numpy opencv matlab eigen SVD结果对比

参考

https://zhuanlan.zhihu.com/p/26306568

https://byjiang.com/2017/11/18/SVD/

http://www.bluebit.gr/matrix-calculator/

https://*.com/questions/3856072/single-value-decomposition-implementation-c

https://*.com/questions/35665090/svd-matlab-implementation

矩阵奇异值分解简介及C++/OpenCV/Eigen的三种实现

https://blog.csdn.net/fengbingchun/article/details/72853757

numpy.linalg.svd 源码

https://docs.scipy.org/doc/numpy/reference/generated/numpy.linalg.svd.html

计算矩阵的特征值与特征向量的方法

https://blog.csdn.net/Junerror/article/details/80222540

https://jingyan.baidu.com/article/27fa7326afb4c146f8271ff3.html

不同的库计算结果不一致

原因在于特征向量不唯一,特征值是唯一的

来源

https://*.com/questions/35665090/svd-matlab-implementation

Both are correct... The rows of the v you got from numpy are the eigenvectors of M.dot(M.T) (the transpose would be a conjugate transpose in the complex case). Eigenvectors are in the general case defined only up to a multiplicative constant, so you could multiply any row of v by a different number, and it will still be an eigenvector matrix.

There is the additional constraint on v that it be a unitary matrix, which loosely translates to its rows being orthonormal. This reduces your available choices for every eigenvector to only 2: the normalized eigenvector pointing in either direction. But you still get to multiply any row by -1 and still have a valid v.

 

A = U * S * V

1 手动计算

给定一个大小为numpy opencv matlab  eigen SVD结果对比的矩阵numpy opencv matlab  eigen SVD结果对比,虽然这个矩阵是方阵,但却不是对称矩阵,我们来看看它的奇异值分解是怎样的。

numpy opencv matlab  eigen SVD结果对比进行对称对角化分解,得到特征值为numpy opencv matlab  eigen SVD结果对比numpy opencv matlab  eigen SVD结果对比,相应地,特征向量为numpy opencv matlab  eigen SVD结果对比numpy opencv matlab  eigen SVD结果对比;由numpy opencv matlab  eigen SVD结果对比进行对称对角化分解,得到特征值为numpy opencv matlab  eigen SVD结果对比numpy opencv matlab  eigen SVD结果对比

当 lamda1 = 32

AA.T  -  lamda1*E = [-7 7

         7 -7]

线性变换 【-1 1

     0 0】

-x1 + x2 = 0

x1 = 1 x2 = 1 特征向量为【1 1】.T  归一化为【1/sqrt(2), 1/sqrt(2)】

x1 = -1 x2 = -1 特征向量为【-1 -1】.T  归一化为【-1/sqrt(2), -1/sqrt(2)】

当 lamda2 = 18

AA.T  -  lamda2*E = [7 7

         7 7]

线性变换 【1 1

     0 0】

x1 + x2 = 0

如果x1 = -1 x2 = 1 特征向量为【-1 1】.T  归一化为【-1/sqrt(2), 1/sqrt(2)】

如果 x1 = 1 x2 = -1 特征向量为【-1 1】.T  归一化为【1/sqrt(2), -1/sqrt(2)】可见特征向量不唯一

特征向量为numpy opencv matlab  eigen SVD结果对比numpy opencv matlab  eigen SVD结果对比。取numpy opencv matlab  eigen SVD结果对比,则矩阵numpy opencv matlab  eigen SVD结果对比的奇异值分解为

numpy opencv matlab  eigen SVD结果对比

numpy opencv matlab  eigen SVD结果对比.

2 MATLAB 结果与手动计算不同

AB =  [[ 4 4 ],
[-3 3 ]]
[U,S,V] = svd(AB);
U
S
V'#需要转置

AB =

4 4
-3 3

U =

-1 0
0 1

S =

5.6569 0
0 4.2426

V =

-0.7071 -0.7071
-0.7071 0.7071

3 NUMPY结果与手动计算不同, 与matlab相同 它们都是调用lapack的svd分解算法。

import numpy as np
import math
import cv2
a = np.array([[4,4],[-3,3]])
# a = np.random.rand(2,2) * 10
print(a)
u, d, v = np.linalg.svd(a)
print(u)
print(d)
print(v)#不需要转置

A = [[ 4 4]
[-3 3]]

U = 
[[-1. 0.]
[ 0. 1.]]

