[MUG] Can Maple krack this kernel?

[MUG] Re: Can Maple krack this kernel?
Author: Maple User Group    Posted: Mon, 11 Nov 2002 15:38:32 -0500
>> From: Maple User Group "maple_gr" -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 6 Nov 2002 19:09:59 -0800 (PST) From: Peter Montgomery "pmontgom" To: "maple-list" Subject: Can Maple krack this kernel? If I understand correctly, you (Jones) are looking for a vector v = f(aijk)*v0 such that 1) Av = 0 for all possible aijk values 2) v0 has real (i.e., constant) entries. If so, then f(aijk) * (A*v0) = 0 is a polynomial identity. Either f(aijk) is the zero polynomial or A*v0 is the zero vector. In the latter case you can set to zero the coefficient of each aijk within the polynomial terms for A*v0, much as you probably did to find v7. Peter Montgomery -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Fri, 8 Nov 2002 14:37:01 -0500 (EST) From: David Linder "dlinder" To: "maple-list" Subject: Can Maple krack this kernel? Hello, You are trying to find the kernel or nullspace of m, that is, vectors ker such that m . ker = 0 # the zero vector You could do an LU decomposition and then a forward and backward substitute, as another way to solve that system. You can also force this LU approach from within the LinearAlgebra:-LinearSolve command. Here's how it might look in Maple 8, with(LinearAlgebra): m := Matrix([[a1111.<ie, your beast>.a2222]],storage=sparse); ker := LinearSolve(m,Vector(RowDimension(m)),'method'='LU'): That took about 60sec and 18Mb of allocated memory on a 1.4Gz P4 (Linux). Checking this solution took another 60sec or so, via multiplication, Norm( m . ker ); There's something different about the result for this approach. The returned "answer" can be just a single Vector, parametrized by, say, 16 new variables. You can see these with, indets(ker); You could do repeated substitutions of the form, eval(ker,{_t0[1]=<value>,..,_t0[16]=<value>}); if you wanted 16 explicit Vectors whose only unknowns were a1112 to a2222. Note that LinearSolve has an option 'free'=<name> which can be used to control how these parametrizing variables are named. If you only have access to a version earlier than Maple 6 you can try that sort of thing using the linalg functions: LUdecomp backsub forwardsub The idea is this. Apologies if this seems obvious already. Use LUdecomp() to get an L and U which are lower and upper triangular respectively. Then you have, in vector notation, with "." used to represent multiplication: m . x = 0 L . U . x = m . x = 0 Let y stand for U . x so that you have L . y = 0 Use forwardsub to solve the above equation. Now you have y, so you can solve U . x = y by using backsub. Here's an example. m := matrix([[1,3,4,0],[3,b,12,0],[2,6,8,0],[6,2*b,24,0]]); U := linalg[LUdecomp](m,'L'=L); y := linalg[forwardsub](L,vector(4,[0,0,0,0])); x := linalg[backsub](U,y); I hope this is of some help to you, regards, Dave Linder

Previous by date: [MUG] solve function's behavior,  Huy Cuong Nguyen
Next by date: [MUG] maximize,  Dragomir Deltchev
Previous thread: [MUG] citation for maple, Nld
Next thread: [MUG] maximize,  Dragomir Deltchev