Convergence of a modified Gauss-Seidel method Given a symmetric, positive definite matrix and a (point decomposition)
of it into the form $A = D - E - F$, we shall examine an iterative method for the
solution of the linear system $Au = b$.
Given an arbitrary vector $u_0$, the sequence $(u_k)$ is defined by,
\begin{align*}
    (D-E)x_{k+\frac{1}{2}}&=Fx_k+b\\
    (D-F)x_{k+1}&=Ex_{k+\frac{1}{2}}+b.
    \end{align*}
I need to prove the convergence of this method.
My attempt:
I founded the iteration matrix. To be, $B=(D-F)^{-1}E(D-E)^{-1}F.$ So, I need to prove that $\rho(B)<1$, where $\rho$ is the spectral radius of the matrix. The book suggests establishing the
relation, $$Bp = \lambda p\quad \Rightarrow \quad \lambda D^{-1}Ap+(\lambda-1)D^{-1}ED^{-1}Fp=0,$$ but I couldn't figure out neither how to prove that relation above or how to use it in order to obtain that $\rho(B)<1$.
 A: The errors satisfy
$$
e_{k+1/2}=(I-M^{-1}A)e_{k}, \quad e_{k+1}=(I-M^{-T}A)e_{k+1/2},
$$
where $M=D-E$ (note that $F=E^T$ since $A$ is symmetric). If we manage to show that a norm of $I-M^{-1}A$ and $I-M^{-T}A$ is strictly smaller than one, this will show the convergence of the combined iteration. Since $A$ is SPD, we can consider the matrix $A$-norm
$$
\|I-M^{-1}A\|_A=\|I-A^{1/2}M^{-1}A^{1/2}\|_2=\|I-A^{1/2}M^{-T}A^{1/2}\|_2=\|I-M^{-T}A\|_A.
$$

Fact $\|I-M^{-1}A\|_A<1$ if and only if $M+M^T-A$ is SPD.

Proof: $\|I-M^{-1}A\|_A<1$ is equivalent to
$$\tag{1}
\|(I-M^{-1}A)x\|_A<\|x\|_A\iff
0<x^TAx-x^T(I-M^{-1}A)^TA(I-M^{-1}A)x
$$
for any nonzero $x$. We can elaborate on (1) a bit and get
$$
\begin{split}
0&<x^TAx-x^T(I-M^{-1}A)^TA(I-M^{-1}A)x
\\&=x^TAx-x^T(A-AM^{-1}A-AM^{-T}A+AM^{-T}AM^{-1}A)x
\\&=x^T(AM^{-1}A+AM^{-T}A-AM^{-T}AM^{-1}A)x
\\&=x^TAM^{-T}(M+M^T-A)M^{-1}Ax
\\&=y^T(M+M^T-A)y. \quad \Box
\end{split}
$$
Now with $M=D-E$, we get that
$$
M+M^T-A=D-E+D-E^T-D+E+E^T=D
$$
which is clearly positive definite (an SPD matrix has SPD diagonal part).
Hence
$$
\|I-M^{-1}A\|_A=\|I-M^{-T}A\|_A=:\alpha<1,
$$
so
$$
\rho(B)\leq\|B\|_A=\|(I-M^{-T}A)(I-M^{-1}A)\|_A\leq\alpha^2<1.
$$
