# When does lattice equal the whole $\mathbb{Z}^n$

I would need to know that if I have $$v_{1},...,v_{n}\in\mathbb{Z}^n$$ being $$\mathbb{Q}$$-linearly independent and constructing a lattice $$L$$ in $$\mathbb{Z}^n$$, if and when does $$L=\mathbb{Z}^n$$.

So, I am very clueless, but I would say that it isn't always the case. Say, we have $$\mathbb{Z}^2$$. Then it would equal the whole $$\mathbb{Z}^2$$ if it would have integer combinations of the vectors (1 0) and (0 1). Visually the regular lattice with vectors which are in rectangular shape, and maybe the one with vectors which form a parallelogram shape wouldn't. Is this true?

I might be completely wrong so I would be very thankful for any sort of help. Thanks in advance.

• You want the matrix whose columns are the $v_i$ to be invertible in $M_n(\mathbb{Z})$, which means it should have determinant $\pm 1$. Commented May 13, 2021 at 18:28
• Thank you for the answer. Can you please give me some background? Why would I want that? I am not aware of the connection. Commented May 13, 2021 at 18:31
• The inverse of the matrix given by @CaptainLama is the one telling the integers $c_{ij}$ such that $e_i = \sum_j c_{ij} v_j$ (where $e_i$ is the vector with only one 1 in $i$-th position). You get that independent vectors generating the full lattice correspond to invertible integer matrices. The adjugate matrix theorem is that it is exactly the integer matrices with determinant invertible in $\Bbb{Z}$ ie. $=\pm1$. Commented May 13, 2021 at 19:18

Let us recall the definition of a lattice. Given a matrix $$B \in \mathbb{R}^{n \times m}$$, the lattice generated by $$B$$ is $$\mathcal{L}(B) = \{Bz | \forall z \in \mathbb{Z}^m\}$$ i.e., it is the set of all the integer combinations of the (column) vectors in $$B$$. In your question, you are assuming that $$\mathcal{L}(V) = \mathbb{Z}^n$$ where $$V = [v_1, \dots, v_n]$$. This implies that there is an integer combination of the vectors in $$V$$ that gives $$e_i$$ where $$e_i$$ is the column vector with 1 at the $$i^{th}$$ index and 0 in the remaining indices. So, we have $$Vu_i = e_i$$ where $$u_i \in \mathbb{Z}^n$$. Since this must be true for all $$i \in \{1,2,\dots, n\}$$, we can write $$VU = I$$ where $$I$$ is the identity matrix. By taking the determinant on both sides of the equation, we get $$det(V)det(U) = 1$$. Since, $$V$$ and $$U$$ are integer matrices, their determinants must also be an integer. Therefore, from the equation $$det(V)det(U) = 1$$, we must have $$det(V) = \pm 1$$ (also $$det(U) = \pm 1$$).