# numpy.linalg.tensorinv¶

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

Find the ‘inverse’ of a N-d array

The result is an inverse corresponding to the operation tensordot(a, b, ind), ie.,

x == tensordot(tensordot(tensorinv(a), a, ind), x, ind)
== tensordot(tensordot(a, tensorinv(a), ind), x, ind)

for all x (up to floating-point accuracy).

Parameters: a : array_like Tensor to ‘invert’. Its shape must ‘square’, ie., prod(a.shape[:ind]) == prod(a.shape[ind:]) ind : integer > 0 How many of the first indices are involved in the inverse sum. b : array, shape a.shape[:ind]+a.shape[ind:] Raises LinAlgError if a is singular or not square :

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
```

#### Previous topic

numpy.linalg.pinv

#### Next topic

numpy.linalg.LinAlgError