Solve the tensor equation a x = b for x.
It is assumed that all indices of x are summed over in the product, together with the rightmost indices of a, as is done in, for example, tensordot(a, x, axes=x.ndim).
Coefficient tensor, of shape b.shape + Q. Q, a tuple, equals the shape of that sub-tensor of a consisting of the appropriate number of its rightmost indices, and must be such that prod(Q) == prod(b.shape) (in which sense a is said to be ‘square’).
Right-hand tensor, which can be of any shape.
Axes in a to reorder to the right, before inversion. If None (default), no reordering is done.
If a is singular or not ‘square’ (in the above sense).
See also
>>> import numpy as np >>> a = np.eye(2*3*4).reshape((2*3, 4, 2, 3, 4)) >>> rng = np.random.default_rng() >>> b = rng.normal(size=(2*3, 4)) >>> x = np.linalg.tensorsolve(a, b) >>> x.shape (2, 3, 4) >>> np.allclose(np.tensordot(a, x, axes=3), b) True
© 2005–2024 NumPy Developers
Licensed under the 3-clause BSD License.
https://numpy.org/doc/2.4/reference/generated/numpy.linalg.tensorsolve.html