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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_valueare set to zero. Broadcasts against the stack of matrices.
- hermitianbool, optional
- 
        If True, ais 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. Ifais amatrixinstance, then so isB.
 
- 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. 
 NotesThe pseudo-inverse of a matrix A, denoted , is defined as: “the matrix that ‘solves’ [the least-squares problem] ,” i.e., if is said solution, then is that matrix such that . It can be shown that if is the singular value decomposition of A, then , where are orthogonal matrices, is a diagonal matrix consisting of A’s so-called singular values, (followed, typically, by zeros), and then 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. 
 ExamplesThe following example checks that a * a+ * a == aanda+ * 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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 https://numpy.org/doc/1.19/reference/generated/numpy.linalg.pinv.html