(I’m not the OP but) absolutely not. The problem is about incomplete matrices, and the idea is to get an upper bound linked to m, i.e. an upper bound linked to how incomplete the matrix is. If you rephrase the question 1 as O(n^3), this is a completely different question, because now you only care about how big the matrix is, not how complete it is.
Also, since question 1 can be achieved with some gaussian pivot in O(n^3), and it also implies being able to tell whether a matrix is PSD or not, I think the best known complexity wrt n is indeed O(n^3).
In Question 1, can I consider that O(n^3)=O(nm) since m=Ω(n) can’t be greater than n^2?
(I’m not the OP but) absolutely not. The problem is about incomplete matrices, and the idea is to get an upper bound linked to m, i.e. an upper bound linked to how incomplete the matrix is. If you rephrase the question 1 as O(n^3), this is a completely different question, because now you only care about how big the matrix is, not how complete it is.
Also, since question 1 can be achieved with some gaussian pivot in O(n^3), and it also implies being able to tell whether a matrix is PSD or not, I think the best known complexity wrt n is indeed O(n^3).