numpy.linalg.slogdet

linalg.slogdet(a)

Compute the sign and (natural) logarithm of the determinant of an array.

If an array has a very small or very large determinant, then a call to det may overflow or underflow. This routine is more robust against such issues, because it computes the logarithm of the determinant rather than the determinant itself.

Parameters:

a(…, M, M) array_like

Input array, has to be a square 2-D array.

Returns:

A namedtuple with the following attributes:

sign(…) array_like

A number representing the sign of the determinant. For a real matrix, this is 1, 0, or -1. For a complex matrix, this is a complex number with absolute value 1 (i.e., it is on the unit circle), or else 0.

logabsdet(…) array_like

The natural log of the absolute value of the determinant.

If the determinant is zero, then sign will be 0 and logabsdet

will be -inf. In all cases, the determinant is equal to

sign * np.exp(logabsdet).

See also

det

Notes

New in version 1.8.0.

Broadcasting rules apply, see the numpy.linalg documentation for details.

New in version 1.6.0.

The determinant is computed via LU factorization using the LAPACK routine z/dgetrf.

Examples

The determinant of a 2-D array [[a, b], [c, d]] is ad - bc:

>>> import numpy as np
>>> a = np.array([[1, 2], [3, 4]])
>>> (sign, logabsdet) = np.linalg.slogdet(a)
>>> (sign, logabsdet)
(-1, 0.69314718055994529) # may vary
>>> sign * np.exp(logabsdet)
-2.0

Computing log-determinants for a stack of matrices:

>>> a = np.array([ [[1, 2], [3, 4]], [[1, 2], [2, 1]], [[1, 3], [3, 1]] ])
>>> a.shape
(3, 2, 2)
>>> sign, logabsdet = np.linalg.slogdet(a)
>>> (sign, logabsdet)
(array([-1., -1., -1.]), array([ 0.69314718,  1.09861229,  2.07944154]))
>>> sign * np.exp(logabsdet)
array([-2., -3., -8.])

This routine succeeds where ordinary det does not:

>>> np.linalg.det(np.eye(500) * 0.1)
0.0
>>> np.linalg.slogdet(np.eye(500) * 0.1)
(1, -1151.2925464970228)