scipy.sparse.linalg.spsolve_triangular#

scipy.sparse.linalg.spsolve_triangular(A, b, lower=True, overwrite_A=False, overwrite_b=False, unit_diagonal=False)[source]#

Solve the equation A x = b for x, assuming A is a triangular matrix.

Parameters:
A(M, M) sparse matrix

A sparse square triangular matrix. Should be in CSR format.

b(M,) or (M, N) array_like

Right-hand side matrix in A x = b

lowerbool, optional

Whether A is a lower or upper triangular matrix. Default is lower triangular matrix.

overwrite_Abool, optional

Allow changing A. The indices of A are going to be sorted and zero entries are going to be removed. Enabling gives a performance gain. Default is False.

overwrite_bbool, optional

Allow overwriting data in b. Enabling gives a performance gain. Default is False. If overwrite_b is True, it should be ensured that b has an appropriate dtype to be able to store the result.

unit_diagonalbool, optional

If True, diagonal elements of a are assumed to be 1 and will not be referenced.

New in version 1.4.0.

Returns:
x(M,) or (M, N) ndarray

Solution to the system A x = b. Shape of return matches shape of b.

Raises:
LinAlgError

If A is singular or not triangular.

ValueError

If shape of A or shape of b do not match the requirements.

Notes

New in version 0.19.0.

Examples

>>> from scipy.sparse import csr_matrix
>>> from scipy.sparse.linalg import spsolve_triangular
>>> A = csr_matrix([[3, 0, 0], [1, -1, 0], [2, 0, 1]], dtype=float)
>>> B = np.array([[2, 0], [-1, 0], [2, 0]], dtype=float)
>>> x = spsolve_triangular(A, B)
>>> np.allclose(A.dot(x), B)
True