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: 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).
See also
tensordot, tensorsolve
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