Hi,
I have an Eigen::MatrixXd m(9,9);
And when I am doing this:

m.inverse();

I am getting the same output.

Is there any other way to invert such a large matrix?

Thanks in advance.

As explained in the doc (http://eigen.tuxfamily.org/dox-devel/cl ... 030f79f9da), mat.inverse() returns the inverse of mat, keeping mat unchanged:

Code: Select all

mat_inv = mat.inverse();


My problem is that when i calculate the inverse of my 9x9 matrix I get "nan" values.

I am having these values in my matrix:

1384.33 0 0 0 0 0 553.1 0 -1707.22
0 673.42 673.42 553.1 -601.406 229.515 0 241.537 0
0 673.42 673.42 553.1 -601.406 229.515 0 241.537 0
0 553.1 553.1 9288.69 771.387 4054.28 0 -672.069 0
0 -601.406 -601.406 771.387 2848.9 781.647 0 -777.88 0
0 229.515 229.515 4054.28 781.647 2986.75 0 -459.042 0
553.1 0 0 0 0 0 771.387 0 -672.069
0 241.537 241.537 -672.069 -777.88 -459.042 0 781.647 0
-1707.22 0 0 0 0 0 -672.069 0 4054.28

And when I printing the inverse I get:

-nan -nan -nan -nan -nan -nan -nan -nan -nan
-nan -nan -nan -nan -nan -nan -nan -nan -nan
-nan -nan -nan -nan -nan -nan -nan -nan -nan
-nan -nan -nan -nan -nan -nan -nan -nan -nan
-nan -nan -nan -nan -nan -nan -nan -nan -nan
-nan -nan -nan -nan -nan -nan -nan -nan -nan
-nan -nan -nan -nan -nan -nan -nan -nan -nan
-nan -inf inf -nan -nan -nan -nan -nan -nan
0.000636602 0 0 0 0 0 -9.36032e-06 0 0.000513168

I am not getting what's going wrong.

Your matrix is singular, so not invertible. Confirmed by matlab:

Code: Select all

inv(M)
warning: inverse: matrix singular to machine precision, rcond = 0
ans =

   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf
   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf
   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf
   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf
   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf
   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf
   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf
   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf
   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf   Inf


If what you what is to solve for A x = b, or equivalently apply the inverse of A to tome matrices (A^-1 * B), then use a rank-revealing decomposition, e.g.:

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ColPivHouseholderQr<MatrixXd> qr(A);
 ... qr.solve(B) ... ; // <=> A^-1 * B


Yeah...
this is what i want to do.

x = bA^-1

where, x = 3x9 mat
b = 3x9 mat
A = 9x9 mat

Is it possible to compute this with LU decomposition as I tried but it is crashing?
Or can I get inverse of A using LU decomposition?

You have many options:

Code: Select all

x = A.lu().solve(b);
x = A.fullPivLu().solve(b);
x = A.colPivHouseholderQr().solve(b);
x = A.jacobiSVD(ComputeThinU|ComputeThinV).solve(b);


If you want to solve for the same matrix multiple times, build the respective factorization object as shown in my previous post.

I get this error when I am doing this:

x = A.lu().solve(b)

where, x = 3x9 mat
b = 3x9 mat
A = 9x9 mat

DeformationDynamics: ./Eigen/src/LU/PartialPivLU.h:449: void Eigen::internal::solve_retval<Eigen::PartialPivLU<_MatrixType>, Rhs>::evalTo(Dest&) const [with Dest = Eigen::Matrix<double, 3, 9>; _MatrixType = Eigen::Matrix<double, 9, 9>; Rhs = Eigen::Matrix<double, 3, 9>]: Assertion `rhs().rows() == dec().matrixLU().rows()' failed.

x.transpose() = A.lu().solve(b.transpose());

Yeah...
I did that but I am still getting some "nan" values in my output matrix.

which solver did you tried? I shown the LU one for reference only, but is well known that a partial pivoting LU requires the matrix to be invertible, so not applicable in your case. This is also explain in our doc: http://eigen.tuxfamily.org/dox-devel/gr ... tions.html.

Have you tried the 3 others options I suggested? They are all working for me:

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#include <iostream>
#include <Eigen/Dense>
using namespace Eigen;
int main() {
  MatrixXd A(9,9);
  A <<  1384.33, 0, 0, 0, 0, 0, 553.1, 0, -1707.22,
        0, 673.42, 673.42, 553.1, -601.406, 229.515, 0, 241.537, 0,
        0, 673.42, 673.42, 553.1, -601.406, 229.515, 0, 241.537, 0,
        0, 553.1, 553.1, 9288.69, 771.387, 4054.28, 0, -672.069, 0,
        0, -601.406, -601.406, 771.387, 2848.9, 781.647, 0, -777.88, 0,
        0, 229.515, 229.515, 4054.28, 781.647, 2986.75, 0, -459.042, 0,
        553.1, 0, 0, 0, 0, 0, 771.387, 0, -672.069,
        0, 241.537, 241.537, -672.069, -777.88, -459.042, 0, 781.647, 0,
        -1707.22, 0, 0, 0, 0, 0, -672.069, 0, 4054.28;
   RowVectorXd x(9), b(9);
   x.setRandom();
   b.transpose() = A * x.transpose(); // make sure a solution exists
   x.setZero();
   
   x.transpose() = A.jacobiSvd(ComputeThinU|ComputeThinV).solve(b.transpose());
   std::cout << "svd: " << (A * x.transpose() - b.transpose()).norm() << "\n";
   
   x.transpose() = A.fullPivLu().solve(b.transpose());
   std::cout << "LU:  " << (A * x.transpose() - b.transpose()).norm() << "\n";
   
   x.transpose() = A.colPivHouseholderQr().solve(b.transpose());
   std::cout << "QR:  " << (A * x.transpose() - b.transpose()).norm() << "\n";
}


Output:

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svd: 2.82792e-12
LU:  4.58286e-13
QR:  1.02002e-12


Thank you for the solution but I don't understand why these really large values (2.82792e-12) are coming and how can I avoid them?

And one more thing rather a stupid question but if the inverse of the matrix doesn't exist then how does the solution is reliable through these decomposition?

This explains the process:

https://www.youtube.com/watch?v=Ih1Vusw_blA