Question

我一直在尝试使用命令A在SymPy中找到符号矩阵A.nullspace的零空间。现在，对于我输入的矩阵，计算没有完成（或者比我等待的时间更长）。

奇怪的是，Mathematica在几分之一秒内完成了这项计算。我将给出我使用的具体示例作为Mathematica输入代码（应该运行）

L = ({{0, -I a Sqrt[b] Conjugate[g] Conjugate[w], 0, I a Sqrt[b] g w, 
 2 g2 + 2 b g z Conjugate[g], 0, 0, 0, 
 2 g2}, {-I a Sqrt[b] g w, -g2 - b g z Conjugate[g] - 
  I (-dD + b d g Conjugate[g]), -I DD, 0, I a Sqrt[b] g w, 0, 0, 
 0, 0}, {0, -I DD, I dD - g2, 0, 0, I a Sqrt[b] g w, 0, 0, 
 0}, {I a Sqrt[b] Conjugate[g] Conjugate[w], 0, 
 0, -g2 - b g z Conjugate[g] - 
  I (dD - b d g Conjugate[g]), -I a Sqrt[b] Conjugate[
   g] Conjugate[w], 0, I DD, 0, 0}, {0, 
 I a Sqrt[b] Conjugate[g] Conjugate[w], 
 0, -I a Sqrt[b] g w, -2 g2 - 2 b g z Conjugate[g], -I DD, 0, 
 I DD, 0}, {0, 0, I a Sqrt[b] Conjugate[g] Conjugate[w], 
 0, -I DD, -2 g2 + I b d g Conjugate[g] - b g z Conjugate[g], 0, 
 0, I DD}, {0, 0, 0, I DD, 0, 
 0, -I dD - g2, -I a Sqrt[b] Conjugate[g] Conjugate[w], 0}, {0, 0,
  0, 0, I DD, 
 0, -I a Sqrt[b] g w, -2 g2 - I b d g Conjugate[g] - 
  b g z Conjugate[g], -I DD}, {0, 0, 0, 0, 0, I DD, 
 0, -I DD, -2 g2}});

MatrixForm[L]
NullSpace[L]

正如同一个矩阵所说，SymPy正在挣扎。这有什么理由吗？有什么其他方法可以在python中解决这个问题吗？

这是我的SymPy代码：

import numpy as np
import scipy
import sympy as sp

a = sp.symbols("a", real=True, positive=True)
b = sp.symbols("b", real=True, positive=True)
g = sp.symbols("g", real=True, positive=True)
g2 = sp.symbols("g2", real=True, positive=True)
DD = sp.symbols("DD", real=True, positive=True)

dD = sp.symbols("dD", real=True)

w = sp.symbols("w")
z = sp.symbols("z")
d = sp.symbols("d")

L = sp.Matrix([[0, -1j*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 0,            1j*a*sp.sqrt(b)*g*w, 
2*g2 + 2*b*g*z*sp.conjugate(g), 0, 0, 0, 
2*g2], [-1j*a*sp.sqrt(b)*g*w, -g2 - b*g*z*sp.conjugate(g) - 
1j*(-dD + b*d*g*sp.conjugate(g)), -1j*DD, 0, 1j*a*sp.sqrt(b)*g*w, 0, 0, 
0, 0], [0, -1j*DD, 1j*dD - g2, 0, 0, 1j*a*sp.sqrt(b)*g*w, 0, 0, 
0], [1j*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 0, 
0, -g2 - b*g*z*sp.conjugate(g) - 
1j*(dD - b*d*g*sp.conjugate(g)), -1j*a*sp.sqrt(b)*sp.conjugate(
g)*sp.conjugate(w), 0, 1j*DD, 0, 0], [0, 
1j*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 
0, -1j*a*sp.sqrt(b)*g*w, -2*g2 - 2*b*g*z*sp.conjugate(g), -1j*DD, 0, 
1j*DD, 0], [0, 0, 1j*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 
0, -1j*DD, -2*g2 + 1j*b*d*g*sp.conjugate(g) - b*g*z*sp.conjugate(g), 0, 
0, 1j*DD], [0, 0, 0, 1j*DD, 0, 
0, -1j*dD - g2, -1j*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 0], [0, 0,
0, 0, 1j*DD, 
0, -1j*a*sp.sqrt(b)*g*w, -2*g2 - 1j*b*d*g*sp.conjugate(g) - 
b*g*z*sp.conjugate(g), -1j*DD], [0, 0, 0, 0, 0, 1j*DD, 
0, -1j*DD, -2*g2]]);

sp.pprint(L)
sp.pprint(L.nullspace())

根据评论，以下信息也可能是相关的：

真实，积极的参数：a, b, g, g2, DD
真实参数：dD
复杂参数：w, z, d

Answer 1

同情的虚构单位是sympy.I（不是1j）。显然，1j是可以接受的，但是，它被解释为数量，在符号计算中应该避免。

以下代码非常快速地给出了nullspace。请注意，我还提供了符号属性。这些通常有助于同情提供更快的答案

a = sp.symbols("a", positive = True)
b = sp.symbols("b", positive = True)
g = sp.symbols("g", positive = True)
w = sp.symbols("w")
z = sp.symbols("z")
g2 = sp.symbols("g2", positive = True)
DD = sp.symbols("DD", positive = True)
dD = sp.symbols("dD", real = True)
d = sp.symbols("d")

L = sp.Matrix([[0, -sp.I*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 0,            sp.I*a*sp.sqrt(b)*g*w, 
2*g2 + 2*b*g*z*sp.conjugate(g), 0, 0, 0, 
2*g2], [-sp.I*a*sp.sqrt(b)*g*w, -g2 - b*g*z*sp.conjugate(g) - 
sp.I*(-dD + b*d*g*sp.conjugate(g)), -sp.I*DD, 0, sp.I*a*sp.sqrt(b)*g*w, 0, 0, 
0, 0], [0, -sp.I*DD, sp.I*dD - g2, 0, 0, sp.I*a*sp.sqrt(b)*g*w, 0, 0, 
0], [sp.I*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 0, 
0, -g2 - b*g*z*sp.conjugate(g) - 
sp.I*(dD - b*d*g*sp.conjugate(g)), -sp.I*a*sp.sqrt(b)*sp.conjugate(
g)*sp.conjugate(w), 0, sp.I*DD, 0, 0], [0, 
sp.I*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 
0, -sp.I*a*sp.sqrt(b)*g*w, -2*g2 - 2*b*g*z*sp.conjugate(g), -sp.I*DD, 0, 
sp.I*DD, 0], [0, 0, sp.I*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 
0, -sp.I*DD, -2*g2 + sp.I*b*d*g*sp.conjugate(g) - b*g*z*sp.conjugate(g), 0, 
0, sp.I*DD], [0, 0, 0, sp.I*DD, 0, 
0, -sp.I*dD - g2, -sp.I*a*sp.sqrt(b)*sp.conjugate(g)*sp.conjugate(w), 0], [0, 0,
0, 0, sp.I*DD, 
0, -sp.I*a*sp.sqrt(b)*g*w, -2*g2 - sp.I*b*d*g*sp.conjugate(g) - 
b*g*z*sp.conjugate(g), -sp.I*DD], [0, 0, 0, 0, 0, sp.I*DD, 
0, -sp.I*DD, -2*g2]]);

Ln = L.nullspace()

当Mathematica轻松成功时，SymPy的零空间挣扎

1 个答案: