matrix for any A. The result is conjugate gradient on the normal equations (CGN or CGNR). A T Ax = A T b
The minimum can be computed using a QR decomposition: find an (n + 1)-by-(n + 1) orthogonal matrix Ω n and an (n + 1)-by-n upper triangular matrix ~ such that ~ = ~. The triangular matrix has one more row than it has columns, so its bottom row consists of zero.
In matrix inversion however, instead of vector b, we have matrix B, where B is an n-by-p matrix, so that we are trying to find a matrix X (also a n-by-p matrix): = =. We can use the same algorithm presented earlier to solve for each column of matrix X. Now suppose that B is the identity matrix of size n.