Singular matrix - python

Question

The following code shows a problem of singularity of a matrix, since working in Pycharm I get

raise LinAlgError("Singular matrix")
numpy.linalg.linalg.LinAlgError: Singular matrix 

I guess the problem is K but I cannot understand exactly how:

from numpy import zeros
from numpy.linalg import linalg
import math

def getA(kappa):
    matrix = zeros((n, n), float)
    for i in range(n):
    for j in range(n):
            matrix[i][j] = 2*math.cos((2*math.pi/n)*(abs(j-i))*kappa)
    return matrix


def getF(csi, a):
    csiInv = linalg.inv(csi)
    valueF = csiInv * a * csiInv * a
    traceF = valueF.trace()
    return 0.5 * traceF


def getG(csi, f, a):
    csiInv = linalg.inv(csi)

    valueG = (csiInv * a * csiInv) / (2 * f)
    return valueG


def getE(g, k):
    KInv = linalg.inv(k)
    Ktrans = linalg.transpose(k)
    KtransInv = linalg.inv(Ktrans)
    e = KtransInv * g * KInv
    return e


file = open('transformed.txt', 'r')
n = 4
transformed = zeros(n)

for counter, line in enumerate(file):
    if counter == n:
        break
    transformed[counter] = float(line)

CSI = zeros((n, n))
for i in range(n):
    for j in range(n):
        CSI[i][j] = transformed[abs(i-j)]

A = getA(1)
F = getF(CSI, A)
G = getG(CSI, F, A)

K = zeros((n, n), float)
for j in range(n):
    K[0][j] = 0.0001

for i in range(1, n):
    for j in range(n):
        K[i][j] = ((3.0*70.0*70.0*0.3)/(2.0*300000.0*300000.0))*((j*(i-j))/i)*(1.0+(70.0/300000.0)*j)



E = getE(G, K)

print G
print K

Does anyone has any suggestions to fix it? Thank you

Answer 1

Inverting matrices that are very "close" to being singular often causes computation problems. A quick hack is to add a very small value to the diagonal of your matrix before inversion.

def getE(g, k):
    m = 10^-6
    KInv = linalg.inv(k + numpy.eye(k.shape[1])*m)
    Ktrans = linalg.transpose(k)
    KtransInv = linalg.inv(Ktrans + + numpy.eye(Ktrans.shape[1])*m)
    e = KtransInv * g * KInv
    return e

I think of that as being good enough for homework. But if you want to really deploy something computationally robust, you should look into alternatives to inverting.

numerically stable inverse of a 2x2 matrix