xorbits.numpy.linalg.tensorsolve(a, b, axes=None)#

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).

  • a (array_like) – 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’).

  • b (array_like) – Right-hand tensor, which can be of any shape.

  • axes (tuple of ints, optional) – Axes in a to reorder to the right, before inversion. If None (default), no reordering is done.



Return type

ndarray, shape Q


LinAlgError – If a is singular or not ‘square’ (in the above sense).

See also

numpy.tensordot, tensorinv, numpy.einsum


>>> a = np.eye(2*3*4)  
>>> a.shape = (2*3, 4, 2, 3, 4)  
>>> b = np.random.randn(2*3, 4)  
>>> x = np.linalg.tensorsolve(a, b)  
>>> x.shape  
(2, 3, 4)
>>> np.allclose(np.tensordot(a, x, axes=3), b)  


This method has not been implemented yet. Xorbits will try to execute it with numpy.linalg.

This docstring was copied from numpy.linalg.