Modify Eigen matrix diagonal

Question

I have an Eigen::MatrixXd and I need to modify the value of the elements in its diagonal. In particular I have another Eigen::MatrixXd with one single column and the same number of rows of the first matrix.

I need to subtract to the diagonal of the first matrix the value of the elements of the second matrix.

Example:

A
 1 2 3
 4 5 6
 7 8 9

B
 1
 1
 1


A'
 0 2 3
 4 4 6
 7 8 8

How can I do?

Zincograph

Subtract some arbitrary diagonal? Or is it always 1's (in other words are you always subtracting the identity matrix) – Haricot

@CoryKramer arbitrary diagonal – Zincograph

Answer 1

This works for me:

A_2=A-B.asDiagonal();

Gossip

Answer 2

The simplest and fastest way to do achieve this is:

Eigen::MatrixXd A1(3,3), B(3,1), A2;
...
A2 = A1;
A2.diagonal() -= B;

of course, better use the VectorXd type for vectors (here for B), and finally if B is constant, then you can use array facilities:

A2.diagonal().array() -= 1;

Developing

How is this simpler and faster than A - B.asDiagonal()? – Rahr

Perhaps because it avoids creating B as a diagonal matrix and only has to do A2.rows() calculations rather than A2.size? Although I suppose Eigen is smart enough when it knows B.asDiagonal() is diagonal. – Menagerie

Answer 3

for(int i = 0; i < matrix1.rows(); ++i)
    matrix1(i, i) -= matrix2(i, 0);

This code iterates over each row of the matrix (matrix1.rows()) and subtracts the corresponding value of matrix2 (matrix2(i, 0)) from the diagonals in matrix1 (matrix1(i, i)).

Cryptogram

While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value. – Slovak

Answer 4

Matrix manipulation in Eigen works very similar to that of arrays. The indexing starts from zero and it is row major. The documentation (Eigen: The Matrix Class is well written and may help you to solve future problems.

For your stated problem the solution is given below.

#include <iostream>
#include <eigen3/Eigen/Dense>
using Eigen::MatrixXd;
int main()
{
  MatrixXd matA(3,3), matB(3,1);
  matA<<1,2,3,
    4,5,6,
    7,8,9;
  matB<<1,1,1;
  for(int i=0; i<3;i++)
    matA(i,i) -= matB(i);
  std::cout<<matA<<std::endl;
  return 0;
}

However, I would use Matrix3d and Vector3d for the first and second matrix, respectively.

Dramatization

It's probably better to use matB.rows() instead of hard-coded number 3, as done in the answer by @otah007 – Dramatization