symjax.tensor.linalg.tensorsolve¶
-
symjax.tensor.linalg.
tensorsolve
(a, b, axes=None)[source]¶ Solve the tensor equation
a x = b
for x.LAX-backend implementation of
tensorsolve()
. Original docstring below.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: - 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 thatprod(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).See also
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
- a (array_like) – Coefficient tensor, of shape