Reducing the Hamiltonian size by using a symmetry to speed computations

Say I have some Hamiltonian $mathcal{H}$ that describes a 1-D spin chain (i.e. Heisenberg model) which has a number of compatible symmetries ${mathcal{O}^i}$ (i.e. total spin…etc). For all $mathcal{O}^i$ we have $[mathcal{H},mathcal{O}^i]=0$ and thus for every $mathcal{O}^i$ there exist a basis ${o_j^i}$ which spans both $mathcal{O}_i$ and $mathcal{H}$. As it seems standard in the lore, in order to make numerics more efficient, one for instance claims that the magnetisation is zero and proceeds to work with $mathcal{H}$ in a “basis” (smaller than ${o_j^i}$) where this is true.

My questions are:

1. This basis is not a basis of the whole Hilbert space, which means we can’t fully decompose $mathcal{H}$ in that basis (no resolution of the id), what is going on here? (In this basis, are the missing matrix elements of $mathcal{H}$ simply zero?)

2. Say I have no problems with 1) and that there is another symmetry I can use, i.e. translational symmetry and picking $k=0$. To import both symmetries in my problem, would I have to choose the intersection between the two basis?

Thank you in advance!