如何使用numpy在Python中找到矩阵的零空间? [英] How to find the Null Space of a matrix in Python using numpy?

问题描述

正如标题所述，如何找到矩阵的空空间 即方程ax = 0的平凡解决方案.

As the title says, how can I find the null space of a matrix i.e. the nontrivial solution to the equation ax=0.

我尝试使用np.linalg.solve(a,b)来解决方程ax = b.因此，将b设置为与矩阵a相同维的零数组，我只会得到平凡的解，即x = 0.

I've tried to use np.linalg.solve(a,b), which solves the equation ax=b. So setting b equal to an array of zeros with the same dimensions as matrix a, I only get the trivial solution i.e. x=0.

推荐答案

来自 SciPy食谱:

import numpy as np
from numpy.linalg import svd

def nullspace(A, atol=1e-13, rtol=0):
    A = np.atleast_2d(A)
    u, s, vh = svd(A)
    tol = max(atol, rtol * s[0])
    nnz = (s >= tol).sum()
    ns = vh[nnz:].conj().T
    return ns

计算A的空空间的近似基础.

Computes an approximate basis for the nullspace of A.

此函数使用的算法基于A的奇异值分解.

The algorithm used by this function is based on the singular value decomposition of A.

参数:

A:ndarray

A最多应为二维.长度为k的一维数组将被视为形状为(1，k)的二维数组

A should be at most 2-D. A 1-D array with length k will be treated as a 2-D with shape (1, k)

atol:浮动

零奇异值的绝对公差.小于atol的奇异值被认为是零.

The absolute tolerance for a zero singular value. Singular values smaller than atol are considered to be zero.

rtol:浮动

相对公差.小于rtol * smax的奇异值被认为是零，其中smax是最大奇异值.

The relative tolerance. Singular values less than rtol*smax are considered to be zero, where smax is the largest singular value.

如果atol和rtol均为正，则组合公差为两者的最大值；即:

If both atol and rtol are positive, the combined tolerance is the maximum of the two; that is:

tol = max(atol, rtol * smax)

小于tol的奇异值被认为是零.

Singular values smaller than tol are considered to be zero.

返回值:

ns:ndarray

如果A是形状为(m，k)的数组，则ns将是形状为(k，n)的数组，其中n是A的空空间的估计维数. ns的列是空空间的基础； numpy.dot(A，ns)中的每个元素将近似为零.

If A is an array with shape (m, k), then ns will be an array with shape (k, n), where n is the estimated dimension of the nullspace of A. The columns of ns are a basis for the nullspace; each element in numpy.dot(A, ns) will be approximately zero.

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