numpy.linalg.pinv¶

numpy.linalg.pinv(a, rcond=1e-15, hermitian=False)[source]¶

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

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.
hermitianbool, optional: If True, a is assumed to be Hermitian (symmetric if real-valued), enabling a more efficient method for finding singular values. Defaults to False.

New in version 1.17.0.

Returns

B(…, N, M) ndarray: The pseudo-inverse of a. If a is a matrix instance, then so is B.

Raises

LinAlgError: If the SVD computation does not converge.

See also

scipy.linalg.pinv: Similar function in SciPy.
scipy.linalg.pinv2: Similar function in SciPy (SVD-based).
scipy.linalg.pinvh: Compute the (Moore-Penrose) pseudo-inverse of a Hermitian matrix.

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

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numpy.linalg.pinv¶