Hi Amit, I submitted a pull request. Thanks for the guidance! Best, Tom On Saturday, August 8, 2015 at 9:07:57 AM UTC-7, AMiT Kumar wrote: > > Hi Tom, > > Sorry for late reply, if you are interested in contributing > code, then you can send us a Pull Request on Github: > https://github.com/sympy/sympy/pulls/ > > BTW, Wolframalpha also doesn't calculates > <http://www.wolframalpha.com/input/?i=LUDecomposition+%28%5B1%2C2%2C3%5D%2C%5B4%2C5%2C6%5D%2C%5B7%2C8%2C9%5D%29> > > the LUDecomposition > of the Matrix you mentioned, since it's singular. > > Though, your code works for singular matrices, but > I am not sure, If the LUDecomposition method in SymPy, > should compute LU Decomposition of a singular matrix. > Anyways, you can send us a pull Request someone would > surely help. > > *AMiT Kumar* > > On Thursday, August 6, 2015 at 11:09:06 PM UTC+5:30, Tom H wrote: >> >> Hi Amit, >> >> Thanks for the speedy response! If I understand the code >> in gauss_jordan_solve correctly, solving A*x=b with multiple right hand >> sides requires a call to gauss_jordan_solve for each right hand side b, >> which means that the same Gaussian elimination steps are performed on all >> but the last column of the augmented matrix [A | b] for each b. Is this >> correct? I have attached what is probably the millionth implementation of >> Gaussian elimination for computing the LU decomposition of a matrix, in >> case someone finds it useful. Let me know if this is not a proper venue for >> sharing such code. Thanks again for your input. >> Regards, >> Tom >> >> On Tuesday, August 4, 2015 at 10:27:39 PM UTC-7, AMiT Kumar wrote: >>> >>> 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 >>>> >>>
-- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/004aca27-9571-4692-90b2-066efdfb4719%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
