How can I calculate the nearest positive semi-definite matrix?
I would submit a non-iterative approach. This is slightly modified from Rebonato and Jackel (1999) (page 7-9). Iterative approaches can take a long time to process on matrices of more than a few hundred variables.
import numpy as np
def nearPSD(A,epsilon=0):
n = A.shape[0]
eigval, eigvec = np.linalg.eig(A)
val = np.matrix(np.maximum(eigval,epsilon))
vec = np.matrix(eigvec)
T = 1/(np.multiply(vec,vec) * val.T)
T = np.matrix(np.sqrt(np.diag(np.array(T).reshape((n)) )))
B = T * vec * np.diag(np.array(np.sqrt(val)).reshape((n)))
out = B*B.T
return(out)
Code is modified from a discussion of this topic here around nonPD/PSD matrices in R.
I don't think there is a library which returns the matrix you want, but here is a "just for fun" coding of neareast positive semi-definite matrix algorithm from Higham (2000)
import numpy as np,numpy.linalg
def _getAplus(A):
eigval, eigvec = np.linalg.eig(A)
Q = np.matrix(eigvec)
xdiag = np.matrix(np.diag(np.maximum(eigval, 0)))
return Q*xdiag*Q.T
def _getPs(A, W=None):
W05 = np.matrix(W**.5)
return W05.I * _getAplus(W05 * A * W05) * W05.I
def _getPu(A, W=None):
Aret = np.array(A.copy())
Aret[W > 0] = np.array(W)[W > 0]
return np.matrix(Aret)
def nearPD(A, nit=10):
n = A.shape[0]
W = np.identity(n)
# W is the matrix used for the norm (assumed to be Identity matrix here)
# the algorithm should work for any diagonal W
deltaS = 0
Yk = A.copy()
for k in range(nit):
Rk = Yk - deltaS
Xk = _getPs(Rk, W=W)
deltaS = Xk - Rk
Yk = _getPu(Xk, W=W)
return Yk
When tested on the example from the paper, it returns the correct answer
print nearPD(np.matrix([[2,-1,0,0],[-1,2,-1,0],[0,-1,2,-1],[0,0,-1,2]]),nit=10)
[[ 1. -0.80842467 0.19157533 0.10677227]
[-0.80842467 1. -0.65626745 0.19157533]
[ 0.19157533 -0.65626745 1. -0.80842467]
[ 0.10677227 0.19157533 -0.80842467 1. ]]