scipy.linalg.lu_factor#
- scipy.linalg.lu_factor(a, overwrite_a=False, check_finite=True)[source]#
Compute pivoted LU decomposition of a matrix.
The decomposition is:
A = P L U
where P is a permutation matrix, L lower triangular with unit diagonal elements, and U upper triangular.
- Parameters:
- a(M, N) array_like
Matrix to decompose
- overwrite_abool, optional
Whether to overwrite data in A (may increase performance)
- check_finitebool, optional
Whether to check that the input matrix contains only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs.
- Returns:
- lu(M, N) ndarray
Matrix containing U in its upper triangle, and L in its lower triangle. The unit diagonal elements of L are not stored.
- piv(K,) ndarray
Pivot indices representing the permutation matrix P: row i of matrix was interchanged with row piv[i]. Of shape
(K,)
, withK = min(M, N)
.
See also
Notes
This is a wrapper to the
*GETRF
routines from LAPACK. Unlikelu
, it outputs the L and U factors into a single array and returns pivot indices instead of a permutation matrix.While the underlying
*GETRF
routines return 1-based pivot indices, thepiv
array returned bylu_factor
contains 0-based indices.Examples
>>> import numpy as np >>> from scipy.linalg import lu_factor >>> A = np.array([[2, 5, 8, 7], [5, 2, 2, 8], [7, 5, 6, 6], [5, 4, 4, 8]]) >>> lu, piv = lu_factor(A) >>> piv array([2, 2, 3, 3], dtype=int32)
Convert LAPACK’s
piv
array to NumPy index and test the permutation>>> def pivot_to_permutation(piv): ... perm = np.arange(len(piv)) ... for i in range(len(piv)): ... perm[i], perm[piv[i]] = perm[piv[i]], perm[i] ... return perm ... >>> p_inv = pivot_to_permutation(piv) >>> p_inv array([2, 0, 3, 1]) >>> L, U = np.tril(lu, k=-1) + np.eye(4), np.triu(lu) >>> np.allclose(A[p_inv] - L @ U, np.zeros((4, 4))) True
The P matrix in P L U is defined by the inverse permutation and can be recovered using argsort:
>>> p = np.argsort(p_inv) >>> p array([1, 3, 0, 2]) >>> np.allclose(A - L[p] @ U, np.zeros((4, 4))) True
or alternatively:
>>> P = np.eye(4)[p] >>> np.allclose(A - P @ L @ U, np.zeros((4, 4))) True