# About the notion of "full rank" in lattices

I'm a newer to the lattice theory, so there is a basic notion about "full rank" confusing me.

In some papers focusing on the lattice theory, they always use the full rank lattice (the number of basis vectors $$m$$ is equal to the dimension $$n$$ of the space $$\mathbb{Z}^n$$) as the research target, such as generating a hard lattice with short basis, solving some lattice problems. However, In some lattice problems, including SIS and LWE, they always introduce a random matrix $$A$$ with $$m=poly(n)$$ as basis vectors.

Therefore, I am confused that why the matrix $$A$$ in SIS and LWE is not full rank?

Although the matrix $$A$$ defines the SIS (or LWE) lattice, it is not a basis of the lattice. A basis $$B$$ can be derived from $$A$$, and $$B$$ will then be square, thus, the lattice will have full rank.
Notice that this is similar to the fact that the public polynomial $$h$$ defines the NTRU lattice, but $$h$$ itself is not a basis of the lattice. Instead, $$h$$ can be used to derive a basis of the lattice.
• Thanks for your answer. However, I am still confused that whether the matrix $A$ in both SIS and LWE problems comes from a full rank matrix $B\in\mathbb{Z}^{m\times m}_q$. In other words, is the matrix $A$ a submatrix of $B$? Commented Aug 13 at 6:33
• Or is $A$ just a simple random matrix? Commented Aug 13 at 6:44
• @X.H.Yue $A$ is random. But given $A$, one can derive a basis $B$. Commented Aug 13 at 14:10