numpy.linalg.slogdet — NumPy v1.11 Manual (original) (raw)
numpy.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 todet 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: | 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. logdet : (...) array_like The natural log of the absolute value of the determinant. If the determinant is zero, then sign will be 0 and logdet will be -Inf. In all cases, the determinant is equal to sign * np.exp(logdet). |
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:
a = np.array([[1, 2], [3, 4]]) (sign, logdet) = np.linalg.slogdet(a) (sign, logdet) (-1, 0.69314718055994529) sign * np.exp(logdet) -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, logdet = np.linalg.slogdet(a) (sign, logdet) (array([-1., -1., -1.]), array([ 0.69314718, 1.09861229, 2.07944154])) sign * np.exp(logdet) 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)