Singular :: View topic - Finding parameters

Author:

malex [ Thu Nov 11, 2010 8:32 pm ]

Post subject:

Re: Finding parameters

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.

Singular https://www.singular.uni-kl.de/forum/

Finding parameters https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=1886	Page 2 of 2

Author:	Michele T [ Thu Nov 11, 2010 5:06 pm ]
Post subject:	Re: Finding parameters
The fact is that I'm not sure on how to interpret the last outcome... thank you!

Author:	Michele T [ Fri Nov 12, 2010 12:28 pm ]
Post subject:	Re: Finding parameters
Thank you very very much, now is more clear! best Michele

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