numpy.linalg.tensorinv

linalg.tensorinv(a, ind=2)[source]

Compute the ‘inverse’ of an N-dimensional array.

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.

Parameters:

aarray_like

Tensor to ‘invert’. Its shape must be ‘square’, i. e., prod(a.shape[:ind]) == prod(a.shape[ind:]).

indint, optional

Number of first indices that are involved in the inverse sum. Must be a positive integer, default is 2.

Returns:

bndarray

a’s tensordot inverse, shape a.shape[ind:] + a.shape[:ind].

Raises:

LinAlgError

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

See also

numpy.tensordot, tensorsolve

Examples

Try it in your browser!

>>> import numpy as np
>>> a = np.eye(4*6).reshape((4, 6, 8, 3))
>>> ainv = np.linalg.tensorinv(a, ind=2)
>>> ainv.shape
(8, 3, 4, 6)
>>> rng = np.random.default_rng()
>>> b = rng.normal(size=(4, 6))
>>> np.allclose(np.tensordot(ainv, b), np.linalg.tensorsolve(a, b))
True

>>> a = np.eye(4*6).reshape((24, 8, 3))
>>> ainv = np.linalg.tensorinv(a, ind=1)
>>> ainv.shape
(8, 3, 24)
>>> rng = np.random.default_rng()
>>> b = rng.normal(size=24)
>>> np.allclose(np.tensordot(ainv, b, 1), np.linalg.tensorsolve(a, b))
True

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