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CVXOPT and portfolio optimization: puzzling issue

I am trying to solve the following simple optimization problem. One seeks to find the global minimum variance portfolio, being the portfolio that minimizes variance with only one constraint : weights must sum to one.
Optimization program
This problem has a well-known closed-form solution: Solution
I'm trying to reproduce the results using CVXopt in Python, and I encounter a puzzling issue.
I have a 10x10 sample covariance matrix.
If I solve the problem with the entire 10x10 matrix, the output is incorrect : the "sum-to-one" constraint is not respected, and weights are different from the closed-form solution
If I solve the problem with a 7x7 subsample of the same matrix, the output is correct : the "sum-to-one" constraint is respected, and weights are equivalent to the closed-form solution
Actually, with any subsample of size equal or lower than 7x7, it works. But for any subsample of size higher or equal to 8x8, it does not work anymore. I can't grasp where the problem is coming from.
Thank you for your help!
Here is a sample code
%reset -sf #Clear Environment
import numpy as np
import cvxopt
from cvxopt import matrix as dmatrix
from cvxopt.solvers import qp, options
from numpy.linalg import inv
from numpy import ones
from numpy import transpose as t
# Sample 10x10 covariance matrix
sigmafull = np.array([[0.01449082, 0.00846992, 0.00846171, 0.00773097, 0.00878925,
0.00748843, 0.00672341, 0.00665912, 0.0068593 , 0.00827341],
[0.00846992, 0.00952205, 0.00766057, 0.00726647, 0.00781524,
0.00672368, 0.00642426, 0.00609368, 0.00617965, 0.00704281],
[0.00846171, 0.00766057, 0.00842194, 0.00700168, 0.00772423,
0.0061137 , 0.00612574, 0.00601041, 0.00621007, 0.00712152],
[0.00773097, 0.00726647, 0.00700168, 0.00687784, 0.00726901,
0.00573606, 0.00567145, 0.00556391, 0.00575279, 0.00660916],
[0.00878925, 0.00781524, 0.00772423, 0.00726901, 0.00860462,
0.00612804, 0.0061301 , 0.00603605, 0.00630947, 0.0075281 ],
[0.00748843, 0.00672368, 0.0061137 , 0.00573606, 0.00612804,
0.00634431, 0.0054793 , 0.00513665, 0.00511852, 0.00575049],
[0.00672341, 0.00642426, 0.00612574, 0.00567145, 0.0061301 ,
0.0054793 , 0.0055722 , 0.0050824 , 0.00512499, 0.00576934],
[0.00665912, 0.00609368, 0.00601041, 0.00556391, 0.00603605,
0.00513665, 0.0050824 , 0.00521583, 0.00510142, 0.00576414],
[0.0068593 , 0.00617965, 0.00621007, 0.00575279, 0.00630947,
0.00511852, 0.00512499, 0.00510142, 0.00547566, 0.00603528],
[0.00827341, 0.00704281, 0.00712152, 0.00660916, 0.0075281 ,
0.00575049, 0.00576934, 0.00576414, 0.00603528, 0.00756009]])
# sigma = sigmafull[0:8,0:8] #With this subsample, output is incorrect. n=8 (and for all n>8)
sigma = sigmafull[0:7,0:7] #With this subsample, output is correct. n=7 (and for all n<7)
n=len(sigma)
sigma = dmatrix(sigma) #Formatting sigma to be a dense matrix for cvxopt
mu = dmatrix(np.zeros(n)) #We just want to minimize variance, hence vector of zeroes
#Format of the equality constraint : Ax = b
#We want the sum of x to be equal to 1
Amatrix = dmatrix(ones(n)).T #Vector of ones
bmatrix = dmatrix(1.0) #Scalar = 1
sol = qp(sigma, mu, None, None, A=Amatrix, b=bmatrix) #No inequality constraint
w_gmv = (inv(sigma)#ones(n))/(t(ones(n))#inv(sigma)#ones(n)) #Analytical solution which indeed does sum to 1
print(t(np.array(sol['x'])) - w_gmv) #If Vector of zeroes -> Weights are equivalent -> OK
print(sum(np.array(sol['x']))) #If equal to 1 -> OK

