Reputation: 2301
I just started with lapack. I thought let's start with LU decomposition, of the following matrix A:
1 1 0 3
2 1 -1 1
3 -1 -1 2
-1 2 3 -1
I wrote the follwinng program,
#include <stdio.h>
#include <lapacke/lapacke.h>
int main (int argc, const char * argv[]) {
float a[4][4]={
{1,1,0,3},
{2,1,-1,1},
{3,-1,-1,2},
{-1,2,3,-1}
};
lapack_int m=4,n=4,lda=4,info;
int i,j;
lapack_int*ipiv=(lapack_int*)malloc(4*4*sizeof(lapack_int));
m = 4;
n = 4;
lda = 4;
info= LAPACKE_sgetrf(LAPACK_ROW_MAJOR,m,n,*a,lda,ipiv);
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
printf("%lf ",a[i][j]);
}
printf("\n");
}
return(info);
}
This should give
L=
1 0 0 0
2 1 0 0
3 4 1 0
-1 3 0 1
and U=
1 1 0 3
0 -1 -1 -5
0 0 3 13
0 0 0 -13
But in the docs I read that L en U are returned in A. How? Maybe skip the 1's on the diagonal of L en merge the two. Is that true?
I do
$ gcc -Wall sgetrf.c -llapacke -llapack
I see a= 3.000000 -1.000000 -1.000000 2.000000 0.666667 1.666667 -0.333333 -0.333333 -0.333333 1.000000 3.000000 0.000000 0.333333 0.800000 0.200000 2.600000
I does not make sense to me.
Upvotes: 2
Views: 564
Reputation: 9817
Your program is perfectly correct and LAPACKE_sgetrf
actually return an LU factorization of the matrix with partial pivoting. It returned :
3 -1 -1 2
2/3 5/3 -1/3 -1/3
-1/3 1 3 0
1/3 0.8 0.2 2.6
The L matrix is:
1 0 0 0
2/3 1 -1/3 -1/3
-1/3 1 1 0
1/3 0.8 0.2 1
The U matrix is:
3 -1 -1 2
0 5/3 -1/3 -1/3
0 0 3 0
0 0 0 2.6
The product is similar to A except for the permutation of the rows:
3 -1 -1 2
2 1 -1 1
-1 2 3 -1
1 1 0 3
The LU factorization you provided is correct (except for the last line of L, which must be -1 -3 0 1 instead of -1 3 0 1). But such a LU factorization is not always possible (See wikipedia). This is the reason why LAPACK compute a PLU factorization, where P is a permutation matrix, returned in ipiv
. Indeed, such a factorization is always possible.
Upvotes: 1