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Solving linear least squares systems

This page describes how to solve linear least squares systems using Eigen. An overdetermined system of equations, say Ax = b, has no solutions. In this case, it makes sense to search for the vector x which is closest to being a solution, in the sense that the difference Ax - b is as small as possible. This x is called the least square solution (if the Euclidean norm is used).

The three methods discussed on this page are the SVD decomposition, the QR decomposition and normal equations. Of these, the SVD decomposition is generally the most accurate but the slowest, normal equations is the fastest but least accurate, and the QR decomposition is in between.

Using the SVD decomposition

The solve() method in the JacobiSVD class can be directly used to solve linear squares systems. It is not enough to compute only the singular values (the default for this class); you also need the singular vectors but the thin SVD decomposition suffices for computing least squares solutions:

#include <iostream>
#include <Eigen/Dense>
using namespace std;
using namespace Eigen;
int main()
cout << "Here is the matrix A:\n" << A << endl;
VectorXf b = VectorXf::Random(3);
cout << "Here is the right hand side b:\n" << b << endl;
cout << "The least-squares solution is:\n"
<< A.jacobiSvd(ComputeThinU | ComputeThinV).solve(b) << endl;
Here is the matrix A:
  0.68  0.597
-0.211  0.823
 0.566 -0.605
Here is the right hand side b:
The least-squares solution is:

This is example from the page Linear algebra and decompositions .

Using the QR decomposition

The solve() method in QR decomposition classes also computes the least squares solution. There are three QR decomposition classes: HouseholderQR (no pivoting, so fast but unstable), ColPivHouseholderQR (column pivoting, thus a bit slower but more accurate) and FullPivHouseholderQR (full pivoting, so slowest and most stable). Here is an example with column pivoting:

VectorXf b = VectorXf::Random(3);
cout << "The solution using the QR decomposition is:\n"
<< A.colPivHouseholderQr().solve(b) << endl;
The solution using the QR decomposition is:

Using normal equations

Finding the least squares solution of Ax = b is equivalent to solving the normal equation ATAx = ATb. This leads to the following code

VectorXf b = VectorXf::Random(3);
cout << "The solution using normal equations is:\n"
<< (A.transpose() * A).ldlt().solve(A.transpose() * b) << endl;
The solution using normal equations is:

If the matrix A is ill-conditioned, then this is not a good method, because the condition number of ATA is the square of the condition number of A. This means that you lose twice as many digits using normal equation than if you use the other methods.