测试矩阵在有限域上是否可逆 [英] Test if matrix is invertible over finite field

问题描述

我想测试一种特定类型的随机矩阵在有限域(尤其是F_2)上是否可逆.我可以使用以下简单代码测试矩阵是否可逆于实数.

I would like to test if a particular type of random matrix is invertible over a finite field, in particular F_2. I can test if a matrix is invertible over the reals using the following simple code.

import random
from scipy.linalg import toeplitz
import numpy as np
n=10
column = [random.choice([0,1]) for x in xrange(n)]
row = [column[0]]+[random.choice([0,1]) for x in xrange(n-1)]
matrix = toeplitz(column, row)
if (np.linalg.matrix_rank(matrix) < n):
    print "Not invertible!"

除了F_2之外，是否有其他方法可以实现相同的目的?

Is there some way to achieve the same thing but over F_2?

推荐答案

为此最好使用Sage或其他一些合适的工具.

It would be better to use Sage or some other proper tool for this.

以下只是做某事的不复杂的非专家尝试，但是枢轴的高斯消去应该给出可逆性的确切结果:

The following is just unsophisticated non-expert attempt at doing something, but pivoted Gaussian elimination should give the exact result for invertibility:

import random
from scipy.linalg import toeplitz
import numpy as np

def is_invertible_F2(a):
    """
    Determine invertibility by Gaussian elimination
    """
    a = np.array(a, dtype=np.bool_)
    n = a.shape[0]
    for i in range(n):
        pivots = np.where(a[i:,i])[0]
        if len(pivots) == 0:
            return False

        # swap pivot
        piv = i + pivots[0]
        row = a[piv,i:].copy()
        a[piv,i:] = a[i,i:]
        a[i,i:] = row

        # eliminate
        a[i+1:,i:] -= a[i+1:,i,None]*row[None,:]

    return True

n = 10
column = [random.choice([0,1]) for x in xrange(n)]
row = [column[0]]+[random.choice([0,1]) for x in xrange(n-1)]
matrix = toeplitz(column, row)

print(is_invertible_F2(matrix))
print(int(np.round(np.linalg.det(matrix))) % 2)

请注意，np.bool_仅在有限的意义上类似于F_2 --- bool的F_2中的二元运算+为-，一元运算符-为+.但是，乘法是相同的.

Note that np.bool_ is analogous to F_2 only in a restricted sense --- the binary operation + in F_2 is - for bool, and the unary op - is +. Multiplication is the same, though.

>>> x = np.array([0, 1], dtype=np.bool_)
>>> x[:,None] - x[None,:]
array([[False,  True],
       [ True, False]], dtype=bool)
>>> x[:,None] * x[None,:]
array([[False, False],
       [False,  True]], dtype=bool)

上面的高斯消除仅使用这些操作，因此可以正常工作.

The gaussian elimination above uses only these operations, so it works.

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