tensorsolve(a, b, axes=None)
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)
.
Coefficient tensor, of shape b.shape + Q
. :None:None:`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').
Right-hand tensor, which can be of any shape.
Axes in a to reorder to the right, before inversion. If None (default), no reordering is done.
If a is singular or not 'square' (in the above sense).
Solve the tensor equation a x = b
for x.
>>> 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) TrueSee :
The following pages refer to to this document either explicitly or contain code examples using this.
numpy.linalg.tensorinv
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