Consider the following rule: $$x_{k+1}=x_k-sign(Ax_k)$$ for $x_k\in\mathbb{R}^2$, $A\in\mathbb{R}^{2*2}$. Let $$A=\begin{bmatrix}3/2&-1/2\\ -1/2&3/2\end{bmatrix}$$ How to prove that for sufficiently large $n$, $x_{n+2}=x_n$?
What I know: $\sigma(A)=\{1,2\}$, then $$x_{k+1}^TAx_{k+1}\leq x_k A x_k -2 ||Ax_k||_1 + 4,$$ then for sufficiently large $n$, we have $ ||Ax_k||_1\leq 2$. Moreover, $x_k$ can only flow along the "grid," so at the end, $x$ must be periodic. But how to show the period is 2?