xorbits.numpy.linalg.tensorsolve#

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)`.

Parameters
• 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.

Returns

x

Return type

ndarray, shape Q

Raises

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

`numpy.tensordot`, `tensorinv`, `numpy.einsum`

Examples

```>>> 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)
True
```

Warning

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

This docstring was copied from numpy.linalg.