Solving natural convection equations (heat and flow) with the shooting method

TL;DR I've been implementing a python program to solve numerically equations for natural convection based on a particular similarity variable using runge-kutta 4 and the shooting method. However I don't get the right solutions when I plot it. Did I make a mistake somewhere ?
Hi !
Starting from a special case of natural convection, we get these similitude equations.
The first describe the fluid flow, the second describe the heat flow.
"Pr" is for Prandtl it's basically a dimensionless number used in fluid dynamics (Prandtl) :
These equations are subjects to the following boundary values such that the temperature near the plate is greater than the temperature outside the boundary layer and such that the fluid velocity is 0 far away from the boundary layer.
I've been trying to resolve these numerically with Runge-Kutta 4 and the shooting method to transform the boundary value problem into an initial value problem. The way the shooting method is implemented is with the newton method.
However, I don't get the right solutions.
As you can see in the following, the temperature (in red) is increasing as we are moving away from the plate whereas it should decrease exponentially.
It's more consistent for the fluid velocity (in blue), however the speed i think it should go up faster then go down faster. Here the curve is smoother.
Now, the fact is that we have a system of 2 coupled ODE. However, right now, I'm only trying to find the one of the two initials values (e.g. f''(0) = a, trying to find a) such that we have a solution to the boundary value problem (shooting method). Once found, I suppose we have the solution for the whole problem.
I guess I should maybe manage the two (f''(0) = a ; theta'(0) = b) but I don't know how to manage these two in parallel.
Last think to mention, if I try to get the initial value of theta' (so theta'(0)) I don't get the right heat profile.
Here is the code :
"""
The goal is to resolve a 3rd order non-linear ODE for the blasius problem.
It's made of 2 equations (flow / heat)
f''' = 3ff'' - 2(f')^2 + theta
3 Pr f theta' + theta'' = 0
RK4 + Shooting Method
"""
import numpy as np
import math
from scipy.integrate import odeint
from scipy.optimize import newton
from edo_solver.plot import plot
from constants import PRECISION
def blasius_edo(y, t, prandtl):
f = y[0:3]
theta = y[3:5]
return np.array([
# flow edo
f[1], # f' = df/dn
f[2], # f'' = d^2f/dn^2
- 3 * f[0] * f[2] + (2 * math.pow(f[1], 2)) - theta[0], # f''' = - 3ff'' + 2(f')^2 - theta,
# heat edo
theta[1], # theta' = dtheta/dn
- 3 * prandtl * f[0] * theta[1], # theta'' = - 3 Pr f theta'
])
def rk4(eta_range, shoot):
prandtl = 0.01
# initial values
f_init = [0, 0, shoot] # f(0), f'(0), f''(0)
theta_init = [1, shoot] # theta(0), theta'(0)
ci = f_init + theta_init # concatenate two ci
# note: tuple with single argument must have "," at the end of the tuple
return odeint(func=blasius_edo, y0=ci, t=eta_range, args=(prandtl,))
"""
if we have :
f'(t_0) = fprime_t0 ; f'(eta -> infty) = fprime_inf
we can transform it into :
f'(t_0) = fprime_t0 ; f''(t_0) = a
we define the function F(a) = f'(infty ; a) - fprime_inf
if F(a) has a root in "a",
then the solutions to the initial value problem with f''(t_0) = a
is also the solution the boundary problem with f'(eta -> infty) = fprime_inf
our goal is to find the root, we have the root...we have the solution.
it can be done with bissection method or newton method.
"""
def shooting(eta_range):
# boundary value
fprimeinf = 0 # f'(eta -> infty) = 0
# initial guess
# as far as I understand
# it has to be the good guess
# otherwise the result can be completely wrong
initial_guess = 10 # guess for f''(0)
# define our function to optimize
# our goal is to take big eta because eta should approach infty
# [-1, 1] : last row, second column => f'(eta_final) ~ f'(eta -> infty)
fun = lambda initial_guess: rk4(eta_range, initial_guess)[-1, 1] - fprimeinf
# newton method resolve the ODE system until eta_final
# then adjust the shoot and resolve again until we have a correct shoot
shoot = newton(func=fun, x0=initial_guess)
# resolve our system of ODE with the good "a"
y = rk4(eta_range, shoot)
return y
def compute_blasius_edo(title, eta_final):
ETA_0 = 0
ETA_INTERVAL = 0.1
ETA_FINAL = eta_final
# default values
title = title
x_label = "$\eta$"
y_label = "profil de vitesse $(f'(\eta))$ / profil de température $(\\theta)$"
legends = ["$f'(\eta)$", "$\\theta$"]
eta_range = np.arange(ETA_0, ETA_FINAL + ETA_INTERVAL, ETA_INTERVAL)
# shoot
y_set = shooting(eta_range)
plot(eta_range, y_set, title, legends, x_label, y_label)
compute_blasius_edo(
title="Convection naturelle - Solution de similitude",
eta_final=10
)

