Hi Tom,
LUdecomposition() doesn't works for under determined systems, you can use ,
linsolve(), It supports all types of systems:
In [1]: from sympy import *
In [2]: from sympy.solvers.solveset import linsolve
In [3]: A = Matrix([[1,2,3],[4,5,6],[7,8,9]])
In [4]: b = Matrix([4, 13, 22])
In [5]: system = (A, b)
In [6]: r, s, t = symbols('r s t')
In [7]: linsolve((A, b), (r, s, t))
Out[8]: {(t + 2, -2⋅t + 1, t)}
If you are interested in Matrix form results, you can
use gauss_jordan_solve()
In [3]: A = Matrix([[1,2,3],[4,5,6],[7,8,9]])
In [4]: b = Matrix([4, 13, 22])
In [5]: sol, params = A.gauss_jordan_solve(b)
In [7]: init_printing()
In [8]: sol
Out[8]:
⎡ τ₀ + 2 ⎤
⎢ ⎥
⎢-2⋅τ₀ + 1⎥
⎢ ⎥
⎣ τ₀ ⎦
Please note that, both these functions are *not available in any release*,
you can use them from the *development version*
: https://github.com/sympy/sympy
--
*AMiT Kumar*
On Wednesday, August 5, 2015 at 4:53:15 AM UTC+5:30, Tom H wrote:
>
> I came across the following issue when trying to use Sympy to compute an
> LU decomposition of a matrix. I'd like to determine the number of solutions
> a system of equations has, for example, if
>
> A = sympy.Matrix([[1,2,3],[4,5,6],[7,8,9]])
> b = sympy.Matrix([4, 13, 22])
>
> then the system A*x=b has infinitely many solutions, all of which can be
> written as
>
> x0 + t*n
>
> where
> x0 = sympy.Matrix([2,1,0])
> n = sympy.Matrix([1,-2,1])
> t = sympy.Symbol('t')
>
> This solution can be computed from the LU decomposition of A, however the
> following code fails to compute the factors L and U.
>
> >>> import sympy
> >>> A = sympy.Matrix([[1,2,3],[4,5,6],[7,8,9]])
> >>> LU, perm = A.LUdecomposition()
> Traceback (most recent call last):
> File "<stdin>", line 1, in <module>
> File "/Library/Python/2.7/site-packages/sympy/matrices/matrices.py",
> line 1297, in LUdecomposition
> combined, p = self.LUdecomposition_Simple(iszerofunc=_iszero)
> File "/Library/Python/2.7/site-packages/sympy/matrices/matrices.py",
> line 1341, in LUdecomposition_Simple
> raise ValueError("No nonzero pivot found; inversion failed.")
> ValueError: No nonzero pivot found; inversion failed.
>
> A is square and noninvertible, which explains the message in the stack
> trace about the elimination algorithm encountering a zero subcolumn below
> the pivot position. Regardless, A can still be written as the product of
> lower triangular L with unit diagonal, and upper triangular U, where
> L = sympy.Matrix([[1, 0, 0], [4, 1, 0], [7, 2, 1]])
> U = sympy.Matrix([[1, 2, 3], [0, -3, -6], [0, 0, 0]])
>
> Is A.LUdecomposition() the wrong function to call if I want to compute the
> LU decomposition of A in the case where A is noninvertible, or not square?
> I wrote my own LU decomposition routine to handle these cases, which I'd be
> happy to contribute, but I'd like to know if this functionality exists in
> Sympy.
>
> Thanks in advance for your input,
> Tom
>
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