Comment #13 on issue 1128 by [email protected]: .rref() sometimes
returns invalid results
http://code.google.com/p/sympy/issues/detail?id=1128
Here's another instance of the problem that my provide insight. Create a
symbolic matrix where each individual column sums to zero. This is clearly
a singular matrix, and you can easily get a row of zeros by adding all the
rows together. rref(), however, returns the identity matrix for the
reduction.
In [1]: import sympy as sym
In [2]: from sympy import Matrix
In [3]: P = Matrix(sym.symbols(tuple('P'+str(ii)+'_0:'+str(4) for ii in
range(4))))
In [4]: for ii in xrange(4):
P[ii,ii] = -(sum(P[:,ii])-P[ii,ii])
In [5]: P
Out[5]:
[-P1_0 - P2_0 - P3_0, P0_1,
P0_2, P0_3]
[ P1_0, -P0_1 - P2_1 - P3_1,
P1_2, P1_3]
[ P2_0, P2_1, -P0_2 - P1_2 -
P3_2, P2_3]
[ P3_0, P3_1, P3_2, -P0_3 -
P1_3 - P2_3]
In [6]: sum(P[:,0]), sum(P[:,1]), sum(P[:,2]), sum(P[:,3])
Out[6]: (0, 0, 0, 0)
In [7]: P.rref()
Out[7]:
([1, 0, 0, 0]
[0, 1, 0, 0]
[0, 0, 1, 0]
[0, 0, 0, 1], [0, 1, 2, 3])
Hope this helps!
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