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Mathc complexes/a230

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Application


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c00a.c
/* ------------------------------------ */
/*  Save as :   c00a.c                  */
/* ------------------------------------ */
#include "w_a.h"  
/* ------------------------------------ */
/* ------------------------------------ */
#define   RA     R3
#define   CA     C4
#define   Cb     C1
/* ------------------------------------ */
void fun(void)
{
double ab[RA*((CA+Cb)*C2)] ={
	                    1,2,  3,4,  5,6,  5,2, 0,0,
                        1,2,  3,4,  5,6,  1,3, 0,0,
                        1,2,  3,4,  1,1,  4,2, 0,0};
                        
double **Ab = ca_A_mZ(ab,i_Abr_Ac_bc_mZ(RA,CA,Cb));
double **A  = c_Ab_A_mZ(Ab,i_mZ(RA,CA));
double **b  = c_Ab_b_mZ(Ab,i_mZ(RA,Cb));

double **B  =              i_mZ(RA,C3) ;
double **BT =              i_mZ(C3,RA) ;
double **BTb =   i_Abr_Ac_bc_mZ(C3,RA,Cb);

  clrscrn();
  printf("Basis for a Column Space by Row Reduction :\n\n");
  printf(" A :");
  p_mZ(A, S3,P0, S3,P0, C8);
  printf(" b :");
  p_mZ(b, S3,P0, S3,P0, C8);
  printf(" Ab :");
  p_mZ(Ab, S3,P0, S3,P0, C8);
  stop();

  clrscrn(); 
  printf(" The leading 1’s of Ab give the position \n"
         " of the columns  of A which form a basis \n"
         " for the column space of A \n\n"
         " A :");
  p_mZ(A, S8,P4, S8,P4, C4);
  printf(" gj_PP_mZ(Ab) :");
  p_mZ(gj_PP_mZ(Ab), S8,P4, S8,P4, C4);
  
  c_c_mZ(A,C1,B,C1);
  c_c_mZ(A,C3,B,C2);
  c_c_mZ(A,C4,B,C3);

  printf(" B :");
  p_mZ(B, S8,P4, S8,P4, C4);
  stop();

  clrscrn();   
  printf(" Check if the columns of B are linearly independent\n\n"
         " BT :");
  p_mZ(transpose_mZ(B,BT), S4,P0, S3,P0, C4);  
  printf(" BTb :");
  p_mZ(c_mZ(BT,BTb), S4,P0, S3,P0, C4); 
  printf(" gj_PP_mZ(BTb) :");
  p_mZ(gj_PP_mZ(BTb), S8,P4, S8,P4, C4); 
        
  stop();   
  
  f_mZ(Ab);
  f_mZ(A);
  f_mZ(b);
  f_mZ(B);
  f_mZ(BT);
  f_mZ(BTb);
  
}
/* ------------------------------------ */
int main(void)
{

  fun();

  return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */


La position des pivots de Ab donne la position des colonnes de A qui forment une base pour l'espace colonnes de A.


Exemple de sortie écran :
 ------------------------------------ 
Basis for a Column Space by Row Reduction :

 A :
 +1 +2i  +3 +4i  +5 +6i  +5 +2i 
 +1 +2i  +3 +4i  +5 +6i  +1 +3i 
 +1 +2i  +3 +4i  +1 +1i  +4 +2i 

 b :
 +0 +0i 
 +0 +0i 
 +0 +0i 

 Ab :
 +1 +2i  +3 +4i  +5 +6i  +5 +2i  +0 +0i 
 +1 +2i  +3 +4i  +5 +6i  +1 +3i  +0 +0i 
 +1 +2i  +3 +4i  +1 +1i  +4 +2i  +0 +0i 

 Press return to continue. 

 ------------------------------------ 
 The leading 1s of Ab give the position 
 of the columns  of A which form a basis 
 for the column space of A 

 A :
 +1.0000 +2.0000i  +3.0000 +4.0000i  +5.0000 +6.0000i  +5.0000 +2.0000i 
 +1.0000 +2.0000i  +3.0000 +4.0000i  +5.0000 +6.0000i  +1.0000 +3.0000i 
 +1.0000 +2.0000i  +3.0000 +4.0000i  +1.0000 +1.0000i  +4.0000 +2.0000i 

 gj_PP_mZ(Ab) :
 +1.0000 +0.0000i  +2.2000 -0.4000i  +3.4000 -0.8000i  +1.8000 -1.6000i 
 -0.0000 +0.0000i  +0.0000 -0.0000i  +1.0000 +0.0000i  +0.0976 -0.1220i 
 +0.0000 -0.0000i  +0.0000 +0.0000i  -0.0000 +0.0000i  +1.0000 +0.0000i 

 +0.0000 +0.0000i 
 -0.0000 +0.0000i 
 +0.0000 -0.0000i 

 B :
 +1.0000 +2.0000i  +5.0000 +6.0000i  +5.0000 +2.0000i 
 +1.0000 +2.0000i  +5.0000 +6.0000i  +1.0000 +3.0000i 
 +1.0000 +2.0000i  +1.0000 +1.0000i  +4.0000 +2.0000i 

 Press return to continue. 

 ------------------------------------ 
 Check if the columns of B are linearly independent

 BT :
  +1 +2i   +1 +2i   +1 +2i 
  +5 +6i   +5 +6i   +1 +1i 
  +5 +2i   +1 +3i   +4 +2i 

 BTb :
  +1 +2i   +1 +2i   +1 +2i   +0 +0i 
  +5 +6i   +5 +6i   +1 +1i   +0 +0i 
  +5 +2i   +1 +3i   +4 +2i   +0 +0i 

 gj_PP_mZ(BTb) :
 +1.0000 +0.0000i  +1.0000 +0.0000i  +0.1803 -0.0164i  +0.0000 +0.0000i 
 -0.0000 +0.0000i  +1.0000 +0.0000i  -0.6201 -0.5853i  +0.0000 -0.0000i 
 +0.0000 -0.0000i  +0.0000 -0.0000i  +1.0000 +0.0000i  +0.0000 +0.0000i 

 Press return to continue.