We want to solve the linear system $Ax=b$, where $A$ is a SPD and approximate the solution $x_*$ by the following iteration. $$\begin{cases}x_{k+1}&=\arg\underset{x\in x_k+span\{e_{i_k}\}}{\min}||x-x_*||_A\\ i_k&=\arg\underset{1\le i\le n}{\max}||<x_k-x,e_i>||_A\end{cases}$$
Then I want to show $$||x_*-x_{k+1}||_A\le \left(1-\dfrac{\lambda_{\min}(A)}{n\lambda_{\max}(A)}\right)^{\frac{1}{2}}||x_*-x_k||_A$$
My Attempt
Since we have $$||x_*-x_{k+1} ||_A=||x_*-x_k||_A-\dfrac{|<A(x_k-x_*),e_{i_k}>|^2}{a_{i_k i_k}}$$ The I will apply the inequality $|e^T_iA(x_k-x_*)|\ge n^{-\frac{1}{2}}|A(x_k-x_*)|_2$
Then we have \begin{align*}||x_*-x_{k+1} ||_A&=||x_*-x_k||_A-\dfrac{||A(x_k-x_*)||_2^2}{na_{i_k i_k}}\\ &=\left<A(x_k-x_*),x_k-x_*\right>-\dfrac{1}{a_{i_ki_k}n}\left<A(x_k-x_*),A(x_k-x_*)\right>\\ &=\left<A(x_k-x_*),x_k-x_*-\dfrac{1}{a_{i_ki_k}n}A(x_k-x_*)\right>\\ &=\left<x_k-x_*,x_k-x_*-\dfrac{1}{a_{i_ki_k}n}A(x_k-x_*)\right>_A \end{align*} I consider a polynomial $p(t)=1-\frac{1}{na_{i_ki_k}}t$. Thus
\begin{align*}||x_*-x_{k+1} ||_A&=\left<x_k-x_*,p(A)(x_k-x_*)\right>_A \end{align*} Then I had no idea for the next step, I did one more step, but I didn't know if it was right.
Consider the operator norm and inequality, so $$||x_*-x_{k+1} ||_A\le |p(A)|||x_*-x_k||_A$$ But I still cant get the result