S=
[5.65685425 4.24264069]

V=
[[-0.70710678 -0.70710678]
[-0.70710678 0.70710678]]

4 opencv结果 与手动计算结果相同

import numpy as np
import math
import cv2
a = np.array([[4,4],[-3,3]],dtype=np.float32)
# a = np.random.rand(2,2) * 10
print(a)
S1, U1, V1 = cv2.SVDecomp(a)
print(U1)
print(S1)
print(V1)#不需要转置

a = [[ 4. 4.]
[-3. 3.]]
U =

[[0.99999994 0. ]
[0. 1. ]]
S = [[5.656854 ]
[4.2426405]]

V =

[[ 0.70710677 0.70710677]
[-0.70710677 0.70710677]]

5  eigen结果与手动计算相同

#include <iostream>
#include <Eigen/SVD>
#include <Eigen/Dense>
#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include "opencv2/imgproc/imgproc.hpp"
#include <iostream> using namespace std;
using namespace cv; //using Eigen::MatrixXf;
using namespace Eigen;
using namespace Eigen::internal;
using namespace Eigen::Architecture; int GetEigenSVD(Mat &Amat, Mat &Umat, Mat &Smat, Mat &Vmat)
{
//-------------------------------svd测试 eigen
Matrix2f A;
A(0,0)=Amat.at<double>(0,0);
A(0,1)=Amat.at<double>(0,1);
A(1,0)=Amat.at<double>(1,0);
A(1,1)=Amat.at<double>(1,1); JacobiSVD<Eigen::MatrixXf> svd(A, ComputeThinU | ComputeThinV );
Matrix2f V = svd.matrixV(), U = svd.matrixU();
Matrix2f S = U.inverse() * A * V.transpose().inverse(); // S = U^-1 * A * VT * -1
//Matrix2f Arestore = U * S * V.transpose();
// printeEigenMat(A ,"/sdcard/220/Ae.txt");
// printeEigenMat(U ,"/sdcard/220/Ue.txt");
// printeEigenMat(S ,"sdcard/220/Se.txt");
// printeEigenMat(V ,"sdcard/220/Ve.txt");
// printeEigenMat(U * S * V.transpose() ,"sdcard/220/U*S*VTe.txt"); Umat.at<double>(0,0) = U(0,0);
Umat.at<double>(0,1) = U(0,1);
Umat.at<double>(1,0) = U(1,0);
Umat.at<double>(1,1) = U(1,1);
Vmat.at<double>(0,0) = V(0,0);
Vmat.at<double>(0,1) = V(0,1);
Vmat.at<double>(1,0) = V(1,0);
Vmat.at<double>(1,1) = V(1,1);
Smat.at<double>(0,0) = S(0,0);
Smat.at<double>(0,1) = S(0,1);
Smat.at<double>(1,0) = S(1,0);
Smat.at<double>(1,1) = S(1,1);
// Smat.at<double>(0,0) = S(0,0);
// Smat.at<double>(0,1) = S(1,1);
//-------------------------------svd测试 eigen return 0;
} int main()
{
// egin();
// opencv3();
//Eigentest();
//opencv();
//similarityTest();
// double data[2][2] = { { 629.70374, 245.4427 },
// { -334.8119 , 862.1787 } }; double data[2][2] = { { 4, 4 },
{-3, 3} };
int dim = 2;
Mat A(dim,dim, CV_64FC1, data);
Mat U(dim, dim, CV_64FC1);
Mat V(dim, dim, CV_64FC1);
Mat S(dim, dim, CV_64FC1);
GetEigenSVD(A, U, S, V);
Mat Arestore = U * S * V.t();
cout <<A<<endl;
cout <<Arestore<< endl;
cout <<"U " << U<<endl;
cout <<"S " <<S<<endl;
cout <<"V " << V.t()<<endl;
cout <<V<<endl; return 0;
}

[4, 4;
-3, 3]

U =

[0.9999999403953552, 0;
0, 0.9999999403953552]
S =

[5.656854629516602, 0;
0, 4.242640972137451]
V =

[0.7071067690849304, 0.7071067690849304;
-0.7071067690849304, 0.7071067690849304]

6 在线计算网站 与手动计算不同

http://www.bluebit.gr/matrix-calculator/calculate.aspx

Input matrix:

 4.000  4.000
-3.000 3.000

Singular Value Decomposition:

U:

-1.000  0.000
0.000 1.000

S:

5.657 0.000
0.000 4.243

VT

-0.707 -0.707
-0.707 0.707
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