Python解非满秩的线性方程组
之前也是需要用的时候发现Numpy只能解满秩的，解不了非满秩有无穷解的，也就是特解+核空间线性组合的方法什么的。后面百度到了结果，下面给出知乎大佬的代码，这个函数复制下就能直接用啦。
Python 如何计算线性方程组 Ax=b 的通解？ - 用我双手乾坤转过的回答 - 知乎 https://www.zhihu.com/question/36865787/answer/2669793744
这个也就是线性代数中增广矩阵高斯消元解线性方程组的规范方法。
import numpy as np
from numpy.linalg import matrix_rank, inv, multi_dot
def mylinalgsolve(A,b):
m, n = A.shape
rank_a = matrix_rank(A)
rank_a_b = matrix_rank(np.concatenate((A, b), axis=1))
print("-----------START-----------")
print("The coefficient matrix A is %s, rank = %d." % (A.shape, rank_a))
if rank_a == m:
print("The column space of A encompasses R^%d." % m)
else:
print("The column space of A does not encompass R^%d." % m)
if rank_a < rank_a_b:
print("rank(A) = %d < %d = rank(A|b), the equation has no solution." % (rank_a, rank_a_b))
else:
if rank_a == n:
print("rank(A) = rank(A|b) = %d, the equation has only one solution." % n)
Ar = A
br = b
if n < m:
Ar = Ar[:n]
br = br[:n]
Ar_inv = inv(Ar)
if n < m:
print("One of the largest sub square of A is T =\n%s" % Ar)
print("T^-1 =\n%s" % Ar_inv)
print("x = T^-1 * b = ", end="")
else:
print("A^-1 =\n%s" % Ar_inv)
print("x = A^-1 * b = ", end="")
x = np.round(multi_dot((Ar_inv, br)), 4).flatten()
print("%s^T" % x)
else:
print("rank(A) = rank(A|b) = %d < %d, the equation has infinitely many solutions." % (rank_a, n))
# find the solution of T*y = b, where T is compounded of linearly independant column vectors of A
Ar = A[:, :rank_a]
br = b
if rank_a < m:
Ar = Ar[:rank_a]
br = br[:rank_a]
Ar_inv = inv(Ar)
# find a specific solution
y = np.concatenate((multi_dot((inv(Ar), br)), np.zeros((n - rank_a, 1))), axis=0)
# find every linear combination of T representating redundant column vectors of A
co_independant = np.zeros((n - rank_a, rank_a))
co_redundant = -np.eye(n - rank_a)
for i in range(rank_a, n):
c = A[:rank_a, i]
co_independant[i - rank_a] = multi_dot((Ar_inv, c)).T
# find the solutions of homogeneous equation Ax=0
z = np.concatenate((co_independant, co_redundant), axis=1).T
y = y
z = z
result = "%s^T" % y.flatten()
for i in range(n - rank_a):
result += " + k_%d * %s^T" % (i + 1, z[:, i])
print("x = %s (k_n ∈ R)" % result)
print("-----------END-----------")
return result下面给出一个算例解AX=b，A不满秩的情况。注意下numpy的结构和转换就好。
x=[2,-3,-4,-6,5,3,-2,4]
y=[3,4,-5,2,3,2,-6,8]
x=np.array(x)
y=np.array(y)
b=np.zeros_like(y)
A=np.array([x**3,x**2*y,x*y**2,y**3,x**2,y**2,x*y,x,y,np.ones_like(y)]).T
mylinalgsolve(A,b.reshape(8,1))
'''
-----------START-----------
The coefficient matrix A is (8, 10), rank = 8.
The column space of A encompasses R^8.
rank(A) = rank(A|b) = 8 < 10, the equation has infinitely many solutions.
x = [0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]^T + k_1 * [-0.11287802 0.46301482 -0.3657235 0.08385066 -1.08818233 -0.46186757
1.56357684 0.89698889 -1. -0. ]^T + k_2 * [-0.0198804 0.05238849 -0.05231127 0.00908104 -0.10905413 -0.01730029
0.17617317 0.38083805 -0. -1. ]^T (k_n ∈ R)
-----------END-----------
'''