I could be completely off base here, but I wrote something similar to solve 1D fluid-reaction-heat equations. Try using solve_ivp and using the RADAU solver method, it helps with more difficult systems.
Also maybe try converting your system of ODES to a system of first order ODEs as that may help.

You are implementing the additional but wrong boundary condition f''(0) = theta'(0), as both slots get the same initial value in the shooting method. You need to hold them separate, giving 2 free variables and thus the need for a 2-dimensional Newton method or any other solver for non-scalar functions.
You could just as well use the solve_bvp routine with a sensible initial guess.

Numerical Stability of Forward Substitution in Python

I am implementing some basic linear equation solvers in Python.
I have currently implemented forward and backward substitution for triangular systems of equations (so very straightforward to solve!), but the precision of the solutions becomes very poor even with systems of about 50 equations (50x50 coefficient matrix).
The following code performs the forward/backward substitution:
FORWARD_SUBSTITUTION = 1
BACKWARD_SUBSTITUTION = 2
def solve_triang_subst(A: np.ndarray, b: np.ndarray,
substitution=FORWARD_SUBSTITUTION) -> np.ndarray:
"""Solves a triangular system via
forward or backward substitution.
A must be triangular. FORWARD_SUBSTITUTION means A should be
lower-triangular, BACKWARD_SUBSTITUTION means A should be upper-triangular.
"""
rows = len(A)
x = np.zeros(rows, dtype=A.dtype)
row_sequence = reversed(range(rows)) if substitution == BACKWARD_SUBSTITUTION else range(rows)
for row in row_sequence:
delta = b[row] - np.dot(A[row], x)
cur_x = delta / A[row][row]
x[row] = cur_x
return x
I am using numpy and 64-bit floats.
Simple Testing Tool
I have set up a simple test suite which generates coefficient matrices and x vectors, computes the b, and then uses forward or backward substitution to recover the x, comparing it to the its known value for validity.
The following code performs these checks:
import numpy as np
import scipy.linalg as sp_la
RANDOM_SEED = 1984
np.random.seed(RANDOM_SEED)
def check(sol: np.ndarray, x_gt: np.ndarray, description: str) -> None:
if not np.allclose(sol, x_gt, rtol=0.1):
print("Found inaccurate solution:")
print(sol)
print("Ground truth (not achieved...):")
print(x_gt)
raise ValueError("{} did not work!".format(description))
def fuzz_test_solving():
N_ITERATIONS = 100
refine_result = True
for mode in [FORWARD_SUBSTITUTION, BACKWARD_SUBSTITUTION]:
print("Starting mode {}".format(mode))
for iteration in range(N_ITERATIONS):
N = np.random.randint(3, 50)
A = np.random.uniform(0.0, 1.0, [N, N]).astype(np.float64)
if mode == BACKWARD_SUBSTITUTION:
A = np.triu(A)
elif mode == FORWARD_SUBSTITUTION:
A = np.tril(A)
else:
raise ValueError()
x_gt = np.random.uniform(0.0, 1.0, N).astype(np.float64)
b = np.dot(A, x_gt)
x_est = solve_triang_subst(A, b, substitution=mode,
refine_result=refine_result)
# TODO report error and count, don't throw!
# Keep track of error norm!!
check(x_est, x_gt,
"Mode {} custom triang iteration {}".format(mode, iteration))
if __name__ == '__main__':
fuzz_test_solving()
Note that the maximum size of a test matrix is 49x49. Even in this case, the system cannot always compute decent solutions, and fails by more than a margin of 0.1. Here's an example of such a failure (this is doing backward substitution, so the biggest error is in the 0th coefficient; all the test data are sampled uniformly from [0, 1[):
Solution found with Mode 2 custom triang iteration 24:
[ 0.27876067 0.55200497 0.49499509 0.3259397 0.62420183 0.47041149
0.63557676 0.41155446 0.47191956 0.74385864 0.03002819 0.4700286
0.37989592 0.56527691 0.15072607 0.05659282 0.52587574 0.82252197
0.65662833 0.50250729 0.74139748 0.10852731 0.27864265 0.42981232
0.16327331 0.74097937 0.24411709 0.96934199 0.890266 0.9183985
0.14842446 0.51806495 0.36966843 0.18227989 0.85399593 0.89615663
0.39819336 0.90445931 0.21430972 0.61212349 0.85205597 0.66758689
0.1793689 0.38067267 0.39104614 0.6765885 0.4118123 ]
Ground truth (not achieved...)
[ 0.20881608 0.71009766 0.44735271 0.31169033 0.63982328 0.49075813
0.59669585 0.43844108 0.47764942 0.72222069 0.03497499 0.4707452
0.37679884 0.56439738 0.15120397 0.05635977 0.52616387 0.82230625
0.65670245 0.50251426 0.74139956 0.10845974 0.27864289 0.42981226
0.1632732 0.74097939 0.24411707 0.96934199 0.89026601 0.91839849
0.14842446 0.51806495 0.36966843 0.18227989 0.85399593 0.89615663
0.39819336 0.90445931 0.21430972 0.61212349 0.85205597 0.66758689
0.1793689 0.38067267 0.39104614 0.6765885 0.4118123 ]
I have also implemented the iterative refinement method described in Section 2.5 of [0], and while it did help a little, the results are still poor for larger matrices.
MATLAB Sanity Check
I also did this experiment in MATLAB, and even there, once there are more than 100 equations, the estimation error shoots up exponentially.
Here is the MATLAB code I used for this experiment:
err_norms = [];
range = 1:3:120;
for size=range
A = rand(size, size);
A = tril(A);
x_gt = rand(size, 1);
b = A * x_gt;
x_sol = A\b;
err_norms = [err_norms, norm(x_gt - x_sol)];
end
plot(range, err_norms);
set(gca, 'YScale', 'log')
And here is the resulting plot:
Main Question
My question is: Is this normal behavior, seeing as there is essentially no structure in the problem, given that I randomly generate the A matrix and x?
What about solving linear systems of 100s of equations for various practical applications? Are these limitations simply an accepted fact, and e.g., optimization algorithms are just naturally robust to these issues? Or am I missing some important facets of this problem?
[0]: Press, William H. Numerical recipes 3rd edition: The art of scientific computing. Cambridge university press, 2007.

