numpy.linalg.slogdet¶

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 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 2D 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)
.See also
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 2D array
[[a, b], [c, d]]
isad  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 logdeterminants 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)