numpy.linalg.tensorinv()

numpy.linalg.tensorinv(a, ind=2) [source]

Compute the ?inverse? of an N-dimensional array.

The result is an inverse for a relative to the tensordot operation tensordot(a, b, ind), i. e., up to floating-point accuracy, tensordot(tensorinv(a), a, ind) is the ?identity? tensor for the tensordot operation.

Parameters:

Parameters:	a : array_like Tensor to ?invert?. Its shape must be ?square?, i. e., `prod(a.shape[:ind]) == prod(a.shape[ind:])`. ind : int, optional Number of first indices that are involved in the inverse sum. Must be a positive integer, default is 2.
Returns:	b : ndarray `a`?s tensordot inverse, shape `a.shape[ind:] + a.shape[:ind]`.
Raises:	LinAlgError If `a` is singular or not ?square? (in the above sense).

a : array_like

Tensor to ?invert?. Its shape must be ?square?, i. e., prod(a.shape[:ind]) == prod(a.shape[ind:]).

ind : int, optional

Number of first indices that are involved in the inverse sum. Must be a positive integer, default is 2.

Returns:

b : ndarray

a?s tensordot inverse, shape a.shape[ind:] + a.shape[:ind].

Raises:

LinAlgError

If a is singular or not ?square? (in the above sense).

Examples

>>> a = np.eye(4*6)
>>> a.shape = (4, 6, 8, 3)
>>> ainv = np.linalg.tensorinv(a, ind=2)
>>> ainv.shape
(8, 3, 4, 6)
>>> b = np.random.randn(4, 6)
>>> np.allclose(np.tensordot(ainv, b), np.linalg.tensorsolve(a, b))
True

>>> a = np.eye(4*6)
>>> a.shape = (24, 8, 3)
>>> ainv = np.linalg.tensorinv(a, ind=1)
>>> ainv.shape
(8, 3, 24)
>>> b = np.random.randn(24)
>>> np.allclose(np.tensordot(ainv, b, 1), np.linalg.tensorsolve(a, b))
True

Links:

https://docs.scipy.org/doc/numpy-1.11.0/reference/generated/numpy.linalg.tensorinv.html

doc_NumPy

2025-01-10 15:47:30

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