numpy.linalg.tensorinv#
- 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
- aarray_like
Tensor to ‘invert’. Its shape must be ‘square’, i. e.,
prod(a.shape[:ind]) == prod(a.shape[ind:])
.- indint, optional
Number of first indices that are involved in the inverse sum. Must be a positive integer, default is 2.
- Returns
- bndarray
a’s tensordot inverse, shape
a.shape[ind:] + a.shape[:ind]
.
- Raises
- LinAlgError
If a is singular or not ‘square’ (in the above sense).
See also
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