Question

这是一个具有相同问题的新创建示例。 2个矩阵＆＃34; firstMultiplyMat＆＃34; ，＆＃34; a＆＃34;具有相同的值，但以不同的方式指定。然而，每一个的倒数是不同的。不能得到＃34; firstMultiplyMat＆＃34;而＆＃34; a＆＃34;具有正确的计算逆矩阵。

#include<iostream>
#include<stdio.h>
#include <Eigen/Core> 
#include <Eigen/Dense>

using namespace std;
using namespace Eigen;
using Eigen::MatrixXd;

struct MemberLocation{
int id;
double x;
double y;
};

int main(){

//Attributes
int M = 2 ;//anchor
int N = 2 ;//nonanchor
int NA = 4;
int p = 0;
int numOfIterations =2;
MemberLocation *member1 = new MemberLocation[M];
MemberLocation *member2 = new MemberLocation[N];
MatrixXd neighbourdistanceEst(M,N);
MatrixXd neighbourdistanceDeriv(M,NA);
MatrixXd distanceDerivTranspose(NA,M);
MatrixXd firstMultiplyMat(NA,NA);
MatrixXd firstMultiplyMatInverse(NA,NA);

//Structs initilaization
member1[0].id = 5;
member1[0].x = 9.31301;
member1[0].y = 19.3955;

member1[1].id = 2;
member1[1].x = 46.6279;
member1[1].y = 0.00571905;

member2[0].id = 4;
member2[0].x = 11.7718;
member2[0].y = 7.99507;

member2[1].id = 6;
member2[1].x = 23.6158;
member2[1].y = 3.80408;

// Filling "neighbourdistanceDeriv" matrix
for (int i = 1 ; i <  numOfIterations ; i++ ){

for (int j = 0 ; j < M ; j++){
for (int k = 0 ; k < N ; k++){

    int id1 = member2[k].id; 
    int id2 = member1[j].id; 
    double xDiff = member2[k].x - member1[j].x; 
    double yDiff = member2[k].y - member1[j].y;
    double distance = pow(xDiff,2) + pow(yDiff,2);           

    neighbourdistanceEst(j,k) = sqrt(distance);

    neighbourdistanceDeriv(j,p) = xDiff / neighbourdistanceEst(j,k);
    neighbourdistanceDeriv(j,p+1) = yDiff / neighbourdistanceEst(j,k);

    p+=2;

}
p = 0;
}
}

// operations on "neighbourdistanceDeriv" matrix
distanceDerivTranspose = neighbourdistanceDeriv.transpose();
firstMultiplyMat = distanceDerivTranspose*neighbourdistanceDeriv;
firstMultiplyMatInverse = firstMultiplyMat.inverse();

// printing "firstMultiplyMat" matrix and its inverse
std::cout << "firstMultiplyMat:\n" << firstMultiplyMat << std::endl;
std::cout << "inverse of firstMultiplyMat:\n" << firstMultiplyMatInverse << std::endl<< std::endl;

// fixed array with same values obtained in "neighbourdistanceDeriv" matrix at runtime
MatrixXd a(4,4);
MatrixXd b(4,4);
a(0,0) =  0.994532;
a(0,1) =   -0.423853;
a(0,2)=  1.10423 ;
a(0,3) =    -0.314096;
a(1,0) = -0.423853 ;
a(1,1) = 1.00547 ;
a(1,2) =   -0.881237 ;
a(1,3) =  0.756726 ;
a(2,0) = 1.10423 ;
a(2,1) =   -0.881237;
a(2,2) =   1.43045;
a(2,3) =   -0.658827;
a(3,0) = -0.314096;
a(3,1) =  0.756726  ;
a(3,2) =  -0.658827;
a(3,3) = 0.569549 ;
b = a.inverse();
std::cout << "matrix a :\n" << a <<std::endl;
std::cout << "inverse of matrix a :\n" << b << std::endl;
//std::cout<<"hi \n";
return 0;
}

输出：

firstMultiplyMat:
0.994534 -0.423857   1.10423 -0.314099
-0.423857   1.00547 -0.881237  0.756725
1.10423 -0.881237   1.43045 -0.658827
-0.314099  0.756725 -0.658827  0.569548
inverse of firstMultiplyMat:
-nan -nan -nan -nan
-nan -nan -nan -nan
-nan -nan -nan -nan
-nan -inf -inf  inf

matrix a :
0.994532 -0.423853   1.10423 -0.314096
-0.423853   1.00547 -0.881237  0.756726
1.10423 -0.881237   1.43045 -0.658827
-0.314096  0.756726 -0.658827  0.569549
inverse of matrix a :
-119414     -1681.81       132361      89489.1
-1681.81       942762      10176.4 -1.24175e+06
132361      10176.4      -146638      -110149
89489.1 -1.24175e+06      -110149  1.57177e+06

Answer 1

矩阵firstMultiplyMat本身记忆力很好;如果你将值放入已经放入的值中，它就会正确反转。你的问题在于firstMultiplyMat中的值，这些值与完全的值相同，因为打印的内容是截断的。

基本上，您的矩阵可能不可逆。您可以通过不计算逆而使用QR求解器来解决这个问题，例如：

HouseholderQR<MatrixXd> qr(A);
x = qr.solve(b); // computes A^-1 * b

有关HouseholderQR的更多信息，请访问here

特征库的矩阵逆函数返回NaN

1 个答案: