Python: computational accuracy when calculating kernel of the matrix

The problem is as follows:
I have an equation: A*x = 0, where A is matrix 8x8, x is vector with 8 elements, and 0 means zero vector. Elements of matrix A contain a parameter E whose values for which the equation is solvable I have to find - and I already did it by using the condition: det(A)=0. And there is the source of my problem - det(A) for found values E is not exactly 0.0 but something very, very near 0.0, for example 9.5e-12. I interpret it as a numerical issue, but I may be wrong?
Next step is to find x. My concept was to find the kernel of the matrix A, but there my problem with non zero det(A) returns, because I do not have E for which my equation can be solved. Is it any way to force Python to operate with approximate values?

Summarizing: I need to find method to determine kernel when A*x is not exactly 0.0 but value is very, very near 0.0.

Edit:
Value of matrix:
[[6.60489454233399, -0.000899873003155720, -1111.26791946547, 0, 0, 0, 0, 0], [8.46748025849121e-8 + 2.5809235665861e-7*I, -9.08063389853990e-10, 0.00112138236223004, 0, 0, 0, 0, 0],
[0, 0.635021913463456, 1.57474880598360, -1.95244188971305, -0.512179135916290, 0, 0, 0],
[0, 6.40801701294518e-7, -1.58908171921515e-6, 2.90298102267184e-6, -7.61531659204450e-7, 0, 0, 0],
[0, 0, 0, 0.512179135916290, 1.95244188971305, -1.57474880598360, -0.635021913463456, 0],
[0, 0, 0, -7.61531659204450e-7, 2.90298102267184e-6, -1.58908171921515e-6, 6.40801701294518e-7, 0],
[0, 0, 0, 0, 0, 784198.968204183, 1.27518657961257e-6, -38.6053002011412],
[0, 0, 0, 0, 0, 0.791336522920740, -1.28679296313874e-12, -1.04862603277821e-5]])

And the code example:

from scipy import *

from numpy.linalg import *

from sympy import *

import sys

import numpy

import cmath

import math



a2= 65e5 #Ae-5

a3= 9e5 #Ae-5

a4= 130e5 #Ae-5



l = -a2-a3/2.0

t = -a3/2.0

f = a3/2.0

g = a4+a3/2.0



print 'Defined thickness'





V1= 0.74015 #eV fabs

V2= 1.1184 #eV fabs

V3= 0.74015 #eV fabs



m= 0.11*0.511e6/(2.99792458e+23)**2 #eV*s**2/A**2



hkr= 6.582119514e-16 #eV*s



x= 0.765705051154







print 'other symbols'



k1=(cmath.sqrt(2.0*(V1-x)*m))/hkr

k2=(cmath.sqrt(2.0*(V2-x)*m))/hkr

k3=(cmath.sqrt(-2.0*x*m))/hkr

k4=(cmath.sqrt(2.0*(V2-x)*m))/hkr

k5=(cmath.sqrt(2.0*(V3-x)*m))/hkr



print 'k-vectors'



a11 = cmath.exp(1.0j*k1*l)

a12 = -1.0*cmath.exp(k2*l)

a13 = -1.0*cmath.exp(-k2*l)



print '1st row'



a21 = 1.0j*k1*cmath.exp(k1*l)

a22 = -k2*cmath.exp(k2*l)

a23 = k2*cmath.exp(-k2*l)



print '2nd row'



a32 = cmath.exp(k2*t)

a33 = cmath.exp(-k2*t)

a34 = -1.0*cmath.exp(1.0j*k3*t)

a35 = -1.0*cmath.exp(-1.0j*k3*t)



print '3rd row'



a42 = k2*cmath.exp(k2*t)

a43 = -k2*cmath.exp(-k2*t)

a44 = -1.0j*k3*cmath.exp(1.0j*k3*t)

a45 = 1.0j*k3*cmath.exp(-1.0j*k3*t)



