I'm attempting to solve a set of equations of the form Ax = 0. A is known 6x6 matrix and I've written the below code using SVD to get the vector x which works to a certain extent. The answer is approximately correct but not good enough to be useful to me, how can I improve the precision of the calculation? Lowering eps below 1.e-4 causes the function to fail.
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)
A
is full rank --- so x
is 0
Since it looks like you need a least-squares solution, i.e. min ||A*x|| s.t. ||x|| = 1
, do the SVD such that [U S V] = svd(A)
and the last column of V
(assuming that the columns are sorted in order of decreasing singular values) is x
.
I.e.,
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
So,
x =
0.0050325
0.0028357
0.00013067
0.0052698
0.98402
0.17788
And, ||A*x|| = 0.00027528
as opposed to your previous solution for x
where ||A*x_old|| = 0.079442
Attention: There might be confusion with the SVD in python vs. matlab-syntax(?): in python, numpy.linalg.svd(A) returns matrices u,s,v such that u*s*v = A (strictly: dot(u, dot(diag(s), v) = A, because s is a vector and not a 2D-matrix in numpy).
The uppermost Answer is correct in that sense that usually you write u*s*vh = A and vh is returned, and this answer discusses v AND NOT vh.
To make a long story short: if you have matrices u,s,v such that u*s*v = A, then the last rows of v, not the last colums of v, describe the nullspace.
Edit: [for people like me:] each of the last rows is a vector v0 such that A*v0 = 0 (if the corresponding singular value is 0)
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