Mathc complexes/08s
Apparence
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c00a.c |
|---|
/* ------------------------------------ */
/* Save as : c00a.c */
/* ------------------------------------ */
#include "w_a.h"
/* ------------------------------------ */
#define RCA RC4
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double a[RCA*RCA] ={
+0.729876318493, +0.169778846788, -0.191923913761, -0.362625471673,
+0.169778846788, +0.893290152659, +0.120628523081, +0.227918315245,
-0.191923913761, +0.120628523081, +0.863637321734, -0.257646791146,
-0.362625471673, +0.227918315245, -0.257646791146, +0.513196207114
};
double v[RCA*RCA] ={
+0.532141930309, -0.579194958348, -0.801954709350, +0.519734241230,
+0.846655163574, +0.000000000000, +0.000000000000, -0.326664732319,
+0.000000000000, +0.815189057964, +0.000000000000, +0.369273175665,
+0.000000000000, +0.000000000000, +0.597384837564, +0.697713259790
};
double **A = ca_A_mRZ(a, i_mZ(RCA,RCA));
double **V = ca_A_mRZ(v, i_mZ(RCA,RCA));
double **invV = invgj_mZ(V, i_mZ(RCA,RCA));
double **EValue = i_mZ(RCA,RCA);
double **T = i_mZ(RCA,RCA);
clrscrn();
printf(" A :");
p_mRZ(A, S8,P6, C4);
printf(" V :");
p_mRZ(V, S9,P6, C4);
printf(" EValue = invV A V");
mul_mZ(invV,A,T);
mul_mZ(T,V,EValue);
p_mRZ(EValue, S9,P6, C4);
stop();
clrscrn();
printf(" A :");
p_mRZ(A, S8,P6, C4);
printf(" A = V EValue invV (Just verify the computation)");
mul_mZ(V,EValue,T);
mul_mZ(T,invV,A);
p_mRZ(A, S8,P6, C4);
stop();
clrscrn();
printf(" V1 V2 V3 V4 :");
p_mRZ(V, S9,P6, C4);
printf(" EValue1 EValue2 EValue3 EValue4 ");
p_mRZ(EValue, S9,P6, C4);
printf(" det(V) = ");p_Z(det_Z(V), S4,P0, S5,P0);
printf("\n det(V) != 0.00 V1,V2,V3 and V4 are linearly independent\n\n");
stop();
clrscrn();
printf(" The matrix A projects the space in the direction\n"
" of the eigenvector V4 on a hyperplan determined\n"
" by the eigenvector V1,V2 and V3 if :\n\n"
" The eigenvector V1 has its eigenvalue equal to one and \n"
" The eigenvector V2 has its eigenvalue equal to one and \n"
" The eigenvector V3 has its eigenvalue equal to one and \n"
" The eigenvector V4 has its eigenvalue equal to zero and \n\n"
" If The vectors V1,V2,V3 and V4 are linearly independent \n\n");
stop();
f_mZ(A);
f_mZ(V);
f_mZ(invV);
f_mZ(T);
f_mZ(EValue);
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
Dans cet exemple nous pouvons voir que nous n'utilisons que des matrices réelles. Les fonctions *_mRZ travail directement sur des valeurs réelles. ex : ca_A_mRZ(); qui copie un tableau de réelles dans une matrice complexes. p_mRZ(); qui affiche uniquement les valeurs réelles.
Exemple de sortie écran :
A :
+0.729876 +0.169779 -0.191924 -0.362625
+0.169779 +0.893290 +0.120629 +0.227918
-0.191924 +0.120629 +0.863637 -0.257647
-0.362625 +0.227918 -0.257647 +0.513196
V :
+0.532142 -0.579195 -0.801955 +0.519734
+0.846655 +0.000000 +0.000000 -0.326665
+0.000000 +0.815189 +0.000000 +0.369273
+0.000000 +0.000000 +0.597385 +0.697713
EValue = invV * A * V
+1.000000 +0.000000 +0.000000 +0.000000
-0.000000 +1.000000 +0.000000 -0.000000
+0.000000 +0.000000 +1.000000 +0.000000
-0.000000 -0.000000 -0.000000 +0.000000
Press return to continue.
A :
+0.729876 +0.169779 -0.191924 -0.362625
+0.169779 +0.893290 +0.120629 +0.227918
-0.191924 +0.120629 +0.863637 -0.257647
-0.362625 +0.227918 -0.257647 +0.513196
A = V * EValue * invV (Just verify the computation)
+0.729876 +0.169779 -0.191924 -0.362625
+0.169779 +0.893290 +0.120629 +0.227918
-0.191924 +0.120629 +0.863637 -0.257647
-0.362625 +0.227918 -0.257647 +0.513196
Press return to continue.
V1 V2 V3 V4 :
+0.532142 -0.579195 -0.801955 +0.519734
+0.846655 +0.000000 +0.000000 -0.326665
+0.000000 +0.815189 +0.000000 +0.369273
+0.000000 +0.000000 +0.597385 +0.697713
EValue1 EValue2 EValue3 EValue4
+1.000000 +0.000000 +0.000000 +0.000000
-0.000000 +1.000000 +0.000000 -0.000000
+0.000000 +0.000000 +1.000000 +0.000000
-0.000000 -0.000000 -0.000000 +0.000000
det(V) = -1 -0i
det(V) != 0.00 V1,V2,V3 and V4 are linearly independent
Press return to continue.
The matrix A projects the space in the direction
of the eigenvector V4 on a hyperplan determined
by the eigenvector V1,V2 and V3 if :
The eigenvector V1 has its eigenvalue equal to one and
The eigenvector V2 has its eigenvalue equal to one and
The eigenvector V3 has its eigenvalue equal to one and
The eigenvector V4 has its eigenvalue equal to zero and
If The vectors V1,V2,V3 and V4 are linearly independent
Press return to continue.