Why does numpy.linalg.solve() offer more precise matrix inversions than numpy.linalg.inv()?

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Question :

Why does numpy.linalg.solve() offer more precise matrix inversions than numpy.linalg.inv()?

I do not quite understand why numpy.linalg.solve() gives the more precise answer, whereas numpy.linalg.inv() breaks down somewhat, giving (what I believe are) estimates.

For a concrete example, I am solving the equation C^{-1} * d where C denotes a matrix, and d is a vector-array. For the sake of discussion, the dimensions of C are shape (1000,1000) and d is shape (1,1000).

numpy.linalg.solve(A, b) solves the equation A*x=b for x, i.e. x = A^{-1} * b. Therefore, I could either solve this equation by


inverse = numpy.linalg.inv(C)
result = inverse * d

or (2)

numpy.linalg.solve(C, d)

Method (2) gives far more precise results. Why is this?

What exactly is happening such that one “works better” than the other?

Answer #1:

np.linalg.solve(A, b) does not compute the inverse of A. Instead it calls one of the gesv LAPACK routines, which first factorizes A using LU decomposition, then solves for x using forward and backward substitution (see here).

np.linalg.inv uses the same method to compute the inverse of A by solving for A-1 in A·A-1 = I where I is the identity*. The factorization step is exactly the same as above, but it takes more floating point operations to solve for A-1 (an n×n matrix) than for x (an n-long vector). Additionally, if you then wanted to obtain x via the identity A-1·b = x then the extra matrix multiplication would incur yet more floating point operations, and therefore slower performance and more numerical error.

There’s no need for the intermediate step of computing A-1 – it is faster and more accurate to obtain x directly.

* The relevant bit of source for inv is here. Unfortunately it’s a bit tricky to understand since it’s templated C. The important thing to note is that an identity matrix is being passed to the LAPACK solver as parameter B.

Answered By: ali_m

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