numpy.linalg.slogdet

linalg.slogdet(a)[source]

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

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

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

Examples

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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)

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