# Gauss-Seidl, Gauss-Southwell, Jacobi...

by onako
Tags: gaussseidl, gausssouthwell, jacobi
 P: 87 Let me clarify what I'm looking for : given a function $$f(x)=\frac{1}{2}x^TAx - b^Tx,$$ and the update $$x_1=D^{-1}(b-Rx_0)$$ where A=D+R, with the being the diagonal matrix with diagonal entries of A (and R containing the off-diagonal entries of A), how could I show $$f(x_0)>f(x_1)$$
 P: 87 Many thanks for the suggestion. Note that following: $$D^{-1}R=D^{-1}(V-D)=D^{-1}V-I$$ In my choice of V, the matrix $$D^{-1}V$$ reduces to a symmetric matrix with the only non-negative elements on the diagonal begin equal to 1: $$D^{-1}V=\left(\begin{array}{ccccc} 1 & & & & \\ & \ddots & & -c_{ij} \\ & & \ddots & & \\ & -c_{ij} & & \ddots & \\ & & & & 1 \end{array}\right) \in\mathbb{R}^{n\times n}$$ such that the sum of the absolute values of all the off-diagonal elements in each row and column sum to 1, which is the corresponding diagonal element. By matrix difference $D^{-1}V-I$, this implies that the diagonal elements of the matrix $D^{-1}R$ are zero, and in terms of the proposed Gershgorin circle theorem, this would correspond to n disks D(0, 1), meaning that eigenvalues of $$eigs(D^{-1}R)\leq 1$$ (correct me if I'm wrong). How to proceed with the proof further?