计算矩阵的零空间

Question

我正在尝试解决一组形式为Ax = 0的方程式.A是已知的6x6矩阵,我使用SVD编写了下面的代码,以获得在某种程度上起作用的向量x.答案大致正确,但不足以对我有用,我怎样才能提高计算的精确度？降低低于1.e-4的eps会导致功能失败.

from numpy.linalg import *
from numpy import *

A = matrix([[0.624010149127497 ,0.020915658603923 ,0.838082638087629 ,62.0778180312547 ,-0.336 ,0],
[0.669649399820597 ,0.344105317421833 ,0.0543868015800246 ,49.0194290212841 ,-0.267 ,0],
[0.473153758252885 ,0.366893577716959 ,0.924972565581684 ,186.071352614705 ,-1 ,0],
[0.0759305208803158 ,0.356365401030535 ,0.126682113674883 ,175.292109352674 ,0 ,-5.201],
[0.91160934274653 ,0.32447818779582 ,0.741382053883291 ,0.11536775372698 ,0 ,-0.034],
[0.480860406786873 ,0.903499596111067 ,0.542581424762866 ,32.782593418975 ,0 ,-1]])

def null(A, eps=1e-3):
  u,s,vh = svd(A,full_matrices=1,compute_uv=1)
  null_space = compress(s <= eps, vh, axis=0)
  return null_space.T

NS = null(A)
print "Null space equals ",NS,"\n"
print dot(A,NS)

Answer 1

A是满级 - 所以x是0

因为看起来你需要一个最小二乘解,即min ||A*x|| s.t. ||x|| = 1做SVD这样[U S V] = svd(A)和最后一列V(假设列按照递减奇异值的顺序排序)是x.

也就是说,

U =

     -0.23024     -0.23241      0.28225     -0.59968     -0.04403     -0.67213
      -0.1818     -0.16426      0.18132      0.39639      0.83929     -0.21343
     -0.69008     -0.59685     -0.18202      0.10908     -0.20664      0.28255
     -0.65033      0.73984    -0.066702     -0.12447     0.088364       0.0442
  -0.00045131    -0.043887      0.71552     -0.32745       0.1436      0.59855
     -0.12164      0.11611       0.5813      0.59046     -0.47173     -0.25029


S =

       269.62            0            0            0            0            0
            0       4.1038            0            0            0            0
            0            0        1.656            0            0            0
            0            0            0       0.6416            0            0
            0            0            0            0      0.49215            0
            0            0            0            0            0   0.00027528


V =

    -0.002597     -0.11341      0.68728     -0.12654      0.70622    0.0050325
   -0.0024567     0.018021       0.4439      0.85217     -0.27644    0.0028357
   -0.0036713      -0.1539      0.55281      -0.4961      -0.6516   0.00013067
      -0.9999    -0.011204   -0.0068651    0.0013713    0.0014128    0.0052698
    0.0030264      0.17515      0.02341    -0.020917   -0.0054032      0.98402
     0.012996     -0.96557     -0.15623      0.10603     0.014754      0.17788

所以,

x =

    0.0050325
    0.0028357
   0.00013067
    0.0052698
      0.98402
      0.17788

而且,||A*x|| = 0.00027528相对于您之前的解决方案x,其中||A*x_old|| = 0.079442

Answer 2

注意:python与matlab语法(？)中的SVD可能存在混淆:在python中,numpy.linalg.svd(A)返回矩阵u,s,v,使得u*s*v = A(严格地说: dot(u,dot(diag(s),v)= A,因为s是矢量而不是numpy中的2D矩阵).

在这个意义上,最上面的答案是正确的,通常你写u*s*vh = A和vh被返回,这个答案讨论v AND NOT vh.

长话短说:如果你有矩阵u,s,v使得u*s*v = A,那么v 的最后一行,而不是 v 的最后一列,描述了零空间.

编辑:[对于像我这样的人:]每个最后一行是向量v0,使得A*v0 = 0(如果相应的奇异值为0)