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Let $A$ be a real $n \times n$ matrix and $A$ is also symmetric definite positive (SPD). Let $b \in \mathbb R^n$. Consider the following iterative procedure to sovle $Ax=b$. $$x_{n+\frac{1}{2}}=x_n +\omega_1 (b-Ax_n)$$ $$x_{n+1}= x_{n+\frac{1}{2}}+\omega_2 (b-Ax_{n+\frac{1}{2}})$$

$\\(a)$ Let $e_n =x-x_n$ be the error. Find a matrix $K$ s.t. $e_{n+1}=Ke_{n}$. I did the calculation and got $K=I-\omega_1 A - \omega_2 A +\omega_1 \omega_2 A^2$.$$ $$ (b) Find the eigenvalues of $K$ in terms of the eigenvalues of $A$, $\omega_1$ and $\omega_2$. From part (a), I have $\lambda(K)=1-\omega_1 \lambda(A) - \omega_2 \lambda(A) +\omega_1 \omega_2 \lambda^2(A)$. My question is about part (c). $$ $$ (c) Show that $\omega_1$ and $\omega_2$ can be chosen so that the convergence method converges for any initial condition. Express the rate of convergence in terms of $\lambda_M$ and $\lambda_m$, where $\lambda_M$ and $\lambda_m$ are the max and min of eigenvalues of $A$, respectively.

I don't know how the condition of $A$ being SPD is used here. I appreciate helpful hints or solutions.

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    SPD guarantees that all eigenvalues of $A$ are positive real values, that is $\lambda(A) \in [\lambda_m, \lambda_M], \lambda_M > \lambda_m > 0$. Also, factored expression for $K$ may be better for this problem: $K = (I - \omega_1 A) (I - \omega_2 A)$ and $\lambda(K) = (1 - \omega_1 \lambda(A))(1 - \omega_2 \lambda(A))$ – 2017-01-02
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    @uranix Thank you! If we want the method converges for any initial value, then it means we need to restrict the spectral radius of $K$ to be smaller than 1, right? ( so that the error will go to zero). However, how to denote the rate of convergence? – 2017-01-02

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