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# In[1]:
import numpy as np
import scipy.linalg as la
# Let's set up some matrices and data for the *rank-one modification*:
# In[6]:
n = 5
A = np.random.randn(n, n)
u = np.random.randn(n)
v = np.random.randn(n)
b = np.random.randn(n)
Ahat = A + np.outer(u, v)
# Let's start by computing the "base" factorization.
#
# We'll use `lu_factor` from `scipy`, which stuffs both `L` and `U` into a single matrix (why can it do that?) and also returns pivoting information:
# In[7]:
LU, piv = la.lu_factor(A)
print(LU)
print(piv)
# Next, we set up a subroutine to solve using that factorization and check that it works:
# In[9]:
def solveA(b):
return la.lu_solve((LU, piv), b)
la.norm(np.dot(A, solveA(b)) - b)
# As a last step, we try the Sherman-Morrison formula:
#
# $$(A+uv^T)^{-1} = A^{-1} - {A^{-1}uv^T A^{-1} \over 1 + v^T A^{-1}u}$$
# To see that we got the right answer, we first compute the right solution of the modified system:
# In[11]:
xhat = la.solve(Ahat, b)
# Next, apply Sherman-Morrison to find `xhat2`:
# In[13]:
xhat2 = solveA(b) - solveA(u)*np.dot(v, solveA(b))/(1+np.dot(v, solveA(u)))
# In[14]:
la.norm(xhat - xhat2)
# * What's the cost of the Sherman-Morrison procedure?