tensorinv(a, ind=2)
The result is an inverse for a relative to the tensordot operation tensordot(a, b, ind)
, i. e., up to floating-point accuracy, tensordot(tensorinv(a), a, ind)
is the "identity" tensor for the tensordot operation.
Tensor to 'invert'. Its shape must be 'square', i. e., prod(a.shape[:ind]) == prod(a.shape[ind:])
.
Number of first indices that are involved in the inverse sum. Must be a positive integer, default is 2.
If a is singular or not 'square' (in the above sense).
Compute the 'inverse' of an N-dimensional array.
>>> a = np.eye(4*6)
... a.shape = (4, 6, 8, 3)
... ainv = np.linalg.tensorinv(a, ind=2)
... ainv.shape (8, 3, 4, 6)
>>> b = np.random.randn(4, 6)
... np.allclose(np.tensordot(ainv, b), np.linalg.tensorsolve(a, b)) True
>>> a = np.eye(4*6)
... a.shape = (24, 8, 3)
... ainv = np.linalg.tensorinv(a, ind=1)
... ainv.shape (8, 3, 24)
>>> b = np.random.randn(24)See :
... np.allclose(np.tensordot(ainv, b, 1), np.linalg.tensorsolve(a, b)) True
The following pages refer to to this document either explicitly or contain code examples using this.
numpy.linalg.tensorsolve
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