Is there a way to understand what happens when a singular operator is discretized and inverted using the pseudoinverse (say using the SVD Moore-Penrose pseudoinverse)?  For example, if we discretize $(\nabla u, \nabla v) $ with a finite element method or something, the operator should usually be singular due to the fact that constants are in the nullspace. Is it known what the pseudoinverse does in such cases - i.e. will it project the solution onto $U/\mathbb{R}$, or does its effect depend other factors such as the choice of basis/discretization? 
Pseudo-inverse of a discretized operator with a null space?
finite element;linear solver;discretization;svd
The Moore-Penrose pseudoinverse of a matrix has the property that $x=A^{\dagger}b$is a least squares solution, and that among all least squares solutions (if there's a nontrivial nullspace of $A$), $x$ will be the least squares solution that minimizes $\| x \|_{2}$.  When it comes to discretizing a PDE and solving the linear system of equations $Ax=b$ that results from that discretization, using the pseudoinverse will get you a least squares solution that minimizes the norm of $x$, but depending on the discretization that you've used (or even a change of basis for the linear system of equations), minimizing the norm of the vector $x$ might not be equivalent to minimizing the norm of the solution.  It's easy to construct examples of bases where this doesn't work out.  If your change of basis is orthogonal, then you'll actually be OK.  For example, consider using the discrete Fourier transform as a change of basis.   