numpy.linalg.tensorsolve¶

numpy.linalg.tensorsolve(a, b, axes=None)[source]¶

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=b.ndim).

Parameters:

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 : ndarray, shape Q
Raises:	LinAlgError If a is singular or not ‘square’ (in the above sense).

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 : ndarray, shape Q

Raises:

LinAlgError

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

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