There are no limitations. This is a very fruitful exercise that we all came to realize; writing linear solvers are not that easy and that's why almost always LAPACK or its cousins in other languages are used with full confidence.
You are hit by almost singular matrices and because you are using matlab's backslash you don't see that matlab is switching to least squares solutions behind the scenes when near singularity is hit. If you just change A\b to linsolve(A,b) hence you restrict the solver to solve square systems you'll probably see lots of warnings on your console.
I didn't test it because I don't have a license anymore but if I write blindly this should show you the condition numbers of the matrices at each step.
err_norms = [];
range = 1:3:120;
for i=1:40
size = range(i);
A = rand(size, size);
A = tril(A);
x_gt = rand(size, 1);
b = A * x_gt;
x_sol = linsolve(A,b);
err_norms = [err_norms, norm(x_gt - x_sol)];
zzz(i) = rcond(A);
end
semilogy(range, err_norms);
figure,semilogy(range,zzz);
Note that because you are picking up numbers from a uniform distribution it becomes more and more likely to hit ill-conditioned matrices (wrt to inversion) as the rows have more probability to have rank deficiency. That's why the error becomes bigger and bigger. Sprinkle some identity matrix times a scalar and all errors should come back to eps*n levels.
But best, leave this to expert algorithms which have been tested through decades. It is really not that trivial to write any of these. You can read the Fortran codes, for example, dtrsm solves the triangular system.
On the Python side, you can use scipy.linalg.solve_triangular which uses ?trtrs routines from LAPACK.

