如何在python中生成随机稀疏Hermitian矩阵？

Question

我想在python中生成给定形状的随机稀疏Hermitian矩阵 。 我如何有效地做到这一点？ 这个任务有内置的python函数吗？

我已经找到了随机稀疏矩阵的解决方案，但我也希望矩阵也是Hermitian。 这是我发现的随机稀疏矩阵的解决方案

import numpy as np
import scipy.stats as stats
import scipy.sparse as sparse
import matplotlib.pyplot as plt
np.random.seed((3,14159))

def sprandsym(n, density):
    rvs = stats.norm().rvs
    X = sparse.random(n, n, density=density, data_rvs=rvs)
    upper_X = sparse.triu(X) 
    result = upper_X + upper_X.T - sparse.diags(X.diagonal())
    return result

M = sprandsym(5000, 0.01)
print(repr(M))
# <5000x5000 sparse matrix of type '<class 'numpy.float64'>'
#   with 249909 stored elements in Compressed Sparse Row format>

# check that the matrix is symmetric. The difference should have no non-zero elements
assert (M - M.T).nnz == 0

statistic, pval = stats.kstest(M.data, 'norm')
# The null hypothesis is that M.data was drawn from a normal distribution.
# A small p-value (say, below 0.05) would indicate reason to reject the null hypothesis.
# Since `pval` below is > 0.05, kstest gives no reason to reject the hypothesis
# that M.data is normally distributed.
print(statistic, pval)
# 0.0015998040114 0.544538788914

fig, ax = plt.subplots(nrows=2)
ax[0].hist(M.data, normed=True, bins=50)
stats.probplot(M.data, dist='norm', plot=ax[1])
plt.show()

Answer 1

我们知道一个矩阵加上它的Hermitian就是一个Hermitian矩阵。 因此，要确保您的最终矩阵B是厄米（hermitian），

B = A + A.conj().T