Condition number of a non-invertible matrix and solving with quantum linear systems algorithms

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In this paper one of the things they do is solve the Poisson equation with periodic BCs by using the finite difference representation then using a quantum linear systems algorithm to solve the resulting matrix equation $L\vec{u}=\vec{f}$.

With the boundary conditions they use the matrix $L$ is non-invertible (u(x) is a solution then so is $u(x)+c, c \in \mathbb{R}$). This is mentioned on page 8, but they then go on to calculate the condition number and I'm a bit confused as to why the condition number isn't just $\infty$? Then discussing solving with a quantum linear systems algorithm whose complexity involves the condition number.

I know the normalisation will make the solution state $\vec{u}$ unique but I'm more confused with the use of the condition number here.

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It is true that Laplace's equation $$\Delta u(x) = 0$$ with homogenous Neumann boundary conditions does not have a unique solution and we expect this property to carry over to discrete approximations $Av = 0$ of the problem. On the other hand, Poisson's equation $$\Delta u(x) = g(x)$$ with Dirichlet boundary conditions, typically has a unique solution and this property carries to useful discrete approximations $Av=f$ of the problem.

In your case, the authors are concerned with the conditioning of the discrete Laplace operator. This means that they are considering Poisson's equation with homogeneous Dirichlet boundary conditions. A nonsingular operator and a finite condition number is expected in this case.