Question: A question about 'completeness'. These two operators are commuting, but I want to know more about their completeness. 

How do you know if {H}, {B}, {H,B} and/or {$H^2$,*B*} are forming (a) Complete Set(s) of Commuting Operators?

[![Two given operators](https://i.stack.imgur.com/aXgmp.png)](https://i.stack.imgur.com/aXgmp.png) .

Answer: 
I'm sorry I first misinterpreted your braces as anticommutators and not as set inclusions! First absorb the constants into the normalizations of **H** and **B** so the (1,1) entry of each is now 1. The common eigenvectors of them both are then the transposes of (1,0,0), (0,1,1), and (0,1,-1); the eigenvalues of each are, under **H**: 1,-1, -1 ; and under **B**: 1, 1, -1, respectively. So eigenvalues for the pair [**H**,**B**]: [1,1], [1,-1], [-1,-1], specify each one of these 3 eigenvectors, uniquely. 

So the set {H,B} is a CSCO, but no other set is. Note that since $H^2=I$, the fourth set is identical to the second. Of the four sets only the 3rd one is CSCO.
Operators form a complete set if specifying eigenvalues of all of them already determines a state uniquely (up to normalization and phase). Since they commute, they will have simultaneous eigenspaces. Eigenstate is uniquely determined if all these eigenspaces are one dimensional.