# numpy.linalg.tensorinv

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

  `numpy.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

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