#### Previous topic

numpy.linalg.pinv

#### Next topic

numpy.linalg.LinAlgError

# numpy.linalg.tensorinv

`numpy.linalg.``tensorinv`(a, ind=2)

  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: | a : array_like Tensor to ‘invert’. Its shape must be ‘square’, i. e., `prod(a.shape[:ind]) == prod(a.shape[ind:])`. ind : int, optional Number of first indices that are involved in the inverse sum. Must be a positive integer, default is 2.
  Returns: | b : ndarray 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

  `tensordot`, `tensorsolve`

  Examples

      >>> 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)
      >>> np.allclose(np.tensordot(ainv, b, 1), np.linalg.tensorsolve(a, b))
      True