Hello,
When using the method image of an FullPivLU object: the columns form a basis of the column space of the original matrix.

My questions is the following: does the columns of FullPivLU.image(A) always are some columns of the original matrix A?

Thanks a lot.

Best,

Rafael

Yes they are. See the respective code by yourself:

Code: Select all

for(Index i = 0; i < rank(); ++i)
      dst.col(i) = originalMatrix().col(dec().permutationQ().indices().coeff(pivots.coeff(i)));


Great, thank you!

BTW, merci pour cette excellente librairie!

Sorry to ask a question again but I'd like to be sure that I understood correctly.

If I want to know the rank() indices of the columns which are linerarly independant I have to take every i such that lu.permutationQ().indices()[i] < rank()

Is it correct?

Thanks!

not exactly, you have to skip the smallest pivots:

Code: Select all

    Matrix<Index, Dynamic, 1, 0, MaxSmallDimAtCompileTime, 1> pivots(rank());
    RealScalar premultiplied_threshold = dec().maxPivot() * dec().threshold();
    Index p = 0;
    for(Index i = 0; i < dec().nonzeroPivots(); ++i)
      if(abs(dec().matrixLU().coeff(i,i)) > premultiplied_threshold)
        pivots.coeffRef(p++) = i;


so what you're looking for would be a "imageIndices()" method?

Yes, that's what I need.

Finally, I took the first rank() entries of lu.permutationQ().indices() and sorted them (I need to preserve the order of my vectors).
This gives me the indices I want.