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.
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 $det(U) = \pm 1$).