numpy.linalg.tensorinv — NumPy v1.13 Manual (original) (raw)
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 operationtensordot(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. |
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| 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