SVM with python and CPLEX, load the quadratic part of the objective function

''In general, it would get better performance creating batches of linear constraints rather than creating them one at a time. I just wondering if it states even with a huge problem.'' - The wise programmer.
To be clear, I have a (35k x 40) dataset, and I want to do SVM on it. I need to produce the Gramm matrix of this dataset, it is fine, but to pass the coefficient to CPLEX is a mess, it takes hours, here my code:
nn = 35000
XXt = np.random.rand(nn,nn) # the gramm matrix of the dataset
yy = np.random.rand(nn) # the label vector of the dataset
temp = ((yy*XXt).T)*yy
xg, yg = np.meshgrid(range(nn), range(nn))
indici = np.dstack([yg,xg])
quadraric_part = []
for ii in xrange(nn):
for indd in indici[ii][ii:]:
quadraric_part.append([indd[0],indd[1],temp[indd[0],indd[1]]])
The 'quadratic_part' is a list of the form [i,j,c_ij] where c_ij is the coefficient stored in temp. It will be passed to the function 'objective.set_quadratic_coefficients()' of the CPLEX Python API.
There is a wiser way to do that?
P.S. I have maybe a Memory problem, so It wold be better, instead store the whole list 'quadratic_part', call several times the function 'objective.set_quadratic_coefficients()'.... you know what I mean?!

Under the hood, objective.set_quadratic makes use of the CPXXcopyquad function in the C Callable Library. Whereas, objective.set_quadratic_coefficients uses CPXXcopyqpsep.
Here is an example (bear in mind that I am not a numpy expert; it's quite possible there's a better way to do that part):
import numpy as np
import cplex
nn = 5 # a small example size here
XXt = np.random.rand(nn,nn) # the gramm matrix of the dataset
yy = np.random.rand(nn) # the label vector of the dataset
temp = ((yy*XXt).T)*yy
# create symetric matrix
tempu = np.triu(temp) # upper triangle
iu1 = np.triu_indices(nn, 1)
tempu.T[iu1] = tempu[iu1] # copy upper into lower
ind = np.array([[x for x in range(nn)] for x in range(nn)])
qmat = []
for i in range(nn):
qmat.append([np.arange(nn), tempu[i]])
c = cplex.Cplex()
c.variables.add(lb=[0]*nn)
c.objective.set_quadratic(qmat)
c.write("test2.lp")
Your Q matrix is completely dense so depending on the amount of memory you have, this technique may not scale. When it's possible, though, you should get better performance initializing your Q matrix with objective.set_quadratic. Perhaps you'll need to use some hybrid technique where you use both set_quadratic and set_quadratic_coefficients.

