Given the system $$Ax=b,$$ where $A\in\mathbb{R}^{n\times n}$, I read that, in case Jacobi iteration is used as a solver, the method will not converge if $b$ has a non-zero component in the null-space of $A$.  So, how could one formally state that, provided that $b$ has a non-zero component spanning the null-space of $A$, Jacobi method is non-convergent? I wonder how could that be mathematically formalized, since part of the solution orthogonal to the null-space does converge.Therefore, by projecting the null-space of $A$ out of each iterate, it converges (or?). .........I'm particularly interested in the case of $$Lx=b,$$where $L$ is a symmetric Laplacian matrix with the null-space spanned by a vector $1_n=[1\dots 1]^T\in\mathbb{R}^n$, and $b$ has a zero component in the null-space of $L$, $$Jb=b,$$ where $J=I-\frac{1}{n}1_n1_n^T$ is the centering matrix. Does that imply that each Jacobi iterate will have the null-space of $L$ projected out, ie., each iterate will be centered? I'm asking this since then there would be no need to project out the null-space of $L$ from Jacobi iterates (or, in other words, to center the iterates).
Projecting out the null-space of $A$ from $b$ in $Ax=b$
linear algebra;optimization;matrices;iterative method;linear solver
The correct condition for solvability has nothing to do with the null space of $A$ (unless $A$ is symmetric) but with the null space of $A^T$. If $A^Tu=0$ then $Ax=b$ implies that $u^Tb=u^TAx=0$, hence $b$ must be orthogonal to any null vector of $A^T$ (otherwise there is no solution, and the Jacobi iteration has no reason to converge).But if this is the case, a solution exists, and in the square case there are infinitely many. In the singular case, as one never knows whether this condition is satisfied (and it would be spoiled by roundoff anyway), one would typically solve the problem as a least squares problem. To find the minimum norm solution, use conjugate gradients on the normal equations; this requires that you code multiplication by $A$ and by $A^T$. (Given only a routine for multiplying with $A$, one could use GMRES instead, with less predictable convergence properties.)