1. ## Question about Residual Norms

Suppose A is 4x4, with 1s in the positions (1,2), (2,3), (3,4), & (4,1), and zeros everywhere else. Let b= $e_{1}$ and $x_{0}=0$. 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!

2. Hello.

Just to be sure, how is x_k defined for successive k? Is x_k=Ar_{k-1} for k=1,2,...?

3. Yes. You are correct.

4. I was just doing some calculations, and it doesn't seem to work out the way it is supposed to.

Are you sure that x_k = Ar_{k-1} for k=1,2,...? Or could it be x_k = Ax_{k-1} for k=1,2,...? If it is this latter option, the problem will turn out to be somewhat easy (and true).

If it is x_k = Ar_{k-1}, then r_0 is equal to -e_1, which has norm 1. Then x_1 = Ar_0 = -Ae_1 = -e_4, and so r_1 = Ax_1-b = -Ae_4 - b = -e_3 - e_1, which has norm sqrt(2).