Why is numpy.linalg.pinv() preferred over numpy.linalg.inv() for creating inverse of a matrix in linear regression

Question

If we want to search for the optimal parameters theta for a linear regression model by using the normal equation with:

theta = inv(X^T * X) * X^T * y

one step is to calculate inv(X^T*X). Therefore numpy provides np.linalg.inv() and np.linalg.pinv()

Though this leads to different results:

X=np.matrix([[1,2104,5,1,45],[1,1416,3,2,40],[1,1534,3,2,30],[1,852,2,1,36]])
y=np.matrix([[460],[232],[315],[178]])

XT=X.T
XTX=XT@X

pinv=np.linalg.pinv(XTX)
theta_pinv=(pinv@XT)@y
print(theta_pinv)

[[188.40031946]
 [  0.3866255 ]
 [-56.13824955]
 [-92.9672536 ]
 [ -3.73781915]]

inv=np.linalg.inv(XTX)
theta_inv=(inv@XT)@y
print(theta_inv)

[[-648.7890625 ]
 [   0.79418945]
 [-110.09375   ]
 [ -74.0703125 ]
 [  -3.69091797]]

The first output, that is the output of pinv is the correct one and additionally recommended in the numpy.linalg.pinv() docs. But why is this and where are the differences / Pros / Cons between inv() and pinv().

Answer 1

If the determinant of the matrix is zero it will not have an inverse and your inv function will not work. This usually happens if your matrix is singular.

But pinv will. This is because pinv returns the inverse of your matrix when it is available and the pseudo inverse when it isn't.

The different results of the functions are because of rounding errors in floating point arithmetic

You can read more about how pseudo inverse works here

Answer 2

inv and pinv are used to compute the (pseudo)-inverse as a standalone matrix. Not to actually use them in the computations.

For such linear system solutions the proper tool to use is numpy.linalg.lstsq (or from scipy) if you have a non invertible coefficient matrix or numpy.linalg.solve (or from scipy) for invertible matrices.

Answer 3

Inverting X^T X is not as numerically stable as computing the pseudo-inverse using numpy's function when X^T X has relatively very small singular values, which can occur when a matrix is "almost" rank-deficient but isn't due to noise. This occurs in particular when the condition number, that is the ratio of max sv to min sv, is large.