Find vectors $x_1, ... , x_k$ such that $D_n = Int(x_1, ... , x_k)$

Question

The following set of points is a lattice in $\mathbb{R}^n$:

$D_n =$ {$(v_1, ... , v_n)| v_1, ... , v_n \in \mathbb{Z}$ and $v_1 + ... + v_n$ is even}

Find vectors $x_1,...,x_k$ such that $D_n = Int(x_1, ..., x_k)$

My solution:

Firstly, I know that if $x_1, ..., x_k \in \mathbb{R}^n$, then we define $Int(x_1,...,x_k) = t_1x_1 + ... + t_kx_k$ where $t_1,...,t_k \in \mathbb{Z}$.

So, I have deduced that the vectors $x_1, ..., x_k \in \mathbb{R}^n$ so that $D_n = Int(x_1, ..., x_k)$ are any vectors such that $t_1x_1 + ... + t_kx_k = (v_1, ..., v_n)$ where $v_1+...+v_n$ are even.

However, I feel my solution is certainly lacking and I was wondering if anyone could help guide me in the right direction.

Thank you.

Answer 1

$D_n$ is known as the checkerboard lattice and a basis for it is

$\begin{align} {\bf x}_1 &= \begin{bmatrix} -1 \\ -1 \\ 0\\ \vdots \\ 0 \end{bmatrix}, \end{align}$ $\begin{align} {\bf x}_2 &= \begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \end{align}$ $\begin{align} {\bf x}_3 &= \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \end{align}$ $\cdots ,$ $\begin{align} {\bf x}_n &= \begin{bmatrix} 0 \\ \vdots \\ 0\\ 1 \\ -1 \end{bmatrix}. \end{align}$