print '4th row'



a54 = cmath.exp(1.0j*k3*f)

a55 = cmath.exp(-1.0j*k3*f)

a56 = -1.0*cmath.exp(k4*f)

a57 = -1.0*cmath.exp(-k4*f)



print '5th row'



a64 = 1.0j*k3*cmath.exp(1.0j*k3*f)

a65 = -1.0j*k3*cmath.exp(-1.0j*k3*f)

a66 = -k4*cmath.exp(k4*f)

a67 = k4*cmath.exp(-k4*f)



print '6th row'



a76 = cmath.exp(k4*g)

a77 = cmath.exp(-k4*g)

a78 = -1.0*cmath.exp(-1.0j*k5*g)



print '7th row'



a86 = k4*cmath.exp(k4*g)

a87 = -k4*cmath.exp(-k4*g)

a88 = 1.0j*k5*cmath.exp(-1.0j*k5*g)

print '8th row'





M = Matrix([[a11,a12,a13,0,0,0,0,0],

        [a21,a22,a23,0,0,0,0,0],

        [0,a32,a33,a34,a35,0,0,0],      

        [0,a42,a43,a44,a45,0,0,0],

        [0,0,0,a54,a55,a56,a57,0],

        [0,0,0,a64,a65,a66,a67,0],            

        [0,0,0,0,0,a76,a77,a78],

        [0,0,0,0,0,a86,a87,a88]])





v=M.nullspace()

m=lcm([val.q for val in v])

PSI=m*v

print M

print PSI

edited Nov 20 '18 at 10:30

asked Nov 19 '18 at 13:53

user3419643

Could you post the value of your matrix A and a code example of what you have done so far?
– MarAja
Nov 19 '18 at 14:17

Have you checked math.isclose(0., det(A))?
– bene
Nov 19 '18 at 14:30

@MarAja done, in edited post.
– user3419643
Nov 20 '18 at 10:33

@bene - no, I suppose it won't help, coz the problem is, that because of a little bit of numerical problems I don't have "right" matrix during calculation of kernel.
– user3419643
Nov 20 '18 at 10:33

add a comment |

Summarizing: I need to find method to determine kernel when A*x is not exactly 0.0 but value is very, very near 0.0.

And the code example:

from scipy import *

from numpy.linalg import *

from sympy import *

import sys

import numpy

import cmath

import math



a2= 65e5 #Ae-5

a3= 9e5 #Ae-5

a4= 130e5 #Ae-5



l = -a2-a3/2.0

t = -a3/2.0

f = a3/2.0

g = a4+a3/2.0



print 'Defined thickness'





V1= 0.74015 #eV fabs

V2= 1.1184 #eV fabs

V3= 0.74015 #eV fabs



m= 0.11*0.511e6/(2.99792458e+23)**2 #eV*s**2/A**2



hkr= 6.582119514e-16 #eV*s



x= 0.765705051154







print 'other symbols'



k1=(cmath.sqrt(2.0*(V1-x)*m))/hkr

k2=(cmath.sqrt(2.0*(V2-x)*m))/hkr

k3=(cmath.sqrt(-2.0*x*m))/hkr

k4=(cmath.sqrt(2.0*(V2-x)*m))/hkr

k5=(cmath.sqrt(2.0*(V3-x)*m))/hkr



print 'k-vectors'



a11 = cmath.exp(1.0j*k1*l)

a12 = -1.0*cmath.exp(k2*l)

a13 = -1.0*cmath.exp(-k2*l)



print '1st row'



a21 = 1.0j*k1*cmath.exp(k1*l)

a22 = -k2*cmath.exp(k2*l)

a23 = k2*cmath.exp(-k2*l)



print '2nd row'



a32 = cmath.exp(k2*t)

a33 = cmath.exp(-k2*t)

a34 = -1.0*cmath.exp(1.0j*k3*t)