Different eigenvalues between scipy.sparse.linalg.eigs and numpy/scipy.eig

Context:
My goal is to create a Python3 program to operate differential operations on a vector V of size N. I did so, test it for basic operation and it works (differentiation, gradient...).
I tried to write with that basis more complex equations (Navier-Stokes, Orr-Sommerfeld,...) and I tried to validate my work by calculating the eigenvalues of these equations.
As these eigenvalues were completely unexpected, I simplify my problem and I am currently trying to calculate the eigenvalues only for the differentiation matrix (see below). But the results seem wrong...
Thanks in advance for your help, because I do not find any solution to my problem...
Definition of DM:
I use Chebyshev spectral method to operate the differentiation of vectors.
I use the following Chebyshev package (translated from Matlab to Python):
http://dip.sun.ac.za/%7Eweideman/research/differ.html
That package allow me to create a differentiation matrix DM, obtained with:
nodes, DM = chebyshev.chebdiff(N, maximal_order)
To obtain the 1st, 2nd, 3th... order differentiation, I write for example:
dVdx1 = np.dot(DM[0,:,:], V)
d2Vdx2 = np.dot(DM[1,:,:], V)
d3Vdx3 = np.dot(DM[2,:,:], V)
I tested that and it works.
I've build different operators based on that differentiation process.
I've tried to validate them by finding their eigenvalues. It didn't go well so I am just trying right now with DM only.
I do not manage to find the right eigenvalues of DM.
I've tried with different functions:
numpy.linalg.eigvals(DM)
scipy.linalg.eig(DM)
scipy.sparse.linalg.eigs(DM)
sympy.solve( (DM-x*np.eye).det(), x) [for snall size only]
Why I use scipy.sparse.LinearOperator:
I do not want to directly use the matrix DM, so I wrapped into a function which operates the differentiation (see code below) like that:
dVdx1 = derivative(V)
The reason why I do that comes from the global project itself.
This is useful for more complicated equations.
Creating such a function prevents me from using directly the matrix DM to find its eigenvalues (because DM stay inside the function).
For that reason, I use a scipy.sparse.LinearOperator to wrap my method derivative() and use it as an input of scipy.sparse.eig().
Code and results:
Here is the code to compute these eigenvalues:
import numpy as np
import scipy
import sympy
from scipy.sparse.linalg import aslinearoperator
from scipy.sparse.linalg import eigs
from scipy.sparse.linalg import LinearOperator
import chebyshev
N = 20 # should be 4, 20, 50, 100, 300
max_order = 4
option = 1
#option 1: building the differentiation matrix DM for a given order
if option == 1:
if 0:
# usage of package chebyshev, but I add a file with the matrix inside
nodes, DM = chebyshev.chebdiff(N, max_order)
order = 1
DM = DM[order-1,:,:]
#outfile = TemporaryFile()
np.save('DM20', DM)
if 1:
# loading the matrix from the file
# uncomment depending on N
#DM = np.load('DM4.npy')
DM = np.load('DM20.npy')
#DM = np.load('DM50.npy')
#DM = np.load('DM100.npy')
#DM = np.load('DM300.npy')
#option 2: building a random matrix
elif option == 2:
j = np.complex(0,1)
np.random.seed(0)
Real = np.random.random((N, N)) - 0.5
Im = np.random.random((N,N)) - 0.5
# If I want DM symmetric:
#Real = np.dot(Real, Real.T)
#Im = np.dot(Im, Im.T)
DM = Real + j*Im
# If I want DM singular:
#DM[0,:] = DM[1,:]
# Test DM symmetric
print('Is DM symmetric ? \n', (DM.transpose() == DM).all() )
# Test DM Hermitian
print('Is DM hermitian ? \n', (DM.transpose().real == DM.real).all() and
(DM.transpose().imag == -DM.imag).all() )
# building a linear operator which wrap matrix DM
def derivative(v):
return np.dot(DM, v)
linop_DM = LinearOperator( (N, N), matvec = derivative)
# building a linear operator directly from a matrix DM with asLinearOperator
aslinop_DM = aslinearoperator(DM)
# comparison of LinearOperator and direct Dot Product
V = np.random.random((N))
diff_lo = linop_DM.matvec(V)
diff_mat = np.dot(DM, V)
# diff_lo and diff_mat are equals
# FINDING EIGENVALUES
#number of eigenvalues to find
k = 1
if 1:
# SCIPY SPARSE LINALG LINEAR OPERATOR
vals_sparse, vecs = scipy.sparse.linalg.eigs(linop_DM, k, which='SR',
maxiter = 10000,
tol = 1E-3)
vals_sparse = np.sort(vals_sparse)
print('\nEigenvalues (scipy.sparse.linalg Linear Operator) : \n', vals_sparse)
if 1:
# SCIPY SPARSE ARRAY
vals_sparse2, vecs2 = scipy.sparse.linalg.eigs(DM, k, which='SR',
maxiter = 10000,
tol = 1E-3)
vals_sparse2 = np.sort(vals_sparse2)
print('\nEigenvalues (scipy.sparse.linalg with matrix DM) : \n', vals_sparse2)
if 1:
# SCIPY SPARSE AS LINEAR OPERATOR
vals_sparse3, vecs3 = scipy.sparse.linalg.eigs(aslinop_DM, k, which='SR',
maxiter = 10000,
tol = 1E-3)
vals_sparse3 = np.sort(vals_sparse3)
print('\nEigenvalues (scipy.sparse.linalg AS linear Operator) : \n', vals_sparse3)
if 0:
# NUMPY LINALG / SAME RESULT AS SCIPY LINALG
vals_np = np.linalg.eigvals(DM)
vals_np = np.sort(vals_np)
print('\nEigenvalues (numpy.linalg) : \n', vals_np)
if 1:
# SCIPY LINALG
vals_sp = scipy.linalg.eig(DM)
vals_sp = np.sort(vals_sp[0])
print('\nEigenvalues (scipy.linalg.eig) : \n', vals_sp)
if 0:
x = sympy.Symbol('x')
D = sympy.Matrix(DM)
print('\ndet D (sympy):', D.det() )
E = D - x*np.eye(DM.shape[0])
eig_sympy = sympy.solve(E.det(), x)
print('\nEigenvalues (sympy) : \n', eig_sympy)
Here are my results (for N=20):
Is DM symmetric ?
False
Is DM hermitian ?
False
Eigenvalues (scipy.sparse.linalg Linear Operator) :
[-2.5838015+0.j]
Eigenvalues (scipy.sparse.linalg with matrix DM) :
[-2.58059801+0.j]
Eigenvalues (scipy.sparse.linalg AS linear Operator) :
[-2.36137671+0.j]
Eigenvalues (scipy.linalg.eig) :
[-2.92933791+0.j -2.72062839-1.01741142j -2.72062839+1.01741142j
-2.15314244-1.84770128j -2.15314244+1.84770128j -1.36473659-2.38021351j
-1.36473659+2.38021351j -0.49536645-2.59716913j -0.49536645+2.59716913j
0.38136094-2.53335888j 0.38136094+2.53335888j 0.55256471-1.68108134j
0.55256471+1.68108134j 1.26425751-2.25101241j 1.26425751+2.25101241j
2.03390489-1.74122287j 2.03390489+1.74122287j 2.57770573-0.95982011j
2.57770573+0.95982011j 2.77749810+0.j ]
The values returned by scipy.sparse should be included in the ones found by scipy/numpy, which is not the case. (idem for sympy)
I've tried with different random matrices instead of DM (see option 2) (symmetric, non-symmetric, real, imaginary, etc...), which had small size N (4,5,6..) and also bigger ones (100,...).
That worked
By changing parameters like 'which' (LM, SM, LR...), 'tol' (10E-3, 10E-6..), 'maxiter', 'sigma' (0) in scipy.sparse... scipy.sparse.linalg.eigs always worked for random matrices but never for my matrix DM. In best cases, found eigenvalues are close to the ones found by scipy, but never match.
I really do not know what is so particular in my matrix.
I also dont know why using scipy.sparse.linagl.eig with a matrix, a LinearOperator or a AsLinearOperator gives different results.
I DO NOT KNOW HOW I COULD INCLUDE MY FILES CONTAINING MATRICES DM...
For N = 4 :
[[ 3.16666667 -4. 1.33333333 -0.5 ]
[ 1. -0.33333333 -1. 0.33333333]
[-0.33333333 1. 0.33333333 -1. ]
[ 0.5 -1.33333333 4. -3.16666667]]
Every idea is welcome.
May a moderator could tag my question with :
scipy.sparse.linalg.eigs / weideman / eigenvalues / scipy.eig /scipy.sparse.lingalg.linearOperator
Geoffroy.

