# Conjugate Gradient for a “slightly” singular system.

Suppose I have a symmetric $N \times N$ matrix A which has a one-dimensional Nullspace $N$. A is positive definite on $N^\bot$. In my case $N$ is the space of constant vectors (i.e. generated by the all-one vector).

I have to solve the problem $Ax = b$, with $b \in R(A)$ which has infinitely many solutions. I am looking for the minimum norm solution. The matrix $A$ is very large and sparse, direct methods aren't really an option. The rank-deficient least squares algorithms I have seen also appear to be prohibitive.

I was solving a non-singular version of this problem with Conjugate Gradient. Is there anyway I can modify the algorithm to solve this particular problem?

EDIT: Boiling the problem down to the bone the question is if $A$ is positive semi-definite with exactly one 0 eigenvalue, does CG work?

Thanks.

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mathoverflow.net/questions/7903 –  J. M. Aug 15 '10 at 13:00

I'd suggest you to shift away the singularity: solve $(A+ee^T)y=b$ instead and then orthogonalize $y$ with respect to $e$ to get $x$. $A+ee^T$ is not sparse but you can compute matrix-vector products cheaply, and that's all you need for CG.

EDIT: forgot to define it, $e$ is the vector of all ones

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This is exactly what I did. With the help of a preconditioner it converges sufficiently fast. In hindsight it was the obvious route to take. Thanks for pointing it out to me. –  RadonNikodym Aug 16 '10 at 8:21

Plain CG, without the regularization, will converge to the pseudo-inverse solution provided you start the iteration with $y=0$. The regularized (preconditioned) solution will be different and the operation count will depend on the sparsity, so it's good to maintain this if you can. I guess partly it must depend on whether you can tolerate a solution that has a nonzero component in the null space.

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You can add a linear constraint, stipulating that your solution should be orthogonal to the null space, assuming it is known to you ahead of time.

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