.. DO NOT EDIT.
.. THIS FILE WAS AUTOMATICALLY GENERATED BY SPHINX-GALLERY.
.. TO MAKE CHANGES, EDIT THE SOURCE PYTHON FILE:
.. "api/gallery/linalg/svd/plot_02_moore_penrose_pseudoinverse.py"
.. LINE NUMBERS ARE GIVEN BELOW.

.. only:: html

    .. note::
        :class: sphx-glr-download-link-note

        :ref:`Go to the end <sphx_glr_download_api_gallery_linalg_svd_plot_02_moore_penrose_pseudoinverse.py>`
        to download the full example code.

.. rst-class:: sphx-glr-example-title

.. _sphx_glr_api_gallery_linalg_svd_plot_02_moore_penrose_pseudoinverse.py:


The Moore-Penrose pseudoinverse
====================================

For a rectangular or rank-deficient ``A`` there is no inverse, but there
is exactly one matrix ``A^+`` satisfying Penrose's four equations, and
``x = A^+ b`` is the shortest vector among all least-squares solutions.

.. GENERATED FROM PYTHON SOURCE LINES 11-16

.. code-block:: Python

    import matplotlib.pyplot as plt
    import numpy as np

    from mathematicskit.linalg import penrose_residuals, pseudoinverse








.. GENERATED FROM PYTHON SOURCE LINES 17-21

A rank-deficient system
-----------------------------
The third column is the sum of the first two, so A has rank 2 and the
least-squares solutions form a line in R^3.

.. GENERATED FROM PYTHON SOURCE LINES 21-46

.. code-block:: Python


    A = np.array([[1.0, 0.0, 1.0], [0.0, 1.0, 1.0], [1.0, 1.0, 2.0], [1.0, -1.0, 0.0]])
    b = np.array([1.0, 2.0, 2.0, 0.0])
    A_plus = pseudoinverse(A)
    x_plus = A_plus @ b
    print("rank(A) =", np.linalg.matrix_rank(A))
    print("Penrose residuals:", np.array2string(penrose_residuals(A, A_plus), precision=1))
    print("x = A^+ b =", np.round(x_plus, 6))

    null = np.array([1.0, 1.0, -1.0]) / np.sqrt(3.0)  # A @ null = 0
    ts = np.linspace(-2, 2, 201)
    norms = [np.linalg.norm(x_plus + t * null) for t in ts]
    residuals = [np.linalg.norm(A @ (x_plus + t * null) - b) for t in ts]
    print(f"residual along the solution line is constant: {min(residuals):.6f} .. {max(residuals):.6f}")

    fig, ax = plt.subplots(figsize=(6, 4))
    ax.plot(ts, norms, label="||x_ls(t)||")
    ax.axvline(0.0, color="k", ls="--", lw=0.8, label="x = A^+ b")
    ax.set_xlabel("t   (x_ls(t) = A^+ b + t n, with A n = 0)")
    ax.set_ylabel("solution norm")
    ax.set_title("A^+ b is the minimum-norm least-squares solution")
    ax.legend()
    fig.tight_layout()

    plt.show()



.. image-sg:: /api/gallery/linalg/svd/images/sphx_glr_plot_02_moore_penrose_pseudoinverse_001.png
   :alt: A^+ b is the minimum-norm least-squares solution
   :srcset: /api/gallery/linalg/svd/images/sphx_glr_plot_02_moore_penrose_pseudoinverse_001.png
   :class: sphx-glr-single-img


.. rst-class:: sphx-glr-script-out

 .. code-block:: none

    rank(A) = 2
    Penrose residuals: [9.1e-16 2.0e-16 2.5e-16 1.8e-16]
    x = A^+ b = [0.222222 0.555556 0.777778]
    residual along the solution line is constant: 0.816497 .. 0.816497





.. rst-class:: sphx-glr-timing

   **Total running time of the script:** (0 minutes 0.042 seconds)


.. _sphx_glr_download_api_gallery_linalg_svd_plot_02_moore_penrose_pseudoinverse.py:

.. only:: html

  .. container:: sphx-glr-footer sphx-glr-footer-example

    .. container:: sphx-glr-download sphx-glr-download-jupyter

      :download:`Download Jupyter notebook: plot_02_moore_penrose_pseudoinverse.ipynb <plot_02_moore_penrose_pseudoinverse.ipynb>`

    .. container:: sphx-glr-download sphx-glr-download-python

      :download:`Download Python source code: plot_02_moore_penrose_pseudoinverse.py <plot_02_moore_penrose_pseudoinverse.py>`

    .. container:: sphx-glr-download sphx-glr-download-zip

      :download:`Download zipped: plot_02_moore_penrose_pseudoinverse.zip <plot_02_moore_penrose_pseudoinverse.zip>`


.. only:: html

 .. rst-class:: sphx-glr-signature

    `Gallery generated by Sphinx-Gallery <https://sphinx-gallery.github.io>`_