I spoke with a few colleague and solve partly my problem.
My conclusion is that my matrix is simply very ill conditioned...
In my project, I can simplify my matrix by imposing boundary condition as follow:
DM[0,:] = 0
DM[:,0] = 0
DM[N-1,:] = 0
DM[:,N-1] = 0
which produces a matrix similar to that for N=4:
[[ 0 0 0 0]
[ 0 -0.33333333 -1. 0]
[ 0 1. 0.33333333 0]
[ 0 0 0 0]]
By using such condition, I obtain eigenvalues for scipy.sparse.linalg.eigs which are equal to the one in scipy.linalg.eig.
I also tried using Matlab, and it return the same values.
To continue my work, I actually needed to use the generalized eigenvalue problem in the standard form
λ B x= DM x
It seems that it does not work in my case because of my matrix B (which represents a Laplacian operator matrix).
If you have a similar problem, I advise you to visit that question:
https://scicomp.stackexchange.com/questions/10940/solving-a-generalised-eigenvalue-problem
(I think that) the matrix B needs to be positive definite to use scipy.sparse.
A solution would be to change B, to use scipy.linalg.eig or to use Matlab.
I will confirm that later.
EDIT:
I wrote a solution to the stack exchange question I post above which explains how I solve my problem.
I appears that scipy.sparse.linalg.eigs has indeed a bug if matrix B is not positive definite, and will return bad eigenvalues.