If a square matrix is not full rank, its determinant is zero, its column vectors are linearly dependent and we can obtain one of the column vectors as a linear combination of the others. Below is a matrix which, according to sympy, has not full rank, and has determinant of 0. However, sympy seems unable to find a linear combination of any two column vectors that is equal to the third column vector. Why is that?
Input:
x = sympy.symbols('x')
D = Matrix([[1,1+x,(1-x)*(1+3*x)],[1, 1-x, (1-x)**2],[1,1,(1-x**2)]])
print(D)
print('rank', D.rank())
print('det', D.det())
D[:,[1,2]].LUsolve(D[:,0]) # gives same result as D[:,[0,1]].LUsolve(D[:,2]) and D[:,[0,2]].LUsolve(D[:,1])
Output:
Matrix([[1, x + 1, (1 - x)*(3*x + 1)], [1, 1 - x, (1 - x)**2], [1, 1, 1 - x**2]])
rank 2
det 0
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-330-4d4f8f3e2cc0> in <module>
5 print('det', D.det())
6
----> 7 D[:,1:].LUsolve(D[:,0])
~\anaconda3\envs\symbolic\lib\site-packages\sympy\matrices\matrices.py in LUsolve(self, rhs, iszerofunc)
2152
2153 def LUsolve(self, rhs, iszerofunc=_iszero):
-> 2154 return _LUsolve(self, rhs, iszerofunc=iszerofunc)
2155
2156 def QRsolve(self, b):
~\anaconda3\envs\symbolic\lib\site-packages\sympy\matrices\solvers.py in _LUsolve(M, rhs, iszerofunc)
357 for j in range(b.cols):
358 if not iszerofunc(b[i, j]):
--> 359 raise ValueError("The system is inconsistent.")
360
361 b = b[0:n, :] # truncate zero rows if consistent
ValueError: The system is inconsistent.
Possibly this is a bug in lusolve.
You can get an answer with linsolve:
In [10]: linsolve((D[:,[1,2]], D[:,0]))
Out[10]:
⎧⎛ -2 -1 ⎞⎫
⎪⎜─────, ────────────⎟⎪
⎨⎜x - 1 2 ⎟⎬
⎪⎝ x - 2⋅x + 1⎠⎪
⎩ ⎭