# Why is the residual vector after one Gauss-Seidel sweep on a red-black ordering equal to 0 in the black-corresponding nodes?

Working with finite difference methods here. One possible ordering of the gridpoints is red-black. This results in a system matrix of the form $$\begin{pmatrix}A_{RR} & A_{RB} \\ A_{BR} & A_{BB} \end{pmatrix}$$

where $$A_{RR} = A_{BB} = 4I_n = \begin{pmatrix} 4 & 0 & \dots & 0 \\ 0 &\ddots & & \vdots\\ \vdots & & & 0\\ 0 & \dots &0 & 4 \end{pmatrix}$$

and $$A_{RB}=A_{BR} =pentadiag[-1 -1 -1 -1 -1] = \begin{pmatrix} -1 & -1 & -1 & 0 &\dots & 0 \\ -1 &\ddots & \ddots& \ddots & & \\ -1 &\ddots & \ddots & & & \vdots \\ 0 & \ddots & & & & \\ & & & & \\ \vdots & & & & \ddots & 0 \\ & & & \ddots & \ddots & -1 \\ & & \ddots& \ddots & \ddots& -1 \\ 0 & \dots &0 & -1 & -1 & -1 \end{pmatrix}$$
(I hope that is understandable)

Now, using the Gauss-Seidel elimination, we want to show that the residual vector after one Gauss-Seidel sweep on such a red-black finite difference ordering is zero in the components that corresponds to the black nodes?

We're provided an answer, but I really do not understand it. It is as follows:

The Gauss-Seidel iteration is such that during the k-th iteration the componentsof the residual vector $$r^k$$ are consecutively made zero. In a red-black orderingthis means that the residual vector in first the red nodes and then in the blacknodes is made equal zero. The decoupling between the black nodes is such thatmaking the residual equal zero in the second black node does not affect theresidual in the first black node. After one red-black Gauss-Seidel sweep theresidual in all the black nodes is equal to zero.

What does he mean?