Linear Algebra and the C Language/a08u
Install and compile this file in your working directory.
/* ------------------------------------ */
/* Save as : c00b.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
#define RA R4
#define CA C6
#define Cb C1
/* ------------------------------------ */
#define CB C2 /* B : a basis for the column space of A */
/* ------------------------------------ */
#define CbFREE Cb+C3
/* ------------------------------------ */
int main(void)
{
double ab[RA*(CA+Cb)]={
+3, -9, +12, -6, +15, +12, +0,
-6, +18, -24, +12, -30, -24, +0,
+7, -21, +28, -14, +35, +28, +0,
-2, +6, -8, +4, -10, -8, +0
};
double **Ab = ca_A_mR(ab, i_Abr_Ac_bc_mR(RA,CA,Cb));
double **A = c_Ab_A_mR(Ab, i_mR(RA,CA));
double **b = c_Ab_b_mR(Ab, i_mR(RA,Cb));
double **AT = transpose_mR(A, i_mR(CA,RA));
double **B = i_mR(RA,CB);
double **BT = i_mR(CB,RA);
double **BTb = i_Abr_Ac_bc_mR(CB,RA,Cb);
double **BTb_free = i_Abr_Ac_bc_mR(RA,RA,CbFREE);
double **b_free = i_mR(RA,CbFREE);
double **AT_bfree = i_mR(CA,CbFREE);
int r;
clrscrn();
printf("Basis for a Column Space by Row Reduction :\n\n");
printf(" A :");
p_mR(A,S6,P1,C10);
printf(" b :");
p_mR(b,S6,P1,C10);
printf(" Ab :");
p_mR(Ab,S6,P1,C10);
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_mR(A,S7,P3,C10);
printf(" gj_PP_mR(Ab,NO) :");
gj_PP_mR(Ab,NO);
p_mR(Ab,S7,P3,C10);
c_c_mR(A,C1,B,C1);
printf(" B : a basis for the column space of A");
p_mR(B,S7,P3,C10);
stop();
/* B has only a column. I copy B into BT. I put the pivot at one.
Instead, check whether the columns of B are linearly independent.
Then I can put the free variables. */
transpose_mR(B,BT);
c_mR(BT,BTb);
gj_PP_mR(BTb,NO);
clrscrn();
put_zeroR_mR(BTb,BTb_free);
printf(" BTb_free : put_zeroR_mR(BTb,BTb_free);");
p_mR(BTb_free,S7,P3,C10);
put_freeV_mR(BTb_free);
printf(" BTb_free : put_freeV_mR(BTb_free);");
p_mR(BTb_free,S7,P3,C10);
stop();
clrscrn();
r = rsize_R(BTb_free);
while(r>R1)
zero_above_pivot_gj1Ab_mR(BTb_free,r--);
printf(" BTb_free : zero_above_pivot_gj1Ab_mR(BTb_free,r--);");
p_mR(BTb_free,S7,P3,C10);
c_Ab_b_mR(BTb_free,b_free);
printf(" b_free : A basis for the null space of AT");
p_mR(b_free,S10,P3,C7);
stop();
clrscrn();
printf(" AT :");
p_mR(AT, S7,P3,C10);
printf(" b_free :");
p_mR(b_free, S7,P3,C10);
printf(" The row vectors of AT"
" are orthogonal to the column vectors of bfree\n");
printf(" AT * bfree :");
p_mR(mul_mR(AT,b_free,AT_bfree), S7,P3,C10);
stop();
f_mR(Ab);
f_mR(A);
f_mR(b);
f_mR(AT);
f_mR(B);
f_mR(BT);
f_mR(BTb);
f_mR(BTb_free);
f_mR(b_free);
f_mR(AT_bfree);
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
The free vectors of the system BTb will be a basis for the orthogonal complement of AT, with B a basis of the column space of A.
Screen output example:
Basis for a Column Space by Row Reduction :
A :
+3.0 -9.0 +12.0 -6.0 +15.0 +12.0
-6.0 +18.0 -24.0 +12.0 -30.0 -24.0
+7.0 -21.0 +28.0 -14.0 +35.0 +28.0
-2.0 +6.0 -8.0 +4.0 -10.0 -8.0
b :
+0.0
+0.0
+0.0
+0.0
Ab :
+3.0 -9.0 +12.0 -6.0 +15.0 +12.0 +0.0
-6.0 +18.0 -24.0 +12.0 -30.0 -24.0 +0.0
+7.0 -21.0 +28.0 -14.0 +35.0 +28.0 +0.0
-2.0 +6.0 -8.0 +4.0 -10.0 -8.0 +0.0
Press return to continue.
The leading 1’s of Ab give the position
of the columns of A which form a basis
for the column space of A
A :
+3.000 -9.000 +12.000 -6.000 +15.000 +12.000
-6.000 +18.000 -24.000 +12.000 -30.000 -24.000
+7.000 -21.000 +28.000 -14.000 +35.000 +28.000
-2.000 +6.000 -8.000 +4.000 -10.000 -8.000
gj_PP_mR(Ab,NO) :
+1.000 -3.000 +4.000 -2.000 +5.000 +4.000 +0.000
+0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000
+0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000
+0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000
B : a basis for the column space of A
+3.000 +0.000
-6.000 +0.000
+7.000 +0.000
-2.000 +0.000
Press return to continue.
BTb_free : put_zeroR_mR(BTb,BTb_free);
+1.000 -2.000 +2.333 -0.667 +0.000 +0.000 +0.000 +0.000
+0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000
+0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000
+0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000
BTb_free : put_freeV_mR(BTb_free);
+1.000 -2.000 +2.333 -0.667 +0.000 +0.000 +0.000 +0.000
+0.000 +1.000 +0.000 +0.000 +0.000 +1.000 +0.000 +0.000
+0.000 +0.000 +1.000 +0.000 +0.000 +0.000 +1.000 +0.000
+0.000 +0.000 +0.000 +1.000 +0.000 +0.000 +0.000 +1.000
Press return to continue.
BTb_free : zero_above_pivot_gj1Ab_mR(BTb_free,r--);
+1.000 +0.000 +0.000 +0.000 +0.000 +2.000 -2.333 +0.667
+0.000 +1.000 +0.000 +0.000 +0.000 +1.000 +0.000 +0.000
+0.000 +0.000 +1.000 +0.000 +0.000 +0.000 +1.000 +0.000
+0.000 +0.000 +0.000 +1.000 +0.000 +0.000 +0.000 +1.000
b_free : A basis for the null space of AT
+0.000 +2.000 -2.333 +0.667
+0.000 +1.000 +0.000 +0.000
+0.000 +0.000 +1.000 +0.000
+0.000 +0.000 +0.000 +1.000
Press return to continue.
AT :
+3.000 -6.000 +7.000 -2.000
-9.000 +18.000 -21.000 +6.000
+12.000 -24.000 +28.000 -8.000
-6.000 +12.000 -14.000 +4.000
+15.000 -30.000 +35.000 -10.000
+12.000 -24.000 +28.000 -8.000
b_free :
+0.000 +2.000 -2.333 +0.667
+0.000 +1.000 +0.000 +0.000
+0.000 +0.000 +1.000 +0.000
+0.000 +0.000 +0.000 +1.000
The row vectors of AT are orthogonal to the column vectors of bfree
AT * bfree :
+0.000 +0.000 +0.000 +0.000
+0.000 +0.000 -0.000 +0.000
+0.000 +0.000 +0.000 +0.000
+0.000 +0.000 -0.000 +0.000
+0.000 +0.000 +0.000 +0.000
+0.000 +0.000 +0.000 +0.000
Press return to continue.