c++ - LAPACK线性方程解中缺少一个元素

我对从线性方程组中获得的解决方案感到非常困惑。我的目标是解决线性方程:通过lapack中的函数A*x = e。这是我的代码:

#include <iostream>
#include "/usr/include/armadillo"
#include "/usr/local/include/lapacke.h"

using namespace std;

int main()
{
    int n = 3;
    arma::fvec alpha( n );//define a vetor alpha with a size 3
    arma::fvec beta( n );//define a vector beta with a size 2
    alpha << 1 << 2 << 3 << arma::endr;//assign 1,2,3 to alpha;
    beta << 0.3 << 0.6 << arma::endr;//assign 0.3, 0.6 to beta;
    float a = 0.1;
    arma::fvec e = arma::zeros<arma::fvec>( n );//define a vector e with all element equal to 0;
    e( n - 1 ) = beta( n - 2 ) * beta( n - 2 ); //the last element of e equals to the square of the last element of vector beta;
    arma::fvec tri_alpha = alpha - a;
    LAPACKE_sgtsv(LAPACK_COL_MAJOR, n, 1, &( beta[ 0 ] ), &( tri_alpha[ 0 ] ), &( beta[ 0 ] ), &( e[ 0 ] ), n );
    cout << e.t() << endl;
    return 0;
}

vector alpha在对角线上， vector beta在次对角线和超对角线上，以构造三对角矩阵(假定为T)。下面是对sgtsv函数的解释。

LAPACKE_sgtsv( int matrix_order, int n, int nrhs, float *dl, float *d, float *du, float *b, int ldb)

和

B is REAL array, dimension (LDB,NRHS)，On exit, if INFO = 0, the N by NRHS solution matrix X

在我的情况下，B = e，我最终输出e，它是(0, -0.0444, 0.1333)，显然，正确的答案应该是(0.0148, -0.0444, 0.1333)，那么第一个元素是错误的或可能缺少，有人可以帮我吗？谢谢。顺便说一句，我正在使用的图书馆是 Armadillo 。

最佳答案

根据sgtsv()的documentation，当求解线性方程时，三对角矩阵DU的上对角线(DL)和下对角线(A)将被覆盖。



和:



之所以出现此问题，是因为beta应该同时是上对角DU和下对角DL。

解决方案是复制beta:

#include <iostream>

#include "/usr/include/armadillo"
#include "/usr/local/include/lapacke.h"

//#include "/usr/local/include/armadillo"
//#include "lapacke.h"

using namespace std;

int main()
{
    int n = 3;
    arma::fvec alpha( n );//define a vetor alpha with a size 3
    arma::fvec beta( n );//define a vector beta with a size 2
    alpha << 1 << 2 << 3 << arma::endr;//assign 1,2,3 to alpha;
    beta << 0.3 << 0.6 << arma::endr;//assign 0.3, 0.6 to beta;
    float a = 0.1;
    arma::fvec e = arma::zeros<arma::fvec>( n );//define a vector e with all element equal to 0;
    e( n - 1 ) = beta( n - 2 ) * beta( n - 2 ); //the last element of e equals to the square of the last element of vector beta;
    arma::fvec tri_alpha = alpha - a;

    arma::fvec betabis( n );//define a vector betabis with a size 2...copy of beta
    betabis=beta;
    LAPACKE_sgtsv(LAPACK_COL_MAJOR, n, 1, &( beta[ 0 ] ), &( tri_alpha[ 0 ] ), &( betabis[ 0 ] ), &( e[ 0 ] ), n );
    cout << e.t() << endl;
    return 0;
}