Suppose A is 4x4, with 1s in the positions (1,2), (2,3), (3,4), & (4,1), and zeros everywhere else. Let b=$\displaystyle e_{1}$ and $\displaystyle x_{0}=0$. How can I show that||r_0||_2 = ||r_1||_2=||r_3||_2 ? r is the residual--that is, r_k=Ax_k-b

I know that the residual norms cannot increase, only decrease or stay the same. I'm trying to show that in this particular case, the residuals stay the same for up to the 3rd iteration. Any help is appreciated. Thanks!