Dear all,

I am doing the most "trivial" thing but cannot get it to work. I am solving a A*x=b type system of 3 equations of 3 unknowns. The matrix is of rank 1, so IIRC a solution must exist. I am comparing

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x = A.colPivHouseholderQr().solve(b);


compiled against eigen 3.1.4 shipped with fc19. For

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MatrixXf sobel(3,3);
sobel <<
    1,0,-1,
    2,0,-2,
    1,0,-1;
VectorXf x(3),b(3);
 b << 1,2,3;


Eigen gives me:

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#from stdout
x =
1.33333
0
0


If I do the same thing with octave (version 3.6.4), I get a different result (that is not even linearly dependent on the eigen result given above):

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sobel = [  1,0,-1 ; 2,0,-2 ; 1,0,-1 ];
b = [1;2;3];
x = b\sobel
x =                                                                                           
                                                                                             
   0.57143  -0.00000  -0.57143


I'd appreciate any hint on what could cause this severe difference.

Best -

Your system of equations has no solutions. If the matrix A is singular, as is the case here, then the equation A*x=0 has nonzero solutions, but the system A*x=b usually has no solutions. Generally, solving systems of equations with singular matrices numerically is not the most trivial thing; non-singular matrices are much easier to handle.

In Octave, the command b\A solves the system x*A=b and indeed the result is a row vector. To solve the system A*x=b you have to compute A\b.

Indeed, using a different matrix:
0.32158 0.08359 0.03690
0.00000 0.47783 0.19056
0.00000 0.00000 0.58779
The solution coincide as they should be.
I had the feeling in the first place, that I oversaw something obvious.

Many thanks for you reply, eigen is a great tool - keep it up.
Peter

Now the question rises, how is b\A mimicked in eigen? I am asking since this is what I actually need.
I know that solving x*A = b can be expressed as A.transpose()*z = b.transpose(). However, the solvers in eigen tend to take column vectors as arguments, not row vectors. Any hints!
Peter

Addendum to the above: x*A = b is equivalent to solving A.transpose()*z = b.transpose() with x = z.transpose()