Hi All,

I'm solving the Ax=b problem (A is mxn where m>>n) by solving the normal equation Bx=c (B=A'A, c=A'b). Sometimes rank(A) can be smaller than n so the B matrix will be singular (but still positive semidefinite). My question is: can I use LDLT to check whether B is singular or not as follows

Code: Select all

Eigen::VectorXd D = B.ldlt().vectorD().cwiseAbs();
bool is_B_singular = D.maxCoeff() > (D.minCoeff() * 1e8);


The Eigen documentation of LDLT says that:

Remember that Cholesky decompositions are not rank-revealing. Also, do not use a Cholesky decomposition to determine whether a system of equations has a solution.

This makes me a bit confused about whether I can do this checking reliably or not.

Best,
Chen

Such a test is indeed not reliable, it will only warn you that the problem might not be reliably addressed by a LDLT decomposition and that using a rank revealing QR of SVD might be safer.

ggael wrote:Such a test is indeed not reliable, it will only warn you that the problem might not be reliably addressed by a LDLT decomposition and that using a rank revealing QR of SVD might be safer.

Thanks a lot! I was using ColPivHouseholderQR but I thought LDLT should be much faster according to the Eigen documentation. But when I test it today, it seems that LDLT is not necessarily faster than ColPivHouseholderQR in my case, so I might eventually pick to use ColPivHouseholderQR.

Is it because my B matrix is small (16x16) so ColPivHouseholderQR and LDLT do not have much speed difference?