- 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=len(b.shape)).
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.
x : ndarray, shape Q
If a is singular or not ‘square’ (in the above sense).
tensordot, tensorinv, 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) True