a35 = -1.0*cmath.exp(-1.0j*k3*t)



print '3rd row'



a42 = k2*cmath.exp(k2*t)

a43 = -k2*cmath.exp(-k2*t)

a44 = -1.0j*k3*cmath.exp(1.0j*k3*t)

a45 = 1.0j*k3*cmath.exp(-1.0j*k3*t)



print '4th row'



a54 = cmath.exp(1.0j*k3*f)

a55 = cmath.exp(-1.0j*k3*f)

a56 = -1.0*cmath.exp(k4*f)

a57 = -1.0*cmath.exp(-k4*f)



print '5th row'



a64 = 1.0j*k3*cmath.exp(1.0j*k3*f)

a65 = -1.0j*k3*cmath.exp(-1.0j*k3*f)

a66 = -k4*cmath.exp(k4*f)

a67 = k4*cmath.exp(-k4*f)



print '6th row'



a76 = cmath.exp(k4*g)

a77 = cmath.exp(-k4*g)

a78 = -1.0*cmath.exp(-1.0j*k5*g)



print '7th row'



a86 = k4*cmath.exp(k4*g)

a87 = -k4*cmath.exp(-k4*g)

a88 = 1.0j*k5*cmath.exp(-1.0j*k5*g)

print '8th row'





M = Matrix([[a11,a12,a13,0,0,0,0,0],

        [a21,a22,a23,0,0,0,0,0],

        [0,a32,a33,a34,a35,0,0,0],      

        [0,a42,a43,a44,a45,0,0,0],

        [0,0,0,a54,a55,a56,a57,0],

        [0,0,0,a64,a65,a66,a67,0],            

        [0,0,0,0,0,a76,a77,a78],

        [0,0,0,0,0,a86,a87,a88]])





v=M.nullspace()

m=lcm([val.q for val in v])

PSI=m*v

print M

print PSI

edited Nov 20 '18 at 10:30

asked Nov 19 '18 at 13:53

user3419643

Could you post the value of your matrix A and a code example of what you have done so far?
– MarAja
Nov 19 '18 at 14:17

Have you checked math.isclose(0., det(A))?
– bene
Nov 19 '18 at 14:30

@MarAja done, in edited post.
– user3419643
Nov 20 '18 at 10:33

@bene - no, I suppose it won't help, coz the problem is, that because of a little bit of numerical problems I don't have "right" matrix during calculation of kernel.
– user3419643
Nov 20 '18 at 10:33

add a comment |

Summarizing: I need to find method to determine kernel when A*x is not exactly 0.0 but value is very, very near 0.0.

And the code example:

from scipy import *

from numpy.linalg import *

from sympy import *

import sys

import numpy

import cmath

import math



a2= 65e5 #Ae-5

a3= 9e5 #Ae-5

a4= 130e5 #Ae-5



l = -a2-a3/2.0

t = -a3/2.0

f = a3/2.0

g = a4+a3/2.0



print 'Defined thickness'





V1= 0.74015 #eV fabs

V2= 1.1184 #eV fabs

V3= 0.74015 #eV fabs



m= 0.11*0.511e6/(2.99792458e+23)**2 #eV*s**2/A**2



hkr= 6.582119514e-16 #eV*s



x= 0.765705051154







print 'other symbols'



k1=(cmath.sqrt(2.0*(V1-x)*m))/hkr

k2=(cmath.sqrt(2.0*(V2-x)*m))/hkr

k3=(cmath.sqrt(-2.0*x*m))/hkr

k4=(cmath.sqrt(2.0*(V2-x)*m))/hkr

k5=(cmath.sqrt(2.0*(V3-x)*m))/hkr



print 'k-vectors'



a11 = cmath.exp(1.0j*k1*l)

a12 = -1.0*cmath.exp(k2*l)

a13 = -1.0*cmath.exp(-k2*l)



print '1st row'



a21 = 1.0j*k1*cmath.exp(k1*l)

