# jax.numpy.linalg.pinv¶

jax.numpy.linalg.pinv = <jax.custom_derivatives.custom_jvp object>[source]

Compute the (Moore-Penrose) pseudo-inverse of a matrix.

LAX-backend implementation of pinv(). It differs only in default value of rcond. In numpy.linalg.pinv, the default rcond is 1e-15. Here the default is 10. * max(num_rows, num_cols) * jnp.finfo(dtype).eps.

Original docstring below.

Calculate the generalized inverse of a matrix using its singular-value decomposition (SVD) and including all large singular values.

Changed in version 1.14: Can now operate on stacks of matrices

Parameters
• a ((.., M, N) array_like) – Matrix or stack of matrices to be pseudo-inverted.

• rcond ((..) array_like of float) – Cutoff for small singular values. Singular values less than or equal to rcond * largest_singular_value are set to zero. Broadcasts against the stack of matrices.

Returns

B – The pseudo-inverse of a. If a is a matrix instance, then so is B.

Return type

(.., N, M) ndarray

Raises

LinAlgError – If the SVD computation does not converge.

Notes

The pseudo-inverse of a matrix A, denoted $$A^+$$, is defined as: “the matrix that ‘solves’ [the least-squares problem] $$Ax = b$$,” i.e., if $$\bar{x}$$ is said solution, then $$A^+$$ is that matrix such that $$\bar{x} = A^+b$$.

It can be shown that if $$Q_1 \Sigma Q_2^T = A$$ is the singular value decomposition of A, then $$A^+ = Q_2 \Sigma^+ Q_1^T$$, where $$Q_{1,2}$$ are orthogonal matrices, $$\Sigma$$ is a diagonal matrix consisting of A’s so-called singular values, (followed, typically, by zeros), and then $$\Sigma^+$$ is simply the diagonal matrix consisting of the reciprocals of A’s singular values (again, followed by zeros). 1

References

1

G. Strang, Linear Algebra and Its Applications, 2nd Ed., Orlando, FL, Academic Press, Inc., 1980, pp. 139-142.

Examples

The following example checks that a * a+ * a == a and a+ * a * a+ == a+:

>>> a = np.random.randn(9, 6)
>>> B = np.linalg.pinv(a)
>>> np.allclose(a, np.dot(a, np.dot(B, a)))
True
>>> np.allclose(B, np.dot(B, np.dot(a, B)))
True