特征库的矩阵逆函数返回NaN。

This is a new created example having the same problem. The 2 matrices "firstMultiplyMat" , "a " are having the same values, but specified in different ways. Nevertheless, the inverse of each one is different. Cannot get inverse for "firstMultiplyMat" whereas "a" has correct calculated inverse matrix.

这是一个有相同问题的新创建的示例。两个矩阵“firstMultiplyMat”，“a”具有相同的值，但以不同的方式指定。然而，每一个的逆都是不同的。而“a”则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;
}

The output :

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

1 个解决方案

#1

The matrix firstMultiplyMat itself is fine memory-wise; if you push the values into it that you had put into a, it'll invert correctly. Your problem lies in the values that are inside firstMultiplyMat, which are not exactly the same values as what get printed, because what is printed is truncated.

矩阵的firstMultiplyMat本身就是一个很好的记忆体;如果你把它的值放到a中，它就会正确地转化。您的问题在于firstMultiplyMat内部的值，这些值与打印的值不完全相同，因为打印的内容被截断了。

Essentially, your matrix may not be invertible. You can work around this by just not computing the inverse and instead using a QR solver, eg:

本质上，你的矩阵可能不是可逆的。你可以通过不计算逆的方法来解决这个问题，而是用一个二维码。

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

More information about HouseholderQR is available here

这里有更多关于家庭住宅的信息。

#1