a22 = -k2*cmath.exp(k2*l)

a23 = k2*cmath.exp(-k2*l)



print '2nd row'



a32 = cmath.exp(k2*t)

a33 = cmath.exp(-k2*t)

a34 = -1.0*cmath.exp(1.0j*k3*t)

a35 = -1.0*cmath.exp(-1.0j*k3*t)



print '3rd row'



a42 = k2*cmath.exp(k2*t)

a43 = -k2*cmath.exp(-k2*t)

a44 = -1.0j*k3*cmath.exp(1.0j*k3*t)

a45 = 1.0j*k3*cmath.exp(-1.0j*k3*t)



print '4th row'



a54 = cmath.exp(1.0j*k3*f)

a55 = cmath.exp(-1.0j*k3*f)

a56 = -1.0*cmath.exp(k4*f)

a57 = -1.0*cmath.exp(-k4*f)



print '5th row'



a64 = 1.0j*k3*cmath.exp(1.0j*k3*f)

a65 = -1.0j*k3*cmath.exp(-1.0j*k3*f)

a66 = -k4*cmath.exp(k4*f)

a67 = k4*cmath.exp(-k4*f)



print '6th row'



a76 = cmath.exp(k4*g)

a77 = cmath.exp(-k4*g)

a78 = -1.0*cmath.exp(-1.0j*k5*g)



print '7th row'



a86 = k4*cmath.exp(k4*g)

a87 = -k4*cmath.exp(-k4*g)

a88 = 1.0j*k5*cmath.exp(-1.0j*k5*g)

print '8th row'





M = Matrix([[a11,a12,a13,0,0,0,0,0],

        [a21,a22,a23,0,0,0,0,0],

        [0,a32,a33,a34,a35,0,0,0],      

        [0,a42,a43,a44,a45,0,0,0],

        [0,0,0,a54,a55,a56,a57,0],

        [0,0,0,a64,a65,a66,a67,0],            

        [0,0,0,0,0,a76,a77,a78],

        [0,0,0,0,0,a86,a87,a88]])





v=M.nullspace()

m=lcm([val.q for val in v])

PSI=m*v

print M

print PSI

edited Nov 20 '18 at 10:30

asked Nov 19 '18 at 13:53

user3419643

Summarizing: I need to find method to determine kernel when A*x is not exactly 0.0 but value is very, very near 0.0.

And the code example:

from scipy import *

from numpy.linalg import *

from sympy import *

import sys

import numpy

import cmath

import math



a2= 65e5 #Ae-5

a3= 9e5 #Ae-5

a4= 130e5 #Ae-5



l = -a2-a3/2.0

t = -a3/2.0

f = a3/2.0

g = a4+a3/2.0



print 'Defined thickness'





V1= 0.74015 #eV fabs

V2= 1.1184 #eV fabs

V3= 0.74015 #eV fabs



m= 0.11*0.511e6/(2.99792458e+23)**2 #eV*s**2/A**2



hkr= 6.582119514e-16 #eV*s



x= 0.765705051154







print 'other symbols'



k1=(cmath.sqrt(2.0*(V1-x)*m))/hkr

k2=(cmath.sqrt(2.0*(V2-x)*m))/hkr

k3=(cmath.sqrt(-2.0*x*m))/hkr

k4=(cmath.sqrt(2.0*(V2-x)*m))/hkr

k5=(cmath.sqrt(2.0*(V3-x)*m))/hkr



print 'k-vectors'



a11 = cmath.exp(1.0j*k1*l)

a12 = -1.0*cmath.exp(k2*l)

a13 = -1.0*cmath.exp(-k2*l)



print '1st row'



a21 = 1.0j*k1*cmath.exp(k1*l)

a22 = -k2*cmath.exp(k2*l)

a23 = k2*cmath.exp(-k2*l)



print '2nd row'



a32 = cmath.exp(k2*t)

