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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?

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  • $\begingroup$ As an aside, for the specific example you give, you ought to be able work through the 2x2, 3x3, ..., etc. cases and come up with a general idea for that operator (say with a small variety of finite elements) using some hand calculations and MATLAB/Octave or Numpy. I know you're looking for a general solution, but that case is easy to hand crank. $\endgroup$ – Bill Barth Sep 1 '13 at 14:46
  • $\begingroup$ That's true. I only tried for one simple cases, but it appeared that Matlab's pinv did produce zero averaged solutions for simple 1D finite elements. I wanted to know if there was a more general result or other examples known by any chance. $\endgroup$ – Jesse Chan Sep 1 '13 at 21:48
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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.

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  • $\begingroup$ Thanks. This was very helpful to clear up some of my misconceptions. $\endgroup$ – Jesse Chan Sep 2 '13 at 0:44
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It depends on how you define your pseudoinverse. If you define it as the operator that produces the least-$L_2$-norm element, then you will get that $u=u_0+c$ that has the smallest norm $\|u_0+c\|_{L_2}$ where $u_0$ is an arbitrarily chosen solution and $c$ is a constant. In general, of course, $u$ will not have zero mean value, but it will be unique.

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If your pseudoinverse is defined through SVD, then the corresponding singular vector will span the nullspace in $U_h$, so it exists. But, given the generality of the question, you will have to answer two more questions in order to get the answer to yours:

  1. Does your discrete space actually contain a nullspace? For instance, you could use a finite element space with vanishing values on the boundary (this is not excluded by your current quesation). In that case, your operator would actually be definite.

  2. Is the discrete nullspace a subspace of the continuous subspace? I can't come up with an example for the Neumann problem, but if you look at the ample literature on divergence conforming discretizations, you see that indeed there might be a discrete nullspace which is not in the kernel of the continuous operator.

If you answer both questions to the affirmative, your finite element is unisolvent, and your discrete subspaces are conforming, the answer is yes. But this does not tell you yet, in what sense it is a projection. That's where the choice of your basis and also of the discrete inner products defining the SVD come into play.

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    $\begingroup$ With vanishing values, I think that would be equivalent to applying Dirichlet conditions. The discrete nullspace is a good point, however, even without div-conforming discretizations - the operator $(Au,Av)$ (where A is the strong form of the wave equation operator) has a continuous nullspace (waves) which are only contained in the asymptotic limit as h->0. $\endgroup$ – Jesse Chan Sep 1 '13 at 21:58

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