Normal Equations vs Pseudoinverse¶

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import numpy as np

import numpy.linalg as la

Here's a simple overdetermined linear system, which we'll solve using both the normal equations and the pseudoinverse:

A = np.random.randn(5, 3)

b = np.random.randn(5)

Normal Equations¶

Solve $Ax\cong b$ using the normal equations:

#clear

x1 = la.solve(A.T@A, A.T@b)

x1

array([ 0.15751782,  0.37883556,  1.02377709])

Pseudoinverse¶

Solve $Ax\cong b$ using the pseudoinverse:

U, sigma, VT = la.svd(A)

print(U)

print(sigma)

print(VT)

[[ 0.40189265  0.43776205 -0.32197721 -0.68693466 -0.26701711]
[-0.60148796 -0.48057566 -0.36776893 -0.32823601 -0.40529791]
[-0.50534567  0.25317047  0.34332919 -0.50930065  0.55069808]
[-0.46931436  0.7148832  -0.11893232  0.37004742 -0.34293739]
[-0.03262382  0.04751987 -0.79313105  0.15511173  0.5861408 ]]
[ 2.48597339  2.0073724   1.09629545]
[[-0.16519593 -0.85534195  0.49101981]
[ 0.90601342 -0.3283188  -0.26710758]
[ 0.38967935  0.40074545  0.82918821]]

Sigma_inv = np.zeros_like(A.T)

Sigma_inv[:3,:3] = np.diag(1/sigma)

Sigma_inv

array([[ 0.40225692,  0.        ,  0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.49816367,  0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.        ,  0.91216287,  0.        ,  0.        ]])

#clear

pinv = VT.T @ Sigma_inv @ U.T

x2 = pinv @ b

x2

array([ 0.15751782,  0.37883556,  1.02377709])

la.norm(x1-x2)

4.7428748402675471e-16