a33 = cmath.exp(-k2*t)

a34 = -1.0*cmath.exp(1.0j*k3*t)

a35 = -1.0*cmath.exp(-1.0j*k3*t)



print '3rd row'



a42 = k2*cmath.exp(k2*t)

a43 = -k2*cmath.exp(-k2*t)

a44 = -1.0j*k3*cmath.exp(1.0j*k3*t)

a45 = 1.0j*k3*cmath.exp(-1.0j*k3*t)



print '4th row'



a54 = cmath.exp(1.0j*k3*f)

a55 = cmath.exp(-1.0j*k3*f)

a56 = -1.0*cmath.exp(k4*f)

a57 = -1.0*cmath.exp(-k4*f)



print '5th row'



a64 = 1.0j*k3*cmath.exp(1.0j*k3*f)

a65 = -1.0j*k3*cmath.exp(-1.0j*k3*f)

a66 = -k4*cmath.exp(k4*f)

a67 = k4*cmath.exp(-k4*f)



print '6th row'



a76 = cmath.exp(k4*g)

a77 = cmath.exp(-k4*g)

a78 = -1.0*cmath.exp(-1.0j*k5*g)



print '7th row'



a86 = k4*cmath.exp(k4*g)

a87 = -k4*cmath.exp(-k4*g)

a88 = 1.0j*k5*cmath.exp(-1.0j*k5*g)

print '8th row'





M = Matrix([[a11,a12,a13,0,0,0,0,0],

        [a21,a22,a23,0,0,0,0,0],

        [0,a32,a33,a34,a35,0,0,0],      

        [0,a42,a43,a44,a45,0,0,0],

        [0,0,0,a54,a55,a56,a57,0],

        [0,0,0,a64,a65,a66,a67,0],            

        [0,0,0,0,0,a76,a77,a78],

        [0,0,0,0,0,a86,a87,a88]])





v=M.nullspace()

m=lcm([val.q for val in v])

PSI=m*v

print M

print PSI

python kernel approximation

edited Nov 20 '18 at 10:30

asked Nov 19 '18 at 13:53

user3419643

edited Nov 20 '18 at 10:30

asked Nov 19 '18 at 13:53

user3419643

edited Nov 20 '18 at 10:30

asked Nov 19 '18 at 13:53

user3419643

asked Nov 19 '18 at 13:53

user3419643

asked Nov 19 '18 at 13:53

user3419643

Could you post the value of your matrix A and a code example of what you have done so far?
– MarAja
Nov 19 '18 at 14:17

Have you checked math.isclose(0., det(A))?
– bene
Nov 19 '18 at 14:30

@MarAja done, in edited post.
– user3419643
Nov 20 '18 at 10:33

@bene - no, I suppose it won't help, coz the problem is, that because of a little bit of numerical problems I don't have "right" matrix during calculation of kernel.
– user3419643
Nov 20 '18 at 10:33

add a comment |

Could you post the value of your matrix A and a code example of what you have done so far?
– MarAja
Nov 19 '18 at 14:17

Have you checked math.isclose(0., det(A))?
– bene
Nov 19 '18 at 14:30

@MarAja done, in edited post.
– user3419643
Nov 20 '18 at 10:33

@bene - no, I suppose it won't help, coz the problem is, that because of a little bit of numerical problems I don't have "right" matrix during calculation of kernel.
– user3419643
Nov 20 '18 at 10:33

Could you post the value of your matrix A and a code example of what you have done so far?
– MarAja
Nov 19 '18 at 14:17

Have you checked math.isclose(0., det(A))?
– bene
Nov 19 '18 at 14:30

@MarAja done, in edited post.
– user3419643
Nov 20 '18 at 10:33

@bene - no, I suppose it won't help, coz the problem is, that because of a little bit of numerical problems I don't have "right" matrix during calculation of kernel.
– user3419643
Nov 20 '18 at 10:33

add a comment |

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