Joined: Tue Jun 23, 2009 10:33 pm
Posts: 51
Location: Kaiserslautern
|
Michele T wrote: just to be sure just look at the second example: This is what I did: Code: > LIB "matrix.lib";
> ring R = 0,(x,y,z), dp;
> matrix B[3][3] = 2y, x, 0, -z,0,y, 0, -z,2z;
> matrix v[3][14]= x,0,0,-2y,2x,0,0,0,z,2y,0,-z,0,2z, 0,2x,0,0,-2y,-x,y,0,0,z,y,0,0,0, 0,0,3x,0,0,-2y,0,2y,0,0,z,0,z,0;
> print(v);
x,0, 0, -2y,2x, 0, 0,0, z,2y,0,-z,0,2z,
0,2x,0, 0, -2y,-x, y,0, 0,z, y,0, 0,0,
0,0, 3x,0, 0, -2y,0,2y,0,0, z,0, z,0
> print( gauss_col (transpose(jacob(B)) + transpose(jacob(-v))) );
1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,1,0,0,0,0,0,0
so does this means that the only solution is the zero one? Should not 'v' look as follows? Code: x,0, 0, -2y,2x, 0, 0,0, z,2y,0, -z,0,2z,0,
0,2x,0, 0, -2y,-x, y,0, 0,z, -y,0, 0,0, -z,
0,0, 3x,0, 0, -2y,0,2y,0,0, z, 0, z,0, 0
Besides, i'm sorry: 1. it's better to construct the system as i show below, and 2. for the intersection you clearly need the row-reduced block-triangular form (use '''gauss_row''' instead!): Code: > def N = transpose(jacob(B)); N = N, transpose(jacob(-v)); print(N);
0, 1, 0,-1,0, 0, 0,-2,0,0, 0, 0, 0, 0, 0,0, 0, 0,
0, 0, 0,0, -2,0, 0,0, 1,0, 0, 0, 0, 0, 0,0, 0, 0,
0, 0, 0,0, 0, -3,0,0, 0,0, 0, 0, 0, 0, 0,0, 0, 0,
2, 0, 0,0, 0, 0, 2,0, 0,0, 0, 0, -2,0, 0,0, 0, 0,
0, 0, 1,0, 0, 0, 0,2, 0,-1,0, 0, 0, 1, 0,0, 0, 0,
0, 0, 0,0, 0, 0, 0,0, 2,0, -2,0, 0, 0, 0,0, 0, 0,
0, 0, 0,0, 0, 0, 0,0, 0,0, 0, -1,0, 0, 1,0, -2,0,
-1,0, 0,0, 0, 0, 0,0, 0,0, 0, 0, -1,0, 0,0, 0, 1,
0, -1,2,0, 0, 0, 0,0, 0,0, 0, 0, 0, -1,0,-1,0, 0
> print( gauss_row (N) );
0, 0, 0, 0, 0, 1,0, 0,0,0, 0,0, 0,0,0, 0,0,0,
0, -1/2,0, 1/2,0, 0,0, 1,0,0, 0,0, 0,0,0, 0,0,0,
0, 0, 0, 0, -2,0,0, 0,1,0, 0,0, 0,0,0, 0,0,0,
0, 0, 0, 0, -2,0,0, 0,0,0, 1,0, 0,0,0, 0,0,0,
-1,0, 0, 0, 0, 0,-1,0,0,0, 0,0, 1,0,0, 0,0,0,
0, 1, 1, -1, 0, 0,0, 0,0,-1,0,0, 0,1,0, 0,0,0,
0, 0, -3,1, 0, 0,0, 0,0,1, 0,0, 0,0,0, 1,0,0,
0, 0, 0, 0, 0, 0,0, 0,0,0, 0,1/2,0,0,-1/2,0,1,0,
-2,0, 0, 0, 0, 0,-1,0,0,0, 0,0, 0,0,0, 0,0,1
this is a homogeneous linear system for your parameters: (rows correspond to the {var*gen} basis), the 1st three cols correspond to the (free) span of B, let's call them: (p,q,r), while the rest 15 - to the parameters from ''v'': (a_1, a_2, a_3, a_4, a_5, a_6, b_2, b_3, b_4, b_5, b_6, c_1, c_3, c_5, c_6 ) Due to that system you have p,q,r,a_1,a_2,a_4,a_6,b_2,b_4,c_1 as free parameters, the 1st row says: a_3 = 0, while all other parameters are linearly expressed via the free parameters: 2nd row => -(-1/2 q + 1/2 a_1) = a_5 3rd row => -(-2 a_2) = a_6 and so on... i hope this helps, O.
|
|
|
Login
Register
Forum FAQ
Forum Search
